Why does a $2×3$ matrix multiplied by a vector in $Bbb{R}^3$ give a vector in $Bbb{R}^2$?
$begingroup$
I'm so confused on how we can have a 2x3 matrix A, multiply it by a vector in $Bbb R^3$ and then end up with a vector in $Bbb R^2$. Is it possible to visualize this at all or do I need to sort of blindly accept this concept as facts that I'll accept and use?
Can someone give a very brief summarization on why this makes sense? Because I just see it as, in a world (dimension) in $Bbb R^3$, we multiply it by a vector in $Bbb R^3$, and out pops a vector in $Bbb R^2$.
Thanks!
linear-algebra matrices vector-spaces vectors
$endgroup$
add a comment |
$begingroup$
I'm so confused on how we can have a 2x3 matrix A, multiply it by a vector in $Bbb R^3$ and then end up with a vector in $Bbb R^2$. Is it possible to visualize this at all or do I need to sort of blindly accept this concept as facts that I'll accept and use?
Can someone give a very brief summarization on why this makes sense? Because I just see it as, in a world (dimension) in $Bbb R^3$, we multiply it by a vector in $Bbb R^3$, and out pops a vector in $Bbb R^2$.
Thanks!
linear-algebra matrices vector-spaces vectors
$endgroup$
1
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
Apr 8 at 1:54
$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
Apr 8 at 2:06
$begingroup$
I think it's just visualizing it that gives me trouble. Like simple vector addition, I can easily say, oh ok just add the $x_1$ unit to the other $x_1$ unit and it stretches towards $x_1$'s side! But in this case, just multiplying a vector by something in one dimension and getting a vector in another dimensions just confuses me. And I do know we can have a function like that, it's just intuitively I guess I don't really understand it
$endgroup$
– ming
Apr 8 at 4:30
$begingroup$
If you think of vectors in $mathbb R^3$ as being three components in the $hat i$, $hat j$ and $hat k$ directions then you can think of the matrix as mapping $hat i$ to the first column vector, $hat j$ to the second column and $hat k$ to the third column. Since each of the column vectors are in two dimensions we end up with a vector in two dimensions.
$endgroup$
– John Douma
2 days ago
$begingroup$
Ohh.. ok thank you!
$endgroup$
– ming
2 days ago
add a comment |
$begingroup$
I'm so confused on how we can have a 2x3 matrix A, multiply it by a vector in $Bbb R^3$ and then end up with a vector in $Bbb R^2$. Is it possible to visualize this at all or do I need to sort of blindly accept this concept as facts that I'll accept and use?
Can someone give a very brief summarization on why this makes sense? Because I just see it as, in a world (dimension) in $Bbb R^3$, we multiply it by a vector in $Bbb R^3$, and out pops a vector in $Bbb R^2$.
Thanks!
linear-algebra matrices vector-spaces vectors
$endgroup$
I'm so confused on how we can have a 2x3 matrix A, multiply it by a vector in $Bbb R^3$ and then end up with a vector in $Bbb R^2$. Is it possible to visualize this at all or do I need to sort of blindly accept this concept as facts that I'll accept and use?
Can someone give a very brief summarization on why this makes sense? Because I just see it as, in a world (dimension) in $Bbb R^3$, we multiply it by a vector in $Bbb R^3$, and out pops a vector in $Bbb R^2$.
Thanks!
linear-algebra matrices vector-spaces vectors
linear-algebra matrices vector-spaces vectors
edited 2 days ago
user21820
40.1k544162
40.1k544162
asked Apr 8 at 1:50
mingming
4606
4606
1
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
Apr 8 at 1:54
$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
Apr 8 at 2:06
$begingroup$
I think it's just visualizing it that gives me trouble. Like simple vector addition, I can easily say, oh ok just add the $x_1$ unit to the other $x_1$ unit and it stretches towards $x_1$'s side! But in this case, just multiplying a vector by something in one dimension and getting a vector in another dimensions just confuses me. And I do know we can have a function like that, it's just intuitively I guess I don't really understand it
$endgroup$
– ming
Apr 8 at 4:30
$begingroup$
If you think of vectors in $mathbb R^3$ as being three components in the $hat i$, $hat j$ and $hat k$ directions then you can think of the matrix as mapping $hat i$ to the first column vector, $hat j$ to the second column and $hat k$ to the third column. Since each of the column vectors are in two dimensions we end up with a vector in two dimensions.
$endgroup$
– John Douma
2 days ago
$begingroup$
Ohh.. ok thank you!
