If $T$ is an invertible linear transformation and $vec{v}$ is an eigenvector of $T$, then $vec{v}$ is an...
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I saw there is a proof for invertible matrices, but I don't know how to put this mathematically for a transformation.
How do I prove an invertible linear transformation has the same eigenvectors as its inverse?
linear-algebra eigenvalues-eigenvectors transformation
edited yesterday
YuiTo Cheng
2,41341037
asked yesterday
DGRDGR
163
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add a comment |
$begingroup$
I saw there is a proof for invertible matrices, but I don't know how to put this mathematically for a transformation.
How do I prove an invertible linear transformation has the same eigenvectors as its inverse?
linear-algebra eigenvalues-eigenvectors transformation
edited yesterday
YuiTo Cheng
2,41341037
asked yesterday
DGRDGR
163
New contributor
DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Can't you just fix bases and consider your linear transformation as a matrix?
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– dcolazin
yesterday
$begingroup$
yup...I suppose that's what should have been done in the first place. I have posted an algebraic solution.
$endgroup$
– DGR
yesterday
$begingroup$
Wouldn’t the proof be exactly the same as the one for matrices?
$endgroup$
– amd
yesterday
add a comment |
$begingroup$
I saw there is a proof for invertible matrices, but I don't know how to put this mathematically for a transformation.
How do I prove an invertible linear transformation has the same eigenvectors as its inverse?
linear-algebra eigenvalues-eigenvectors transformation
edited yesterday
YuiTo Cheng
2,41341037
asked yesterday
DGRDGR
163
New contributor
DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I saw there is a proof for invertible matrices, but I don't know how to put this mathematically for a transformation.
How do I prove an invertible linear transformation has the same eigenvectors as its inverse?
linear-algebra eigenvalues-eigenvectors transformation
linear-algebra eigenvalues-eigenvectors transformation
edited yesterday
YuiTo Cheng
2,41341037
asked yesterday
DGRDGR
163
New contributor
DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
YuiTo Cheng
2,41341037
asked yesterday
DGRDGR
163
New contributor
DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
YuiTo Cheng
2,41341037
edited yesterday
YuiTo Cheng
2,41341037
edited yesterday
YuiTo Cheng
2,41341037
2,41341037
asked yesterday
DGRDGR
163
New contributor
DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
DGRDGR
163
asked yesterday
DGRDGR
163
163
New contributor
DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$begingroup$
Can't you just fix bases and consider your linear transformation as a matrix?
$endgroup$
– dcolazin
yesterday
$begingroup$
yup...I suppose that's what should have been done in the first place. I have posted an algebraic solution.
$endgroup$
– DGR
yesterday
$begingroup$
Wouldn’t the proof be exactly the same as the one for matrices?
$endgroup$
– amd
yesterday
add a comment |
$begingroup$
Can't you just fix bases and consider your linear transformation as a matrix?
$endgroup$
– dcolazin
yesterday
$begingroup$
yup...I suppose that's what should have been done in the first place. I have posted an algebraic solution.
$endgroup$
– DGR
yesterday
$begingroup$
Wouldn’t the proof be exactly the same as the one for matrices?
$endgroup$
– amd
yesterday
$begingroup$
Can't you just fix bases and consider your linear transformation as a matrix?
$endgroup$
– dcolazin
yesterday
$begingroup$
Can't you just fix bases and consider your linear transformation as a matrix?
$endgroup$
– dcolazin
yesterday
$begingroup$
yup...I suppose that's what should have been done in the first place. I have posted an algebraic solution.
$endgroup$
– DGR
yesterday
$begingroup$
yup...I suppose that's what should have been done in the first place. I have posted an algebraic solution.
$endgroup$
– DGR
yesterday
$begingroup$
Wouldn’t the proof be exactly the same as the one for matrices?
$endgroup$
– amd
yesterday
$begingroup$
Wouldn’t the proof be exactly the same as the one for matrices?
$endgroup$
– amd
yesterday
add a comment |
2 Answers
2
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oldest
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Observe that for an invertible matrix $A$ with eigenvector $mathbf v$ and corresponding eigenvalue $lambdaneq 0$, you have that
begin{align*}
mathbf v &= I mathbf v\&=A^{-1}A mathbf v\&=lambda A^{-1} mathbf v
end{align*}
Hence $A^{-1} mathbf v = lambda^{-1}mathbf v$
answered yesterday
P. QuintonP. Quinton
2,0001214
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add a comment |
$begingroup$
OK, so Iv'e found a solution!
begin{align*}\
Tv &= lambda v vert *T^{-1} \
T^{-1}Tv &= T^{-1}lambda v \
v &= lambda T^{-1}v vert *lambda^{-1} & text{(Invertible transformation, $lambdaneq 0$)} \
lambda^{-1}v &= T^{-1}v
end{align*}
For $T^{-1}$, eigenvalue $lambda^{-1}$, eigenvector is $v$ (same as eigenvector of $T$).