$endgroup$
– ming
2 days ago
add a comment |
1
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
Apr 8 at 1:54
$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
Apr 8 at 2:06
$begingroup$
I think it's just visualizing it that gives me trouble. Like simple vector addition, I can easily say, oh ok just add the $x_1$ unit to the other $x_1$ unit and it stretches towards $x_1$'s side! But in this case, just multiplying a vector by something in one dimension and getting a vector in another dimensions just confuses me. And I do know we can have a function like that, it's just intuitively I guess I don't really understand it
$endgroup$
– ming
Apr 8 at 4:30
$begingroup$
If you think of vectors in $mathbb R^3$ as being three components in the $hat i$, $hat j$ and $hat k$ directions then you can think of the matrix as mapping $hat i$ to the first column vector, $hat j$ to the second column and $hat k$ to the third column. Since each of the column vectors are in two dimensions we end up with a vector in two dimensions.
$endgroup$
– John Douma
2 days ago
$begingroup$
Ohh.. ok thank you!
$endgroup$
– ming
2 days ago
1
1
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
Apr 8 at 1:54
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
Apr 8 at 1:54
$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
Apr 8 at 2:06
$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
Apr 8 at 2:06
$begingroup$
I think it's just visualizing it that gives me trouble. Like simple vector addition, I can easily say, oh ok just add the $x_1$ unit to the other $x_1$ unit and it stretches towards $x_1$'s side! But in this case, just multiplying a vector by something in one dimension and getting a vector in another dimensions just confuses me. And I do know we can have a function like that, it's just intuitively I guess I don't really understand it
$endgroup$
– ming
Apr 8 at 4:30
$begingroup$
I think it's just visualizing it that gives me trouble. Like simple vector addition, I can easily say, oh ok just add the $x_1$ unit to the other $x_1$ unit and it stretches towards $x_1$'s side! But in this case, just multiplying a vector by something in one dimension and getting a vector in another dimensions just confuses me. And I do know we can have a function like that, it's just intuitively I guess I don't really understand it
$endgroup$
– ming
Apr 8 at 4:30
$begingroup$
If you think of vectors in $mathbb R^3$ as being three components in the $hat i$, $hat j$ and $hat k$ directions then you can think of the matrix as mapping $hat i$ to the first column vector, $hat j$ to the second column and $hat k$ to the third column. Since each of the column vectors are in two dimensions we end up with a vector in two dimensions.
$endgroup$
– John Douma
2 days ago
$begingroup$
If you think of vectors in $mathbb R^3$ as being three components in the $hat i$, $hat j$ and $hat k$ directions then you can think of the matrix as mapping $hat i$ to the first column vector, $hat j$ to the second column and $hat k$ to the third column. Since each of the column vectors are in two dimensions we end up with a vector in two dimensions.
$endgroup$
– John Douma
2 days ago
$begingroup$
Ohh.. ok thank you!
$endgroup$
– ming
2 days ago
$begingroup$
Ohh.. ok thank you!
$endgroup$
– ming
2 days ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
A more intuitive way is to think of a matrix "performing" on a vector, instead of a matrix "multiplying" with a vector.
Let's give an example. You have some triples of real numbers:
(1,2,3), (2,5,1), (3,5,9), (2,9,8)
and you "forget" the third coordinate:
(1,2), (2,5), (3,5), (2,9)
Surprisingly, this is an example of "matrix performance." Can you find
a matrix $M$ that "forgets" the third coordinate?
Answer:
The matrix is $$left(begin{array}{l}1 & 0 & 0 \ 0 & 1 & 0 end{array}right)$$
Explanation:
To get the first column, think about what happens under matrix multiplication to the vector $(1,0,0)$. The next two columns are similar.
We call such a matrix $M$ a projection.
We may visualize the projection as such.
Can you see what it means to "forget" the
third coordinate?
The important part of
a projection is linearity:
- You may project the addition of two vectors, or you may
add the projection of two vectors and you get the same result. - Similarly, you may project a scaled vector, or scale the vector
and then project it, and you get the same result.
We call a function with the linearity property a linear function.
In symbols, for any linear $f$,
- $f(v + w) = f(v) + f(w)$
- $f(cv) = cf(v)$
We see that the projection defined above is a
linear function.
Actually, you can check that every matrix is a linear function.
Perhaps it is more surprising that every linear function is a matrix. You may think of a matrix as a way to represent some linear function.
$endgroup$
2
$begingroup$
This was really helpful, thanks! Now if we "forget" about that third element, does that mean that the third dimension just totally disappears? The z-axis is just removed completely?
$endgroup$
– ming
Apr 8 at 4:32
$begingroup$
Yes, the third dimension "totally disappears!"
$endgroup$
– user156213
yesterday
add a comment |
$begingroup$
For the moment don't think about multiplication and matrices.
You can imagine starting from a vector $(x,y,z)$ in $mathbb{R}^3$ and mapping it to a vector in $mathbb{R}^2$ this way, for example:
$$
(x, y, z) mapsto (2x+ z, 3x+ 4y).
$$
Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
$$
begin{bmatrix}
2 & 0 & 1 \
3 & 4 & 0
end{bmatrix}
begin{bmatrix}
1 \
2 \
3
end{bmatrix}
=
begin{bmatrix}
5\
11
end{bmatrix}.
$$
You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.
Edit in response to a comment.
No, this does not make $(5,11)$ "look like" $(1,2,3)$. Here is a toy example that suggests where you might find this kind of calculation. Suppose you run a business that builds three products. Call them A, B and C. To make an A you need $2$ widgets and $3$ gadgets. To make a B you need just $4$ gadgets. For a C you need just a single widget. How many widgets and gadgets should you order to make $1$ A, $2$ B's and $3$ C's? The matrix product above provides the answer. You could also use that $2 times 3$ matrix to figure out what orders you might fill if you knew how many widgets and gadgets you had in stock.
Matrices are helpful in geometry too. In a linear algebra course you learn how to see that when you use the matrix
$$
begin{bmatrix}
3 & -1 \
-1 & 3
end{bmatrix}
$$
to map the coordinate plane (pairs of numbers) to itself what you have done is stretch circles centered at the origin into ellipses by changing the scales along the diagonal lines $y=x$ and $y=-x$ m
$endgroup$
$begingroup$
So in really simple terms, is (5, 11) a vector in 2 dimensions, that just "looks" like the vector (1, 2, 3) in 3 dimensions?
$endgroup$
– ming
Apr 8 at 4:33
$begingroup$
@ming No. See my edit.
$endgroup$
– Ethan Bolker
2 days ago
add a comment |
$begingroup$
A linear mapping has the property that it maps subspaces to subspaces.
So it will map a line to a line or ${0}$, a plane to a plane, a line, or ${0}$, and so on.
By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.
$endgroup$
add a comment |
$begingroup$
Suppose you have a green tank, a blue tank, and a red tank, and suppose each liter in the blue tank contain .2 L of water and .1 L alcohol. For the blue tank, it's .3 L water .6 L alcohol. The red tank is .4 L water and .5 L alcohol. Now suppose we take $b$ liters from the blue tank, $g$ from the green, and $r$ from the red, and we look at how much total water ($w$) and alcohol ($a$) we have. We are starting with a three dimensional vector (how much from the green, blue, and red tanks), and ending up with a two dimensional vector (how much of water and alcohol). We can write this as:
$.2 g + .3 b + .4 r = w$
$.1 g + .6 b + .5 r = a$
In matrix form, that's
$$
begin{bmatrix}
.2 & .3 & .4 \
.1 & .6 & .5
end{bmatrix}
begin{bmatrix}
g \
b \
r
end{bmatrix}
=
begin{bmatrix}
w\
a
end{bmatrix}.
$$
Multiplying a vector be a matrix is simply a compact form of saying "take this much of each element"; in this case the $.2$ saying "take 20% of $b$ to get $w$", the $.1$ is saying "take 10% of $g$ to get $a$", etc. The column a number is in tells you which input number it's being multiplied with, and the row tells you what output it's contributing to.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3179032%2fwhy-does-a-2%25c3%25973-matrix-multiplied-by-a-vector-in-bbbr3-give-a-vector-in%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A more intuitive way is to think of a matrix "performing" on a vector, instead of a matrix "multiplying" with a vector.
Let's give an example. You have some triples of real numbers:
(1,2,3), (2,5,1), (3,5,9), (2,9,8)
and you "forget" the third coordinate:
(1,2), (2,5), (3,5), (2,9)
Surprisingly, this is an example of "matrix performance." Can you find
a matrix $M$ that "forgets" the third coordinate?
Answer:
The matrix is $$left(begin{array}{l}1 & 0 & 0 \ 0 & 1 & 0 end{array}right)$$
Explanation:
To get the first column, think about what happens under matrix multiplication to the vector $(1,0,0)$. The next two columns are similar.
We call such a matrix $M$ a projection.