Hope this helps!
edited yesterday
Widawensen
4,80531447
answered yesterday
DGRDGR
163
New contributor
DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Observe that for an invertible matrix $A$ with eigenvector $mathbf v$ and corresponding eigenvalue $lambdaneq 0$, you have that
begin{align*}
mathbf v &= I mathbf v\&=A^{-1}A mathbf v\&=lambda A^{-1} mathbf v
end{align*}
Hence $A^{-1} mathbf v = lambda^{-1}mathbf v$
answered yesterday
P. QuintonP. Quinton
2,0001214
$endgroup$
add a comment |
$begingroup$
Observe that for an invertible matrix $A$ with eigenvector $mathbf v$ and corresponding eigenvalue $lambdaneq 0$, you have that
begin{align*}
mathbf v &= I mathbf v\&=A^{-1}A mathbf v\&=lambda A^{-1} mathbf v
end{align*}
Hence $A^{-1} mathbf v = lambda^{-1}mathbf v$
answered yesterday
P. QuintonP. Quinton
2,0001214
$endgroup$
add a comment |
$begingroup$
Observe that for an invertible matrix $A$ with eigenvector $mathbf v$ and corresponding eigenvalue $lambdaneq 0$, you have that
begin{align*}
mathbf v &= I mathbf v\&=A^{-1}A mathbf v\&=lambda A^{-1} mathbf v
end{align*}
Hence $A^{-1} mathbf v = lambda^{-1}mathbf v$
answered yesterday
P. QuintonP. Quinton
2,0001214
$endgroup$
Observe that for an invertible matrix $A$ with eigenvector $mathbf v$ and corresponding eigenvalue $lambdaneq 0$, you have that
begin{align*}
mathbf v &= I mathbf v\&=A^{-1}A mathbf v\&=lambda A^{-1} mathbf v
end{align*}
Hence $A^{-1} mathbf v = lambda^{-1}mathbf v$
answered yesterday
P. QuintonP. Quinton
2,0001214
answered yesterday
P. QuintonP. Quinton
2,0001214
answered yesterday
P. QuintonP. Quinton
2,0001214
answered yesterday
P. QuintonP. Quinton
2,0001214
2,0001214
add a comment |
add a comment |
$begingroup$
OK, so Iv'e found a solution!
begin{align*}\
Tv &= lambda v vert *T^{-1} \
T^{-1}Tv &= T^{-1}lambda v \
v &= lambda T^{-1}v vert *lambda^{-1} & text{(Invertible transformation, $lambdaneq 0$)} \
lambda^{-1}v &= T^{-1}v
end{align*}
For $T^{-1}$, eigenvalue $lambda^{-1}$, eigenvector is $v$ (same as eigenvector of $T$).
Hope this helps!
edited yesterday
Widawensen
4,80531447
answered yesterday
DGRDGR
163
New contributor
DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
OK, so Iv'e found a solution!
begin{align*}\
Tv &= lambda v vert *T^{-1} \
T^{-1}Tv &= T^{-1}lambda v \
v &= lambda T^{-1}v vert *lambda^{-1} & text{(Invertible transformation, $lambdaneq 0$)} \
lambda^{-1}v &= T^{-1}v
end{align*}
For $T^{-1}$, eigenvalue $lambda^{-1}$, eigenvector is $v$ (same as eigenvector of $T$).
Hope this helps!
edited yesterday
Widawensen
4,80531447
answered yesterday
DGRDGR
163
New contributor
DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
OK, so Iv'e found a solution!
begin{align*}\
Tv &= lambda v vert *T^{-1} \
T^{-1}Tv &= T^{-1}lambda v \
v &= lambda T^{-1}v vert *lambda^{-1} & text{(Invertible transformation, $lambdaneq 0$)} \
lambda^{-1}v &= T^{-1}v
end{align*}
For $T^{-1}$, eigenvalue $lambda^{-1}$, eigenvector is $v$ (same as eigenvector of $T$).
Hope this helps!
edited yesterday
Widawensen
4,80531447
answered yesterday
DGRDGR
163
New contributor
DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
OK, so Iv'e found a solution!
begin{align*}\
Tv &= lambda v vert *T^{-1} \
T^{-1}Tv &= T^{-1}lambda v \
v &= lambda T^{-1}v vert *lambda^{-1} & text{(Invertible transformation, $lambdaneq 0$)} \
lambda^{-1}v &= T^{-1}v
end{align*}
For $T^{-1}$, eigenvalue $lambda^{-1}$, eigenvector is $v$ (same as eigenvector of $T$).
Hope this helps!
edited yesterday
Widawensen
4,80531447
answered yesterday
DGRDGR
163
New contributor
DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Widawensen
4,80531447
edited yesterday
Widawensen
4,80531447
edited yesterday
Widawensen
4,80531447
4,80531447
answered yesterday
DGRDGR
163
New contributor
DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
DGRDGR
163
answered yesterday
DGRDGR
163
163
New contributor
DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
DGR is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Can't you just fix bases and consider your linear transformation as a matrix?
$endgroup$
– dcolazin
yesterday
$begingroup$
yup...I suppose that's what should have been done in the first place. I have posted an algebraic solution.
$endgroup$
– DGR
yesterday
$begingroup$
Wouldn’t the proof be exactly the same as the one for matrices?
$endgroup$
– amd
yesterday