We may visualize the projection as such.
Can you see what it means to "forget" the
third coordinate?
The important part of
a projection is linearity:
- You may project the addition of two vectors, or you may
add the projection of two vectors and you get the same result. - Similarly, you may project a scaled vector, or scale the vector
and then project it, and you get the same result.
We call a function with the linearity property a linear function.
In symbols, for any linear $f$,
- $f(v + w) = f(v) + f(w)$
- $f(cv) = cf(v)$
We see that the projection defined above is a
linear function.
Actually, you can check that every matrix is a linear function.
Perhaps it is more surprising that every linear function is a matrix. You may think of a matrix as a way to represent some linear function.
$endgroup$
2
$begingroup$
This was really helpful, thanks! Now if we "forget" about that third element, does that mean that the third dimension just totally disappears? The z-axis is just removed completely?
$endgroup$
– ming
Apr 8 at 4:32
$begingroup$
Yes, the third dimension "totally disappears!"
$endgroup$
– user156213
yesterday
add a comment |
$begingroup$
A more intuitive way is to think of a matrix "performing" on a vector, instead of a matrix "multiplying" with a vector.
Let's give an example. You have some triples of real numbers:
(1,2,3), (2,5,1), (3,5,9), (2,9,8)
and you "forget" the third coordinate:
(1,2), (2,5), (3,5), (2,9)
Surprisingly, this is an example of "matrix performance." Can you find
a matrix $M$ that "forgets" the third coordinate?
Answer:
The matrix is $$left(begin{array}{l}1 & 0 & 0 \ 0 & 1 & 0 end{array}right)$$
Explanation:
To get the first column, think about what happens under matrix multiplication to the vector $(1,0,0)$. The next two columns are similar.
We call such a matrix $M$ a projection.
We may visualize the projection as such.
Can you see what it means to "forget" the
third coordinate?
The important part of
a projection is linearity:
- You may project the addition of two vectors, or you may
add the projection of two vectors and you get the same result. - Similarly, you may project a scaled vector, or scale the vector
and then project it, and you get the same result.
We call a function with the linearity property a linear function.
In symbols, for any linear $f$,
- $f(v + w) = f(v) + f(w)$
- $f(cv) = cf(v)$
We see that the projection defined above is a
linear function.
Actually, you can check that every matrix is a linear function.
Perhaps it is more surprising that every linear function is a matrix. You may think of a matrix as a way to represent some linear function.
$endgroup$
2
$begingroup$
This was really helpful, thanks! Now if we "forget" about that third element, does that mean that the third dimension just totally disappears? The z-axis is just removed completely?
$endgroup$
– ming
Apr 8 at 4:32
$begingroup$
Yes, the third dimension "totally disappears!"
$endgroup$
– user156213
yesterday
add a comment |
$begingroup$
A more intuitive way is to think of a matrix "performing" on a vector, instead of a matrix "multiplying" with a vector.
Let's give an example. You have some triples of real numbers:
(1,2,3), (2,5,1), (3,5,9), (2,9,8)
and you "forget" the third coordinate:
(1,2), (2,5), (3,5), (2,9)
Surprisingly, this is an example of "matrix performance." Can you find
a matrix $M$ that "forgets" the third coordinate?
Answer:
The matrix is $$left(begin{array}{l}1 & 0 & 0 \ 0 & 1 & 0 end{array}right)$$
Explanation:
To get the first column, think about what happens under matrix multiplication to the vector $(1,0,0)$. The next two columns are similar.
We call such a matrix $M$ a projection.
We may visualize the projection as such.
Can you see what it means to "forget" the
third coordinate?
The important part of
a projection is linearity:
- You may project the addition of two vectors, or you may
add the projection of two vectors and you get the same result. - Similarly, you may project a scaled vector, or scale the vector
and then project it, and you get the same result.
We call a function with the linearity property a linear function.
In symbols, for any linear $f$,
- $f(v + w) = f(v) + f(w)$
- $f(cv) = cf(v)$
We see that the projection defined above is a
linear function.
Actually, you can check that every matrix is a linear function.
Perhaps it is more surprising that every linear function is a matrix. You may think of a matrix as a way to represent some linear function.
$endgroup$
A more intuitive way is to think of a matrix "performing" on a vector, instead of a matrix "multiplying" with a vector.
Let's give an example. You have some triples of real numbers:
(1,2,3), (2,5,1), (3,5,9), (2,9,8)
and you "forget" the third coordinate:
(1,2), (2,5), (3,5), (2,9)
Surprisingly, this is an example of "matrix performance." Can you find
a matrix $M$ that "forgets" the third coordinate?
Answer:
The matrix is $$left(begin{array}{l}1 & 0 & 0 \ 0 & 1 & 0 end{array}right)$$
Explanation:
To get the first column, think about what happens under matrix multiplication to the vector $(1,0,0)$. The next two columns are similar.
We call such a matrix $M$ a projection.
We may visualize the projection as such.
Can you see what it means to "forget" the
third coordinate?
The important part of
a projection is linearity:
- You may project the addition of two vectors, or you may
add the projection of two vectors and you get the same result. - Similarly, you may project a scaled vector, or scale the vector
and then project it, and you get the same result.
We call a function with the linearity property a linear function.
In symbols, for any linear $f$,
- $f(v + w) = f(v) + f(w)$
- $f(cv) = cf(v)$
We see that the projection defined above is a
linear function.
Actually, you can check that every matrix is a linear function.
Perhaps it is more surprising that every linear function is a matrix. You may think of a matrix as a way to represent some linear function.
edited 2 days ago
answered Apr 8 at 3:32
user156213user156213
69238
69238
2
$begingroup$
This was really helpful, thanks! Now if we "forget" about that third element, does that mean that the third dimension just totally disappears? The z-axis is just removed completely?
$endgroup$
– ming
Apr 8 at 4:32
$begingroup$
Yes, the third dimension "totally disappears!"
$endgroup$
– user156213
yesterday
add a comment |
2
$begingroup$
This was really helpful, thanks! Now if we "forget" about that third element, does that mean that the third dimension just totally disappears? The z-axis is just removed completely?
$endgroup$
– ming
Apr 8 at 4:32
$begingroup$
Yes, the third dimension "totally disappears!"
$endgroup$
– user156213
yesterday
2
2
$begingroup$
This was really helpful, thanks! Now if we "forget" about that third element, does that mean that the third dimension just totally disappears? The z-axis is just removed completely?
$endgroup$
– ming
Apr 8 at 4:32
$begingroup$
This was really helpful, thanks! Now if we "forget" about that third element, does that mean that the third dimension just totally disappears? The z-axis is just removed completely?
$endgroup$
– ming
Apr 8 at 4:32
$begingroup$
Yes, the third dimension "totally disappears!"
$endgroup$
– user156213
yesterday
$begingroup$
Yes, the third dimension "totally disappears!"
$endgroup$
– user156213
yesterday
add a comment |
$begingroup$
For the moment don't think about multiplication and matrices.
You can imagine starting from a vector $(x,y,z)$ in $mathbb{R}^3$ and mapping it to a vector in $mathbb{R}^2$ this way, for example:
$$
(x, y, z) mapsto (2x+ z, 3x+ 4y).
$$
Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
$$
begin{bmatrix}
2 & 0 & 1 \
3 & 4 & 0
end{bmatrix}
begin{bmatrix}
1 \
2 \
3
end{bmatrix}
=
begin{bmatrix}
5\
11
end{bmatrix}.
$$
You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.
Edit in response to a comment.
No, this does not make $(5,11)$ "look like" $(1,2,3)$. Here is a toy example that suggests where you might find this kind of calculation. Suppose you run a business that builds three products. Call them A, B and C. To make an A you need $2$ widgets and $3$ gadgets. To make a B you need just $4$ gadgets. For a C you need just a single widget. How many widgets and gadgets should you order to make $1$ A, $2$ B's and $3$ C's? The matrix product above provides the answer. You could also use that $2 times 3$ matrix to figure out what orders you might fill if you knew how many widgets and gadgets you had in stock.
Matrices are helpful in geometry too. In a linear algebra course you learn how to see that when you use the matrix
$$
begin{bmatrix}
3 & -1 \
-1 & 3
end{bmatrix}
$$
to map the coordinate plane (pairs of numbers) to itself what you have done is stretch circles centered at the origin into ellipses by changing the scales along the diagonal lines $y=x$ and $y=-x$ m
$endgroup$
$begingroup$
So in really simple terms, is (5, 11) a vector in 2 dimensions, that just "looks" like the vector (1, 2, 3) in 3 dimensions?
$endgroup$
– ming
Apr 8 at 4:33
$begingroup$
@ming No. See my edit.
$endgroup$
– Ethan Bolker
2 days ago
add a comment |
$begingroup$
For the moment don't think about multiplication and matrices.
You can imagine starting from a vector $(x,y,z)$ in $mathbb{R}^3$ and mapping it to a vector in $mathbb{R}^2$ this way, for example:
$$
(x, y, z) mapsto (2x+ z, 3x+ 4y).
$$
Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
$$
begin{bmatrix}
2 & 0 & 1 \
3 & 4 & 0
end{bmatrix}
begin{bmatrix}
1 \
2 \
3
end{bmatrix}
=
begin{bmatrix}
5\
11
end{bmatrix}.
$$
You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.
Edit in response to a comment.
No, this does not make $(5,11)$ "look like" $(1,2,3)$. Here is a toy example that suggests where you might find this kind of calculation. Suppose you run a business that builds three products. Call them A, B and C. To make an A you need $2$ widgets and $3$ gadgets. To make a B you need just $4$ gadgets. For a C you need just a single widget. How many widgets and gadgets should you order to make $1$ A, $2$ B's and $3$ C's? The matrix product above provides the answer. You could also use that $2 times 3$ matrix to figure out what orders you might fill if you knew how many widgets and gadgets you had in stock.
Matrices are helpful in geometry too. In a linear algebra course you learn how to see that when you use the matrix
$$
begin{bmatrix}
3 & -1 \
-1 & 3
end{bmatrix}
$$
to map the coordinate plane (pairs of numbers) to itself what you have done is stretch circles centered at the origin into ellipses by changing the scales along the diagonal lines $y=x$ and $y=-x$ m
$endgroup$
$begingroup$
So in really simple terms, is (5, 11) a vector in 2 dimensions, that just "looks" like the vector (1, 2, 3) in 3 dimensions?
$endgroup$
– ming
Apr 8 at 4:33
$begingroup$
@ming No. See my edit.
$endgroup$
– Ethan Bolker
2 days ago
add a comment |
$begingroup$
For the moment don't think about multiplication and matrices.
You can imagine starting from a vector $(x,y,z)$ in $mathbb{R}^3$ and mapping it to a vector in $mathbb{R}^2$ this way, for example:
$$
(x, y, z) mapsto (2x+ z, 3x+ 4y).
$$
Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
$$
begin{bmatrix}
2 & 0 & 1 \
3 & 4 & 0
end{bmatrix}
begin{bmatrix}
1 \
2 \
3
end{bmatrix}
=
begin{bmatrix}
5\
11
end{bmatrix}.
$$
You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.
Edit in response to a comment.
No, this does not make $(5,11)$ "look like" $(1,2,3)$. Here is a toy example that suggests where you might find this kind of calculation. Suppose you run a business that builds three products. Call them A, B and C. To make an A you need $2$ widgets and $3$ gadgets. To make a B you need just $4$ gadgets. For a C you need just a single widget. How many widgets and gadgets should you order to make $1$ A, $2$ B's and $3$ C's? The matrix product above provides the answer. You could also use that $2 times 3$ matrix to figure out what orders you might fill if you knew how many widgets and gadgets you had in stock.
Matrices are helpful in geometry too. In a linear algebra course you learn how to see that when you use the matrix
$$
begin{bmatrix}
3 & -1 \
-1 & 3
end{bmatrix}
$$
to map the coordinate plane (pairs of numbers) to itself what you have done is stretch circles centered at the origin into ellipses by changing the scales along the diagonal lines $y=x$ and $y=-x$ m
$endgroup$
For the moment don't think about multiplication and matrices.
You can imagine starting from a vector $(x,y,z)$ in $mathbb{R}^3$ and mapping it to a vector in $mathbb{R}^2$ this way, for example:
$$
(x, y, z) mapsto (2x+ z, 3x+ 4y).
$$
Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
$$
begin{bmatrix}
2 & 0 & 1 \
3 & 4 & 0
end{bmatrix}
begin{bmatrix}
1 \
2 \
3
end{bmatrix}
=
begin{bmatrix}
5\
11
end{bmatrix}.
$$
You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.
Edit in response to a comment.
No, this does not make $(5,11)$ "look like" $(1,2,3)$. Here is a toy example that suggests where you might find this kind of calculation. Suppose you run a business that builds three products. Call them A, B and C. To make an A you need $2$ widgets and $3$ gadgets. To make a B you need just $4$ gadgets. For a C you need just a single widget. How many widgets and gadgets should you order to make $1$ A, $2$ B's and $3$ C's? The matrix product above provides the answer. You could also use that $2 times 3$ matrix to figure out what orders you might fill if you knew how many widgets and gadgets you had in stock.
Matrices are helpful in geometry too. In a linear algebra course you learn how to see that when you use the matrix
$$
begin{bmatrix}
3 & -1 \
-1 & 3
end{bmatrix}
$$
to map the coordinate plane (pairs of numbers) to itself what you have done is stretch circles centered at the origin into ellipses by changing the scales along the diagonal lines $y=x$ and $y=-x$ m
edited 2 days ago
answered Apr 8 at 2:06
Ethan BolkerEthan Bolker
46k553120
46k553120
$begingroup$
So in really simple terms, is (5, 11) a vector in 2 dimensions, that just "looks" like the vector (1, 2, 3) in 3 dimensions?
$endgroup$
– ming
Apr 8 at 4:33
$begingroup$
@ming No. See my edit.
$endgroup$
– Ethan Bolker
2 days ago
add a comment |
$begingroup$
So in really simple terms, is (5, 11) a vector in 2 dimensions, that just "looks" like the vector (1, 2, 3) in 3 dimensions?
$endgroup$
– ming
Apr 8 at 4:33
$begingroup$
@ming No. See my edit.
$endgroup$
– Ethan Bolker
2 days ago
$begingroup$
So in really simple terms, is (5, 11) a vector in 2 dimensions, that just "looks" like the vector (1, 2, 3) in 3 dimensions?
$endgroup$
– ming
Apr 8 at 4:33
$begingroup$
So in really simple terms, is (5, 11) a vector in 2 dimensions, that just "looks" like the vector (1, 2, 3) in 3 dimensions?
$endgroup$
– ming
Apr 8 at 4:33
$begingroup$
@ming No. See my edit.
$endgroup$
– Ethan Bolker
2 days ago
$begingroup$
@ming No. See my edit.
$endgroup$
– Ethan Bolker
2 days ago
add a comment |
$begingroup$
A linear mapping has the property that it maps subspaces to subspaces.
So it will map a line to a line or ${0}$, a plane to a plane, a line, or ${0}$, and so on.
By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.
$endgroup$
add a comment |
$begingroup$
A linear mapping has the property that it maps subspaces to subspaces.
So it will map a line to a line or ${0}$, a plane to a plane, a line, or ${0}$, and so on.
By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.
$endgroup$
add a comment |
$begingroup$
A linear mapping has the property that it maps subspaces to subspaces.
So it will map a line to a line or ${0}$, a plane to a plane, a line, or ${0}$, and so on.
By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.
$endgroup$
A linear mapping has the property that it maps subspaces to subspaces.
So it will map a line to a line or ${0}$, a plane to a plane, a line, or ${0}$, and so on.
By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.
answered Apr 8 at 2:15
rschwiebrschwieb
108k12103253
108k12103253
add a comment |
add a comment |
$begingroup$
Suppose you have a green tank, a blue tank, and a red tank, and suppose each liter in the blue tank contain .2 L of water and .1 L alcohol. For the blue tank, it's .3 L water .6 L alcohol. The red tank is .4 L water and .5 L alcohol. Now suppose we take $b$ liters from the blue tank, $g$ from the green, and $r$ from the red, and we look at how much total water ($w$) and alcohol ($a$) we have. We are starting with a three dimensional vector (how much from the green, blue, and red tanks), and ending up with a two dimensional vector (how much of water and alcohol). We can write this as:
$.2 g + .3 b + .4 r = w$
$.1 g + .6 b + .5 r = a$
In matrix form, that's
$$
begin{bmatrix}
.2 & .3 & .4 \
.1 & .6 & .5
end{bmatrix}
begin{bmatrix}
g \
b \
r
end{bmatrix}
=
begin{bmatrix}
w\
a
end{bmatrix}.
$$
Multiplying a vector be a matrix is simply a compact form of saying "take this much of each element"; in this case the $.2$ saying "take 20% of $b$ to get $w$", the $.1$ is saying "take 10% of $g$ to get $a$", etc. The column a number is in tells you which input number it's being multiplied with, and the row tells you what output it's contributing to.
$endgroup$
add a comment |
$begingroup$
Suppose you have a green tank, a blue tank, and a red tank, and suppose each liter in the blue tank contain .2 L of water and .1 L alcohol. For the blue tank, it's .3 L water .6 L alcohol. The red tank is .4 L water and .5 L alcohol. Now suppose we take $b$ liters from the blue tank, $g$ from the green, and $r$ from the red, and we look at how much total water ($w$) and alcohol ($a$) we have. We are starting with a three dimensional vector (how much from the green, blue, and red tanks), and ending up with a two dimensional vector (how much of water and alcohol). We can write this as:
$.2 g + .3 b + .4 r = w$
$.1 g + .6 b + .5 r = a$
In matrix form, that's
$$
begin{bmatrix}
.2 & .3 & .4 \
.1 & .6 & .5
end{bmatrix}
begin{bmatrix}
g \
b \
r
end{bmatrix}
=
begin{bmatrix}
w\
a
end{bmatrix}.
$$
Multiplying a vector be a matrix is simply a compact form of saying "take this much of each element"; in this case the $.2$ saying "take 20% of $b$ to get $w$", the $.1$ is saying "take 10% of $g$ to get $a$", etc. The column a number is in tells you which input number it's being multiplied with, and the row tells you what output it's contributing to.
$endgroup$
add a comment |
$begingroup$
Suppose you have a green tank, a blue tank, and a red tank, and suppose each liter in the blue tank contain .2 L of water and .1 L alcohol. For the blue tank, it's .3 L water .6 L alcohol. The red tank is .4 L water and .5 L alcohol. Now suppose we take $b$ liters from the blue tank, $g$ from the green, and $r$ from the red, and we look at how much total water ($w$) and alcohol ($a$) we have. We are starting with a three dimensional vector (how much from the green, blue, and red tanks), and ending up with a two dimensional vector (how much of water and alcohol). We can write this as:
$.2 g + .3 b + .4 r = w$
$.1 g + .6 b + .5 r = a$
In matrix form, that's
$$
begin{bmatrix}
.2 & .3 & .4 \
.1 & .6 & .5
end{bmatrix}
begin{bmatrix}
g \
b \
r
end{bmatrix}
=
begin{bmatrix}
w\
a
end{bmatrix}.
$$
Multiplying a vector be a matrix is simply a compact form of saying "take this much of each element"; in this case the $.2$ saying "take 20% of $b$ to get $w$", the $.1$ is saying "take 10% of $g$ to get $a$", etc. The column a number is in tells you which input number it's being multiplied with, and the row tells you what output it's contributing to.
$endgroup$
Suppose you have a green tank, a blue tank, and a red tank, and suppose each liter in the blue tank contain .2 L of water and .1 L alcohol. For the blue tank, it's .3 L water .6 L alcohol. The red tank is .4 L water and .5 L alcohol. Now suppose we take $b$ liters from the blue tank, $g$ from the green, and $r$ from the red, and we look at how much total water ($w$) and alcohol ($a$) we have. We are starting with a three dimensional vector (how much from the green, blue, and red tanks), and ending up with a two dimensional vector (how much of water and alcohol). We can write this as:
$.2 g + .3 b + .4 r = w$
$.1 g + .6 b + .5 r = a$
In matrix form, that's
$$
begin{bmatrix}
.2 & .3 & .4 \
.1 & .6 & .5
end{bmatrix}
begin{bmatrix}
g \
b \
r
end{bmatrix}
=
begin{bmatrix}
w\
a
end{bmatrix}.
$$
Multiplying a vector be a matrix is simply a compact form of saying "take this much of each element"; in this case the $.2$ saying "take 20% of $b$ to get $w$", the $.1$ is saying "take 10% of $g$ to get $a$", etc. The column a number is in tells you which input number it's being multiplied with, and the row tells you what output it's contributing to.
answered 2 days ago
AcccumulationAcccumulation
7,3052619
7,3052619
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3179032%2fwhy-does-a-2%25c3%25973-matrix-multiplied-by-a-vector-in-bbbr3-give-a-vector-in%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
Apr 8 at 1:54
$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
Apr 8 at 2:06
$begingroup$
I think it's just visualizing it that gives me trouble. Like simple vector addition, I can easily say, oh ok just add the $x_1$ unit to the other $x_1$ unit and it stretches towards $x_1$'s side! But in this case, just multiplying a vector by something in one dimension and getting a vector in another dimensions just confuses me. And I do know we can have a function like that, it's just intuitively I guess I don't really understand it
$endgroup$
– ming
Apr 8 at 4:30
$begingroup$
If you think of vectors in $mathbb R^3$ as being three components in the $hat i$, $hat j$ and $hat k$ directions then you can think of the matrix as mapping $hat i$ to the first column vector, $hat j$ to the second column and $hat k$ to the third column. Since each of the column vectors are in two dimensions we end up with a vector in two dimensions.
$endgroup$
– John Douma
2 days ago
$begingroup$
Ohh.. ok thank you!
$endgroup$
– ming
2 days ago