If a square matrix $A$ has eigenvalue $1$, then $I-A$ is singular?
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I am a bit confused about the title. Consider the following
$$I-A = I - U^{-1}JU = U^{-1}U - U^{-1}JU = U^{-1}(I-J)U$$
However, for example if $$J = begin{bmatrix}1 & 1 \ 0 & 1 end{bmatrix}$$ then $$I-J = begin{bmatrix}0 & -1 \ 0 & 0 end{bmatrix} $$
this is not a Jordan block and how to deal with $-1$? It seems we cannot include $-1$ to $U$; otherwise, we cannot get $U$ and $U^{-1}$.
How to show this? thanks!
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 16 '18 at 22:34
sleeve chensleeve chen
3,15042053
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add a comment |
$begingroup$
I am a bit confused about the title. Consider the following
$$I-A = I - U^{-1}JU = U^{-1}U - U^{-1}JU = U^{-1}(I-J)U$$
However, for example if $$J = begin{bmatrix}1 & 1 \ 0 & 1 end{bmatrix}$$ then $$I-J = begin{bmatrix}0 & -1 \ 0 & 0 end{bmatrix} $$
this is not a Jordan block and how to deal with $-1$? It seems we cannot include $-1$ to $U$; otherwise, we cannot get $U$ and $U^{-1}$.
How to show this? thanks!
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 16 '18 at 22:34
sleeve chensleeve chen
3,15042053
$endgroup$
add a comment |
$begingroup$
I am a bit confused about the title. Consider the following
$$I-A = I - U^{-1}JU = U^{-1}U - U^{-1}JU = U^{-1}(I-J)U$$
However, for example if $$J = begin{bmatrix}1 & 1 \ 0 & 1 end{bmatrix}$$ then $$I-J = begin{bmatrix}0 & -1 \ 0 & 0 end{bmatrix} $$
this is not a Jordan block and how to deal with $-1$? It seems we cannot include $-1$ to $U$; otherwise, we cannot get $U$ and $U^{-1}$.
How to show this? thanks!
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 16 '18 at 22:34
sleeve chensleeve chen
3,15042053
$endgroup$
I am a bit confused about the title. Consider the following
$$I-A = I - U^{-1}JU = U^{-1}U - U^{-1}JU = U^{-1}(I-J)U$$
However, for example if $$J = begin{bmatrix}1 & 1 \ 0 & 1 end{bmatrix}$$ then $$I-J = begin{bmatrix}0 & -1 \ 0 & 0 end{bmatrix} $$
this is not a Jordan block and how to deal with $-1$? It seems we cannot include $-1$ to $U$; otherwise, we cannot get $U$ and $U^{-1}$.
How to show this? thanks!
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 16 '18 at 22:34
sleeve chensleeve chen
3,15042053
asked Dec 16 '18 at 22:34
sleeve chensleeve chen
3,15042053
asked Dec 16 '18 at 22:34
sleeve chensleeve chen
3,15042053
asked Dec 16 '18 at 22:34
sleeve chensleeve chen
3,15042053
asked Dec 16 '18 at 22:34
sleeve chensleeve chen
3,15042053
3,15042053
add a comment |
add a comment |
1 Answer
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Your thinking is too advanced. If $v$ is an eigenvector of $A$ with eigenvalue $1$, then $(I-A)v=0$, meaning $I-A$ has non-trivial kernel and is singular.
answered Dec 16 '18 at 22:39
ArthurArthur
118k7117200
$endgroup$
$begingroup$
I get it. This proof is nice and really consider the repeated eigenvalues, Jordan block..etc. Thanks!
$endgroup$
– sleeve chen
Dec 16 '18 at 22:43
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Your thinking is too advanced. If $v$ is an eigenvector of $A$ with eigenvalue $1$, then $(I-A)v=0$, meaning $I-A$ has non-trivial kernel and is singular.
answered Dec 16 '18 at 22:39
ArthurArthur
118k7117200
$endgroup$
$begingroup$
I get it. This proof is nice and really consider the repeated eigenvalues, Jordan block..etc. Thanks!
$endgroup$
– sleeve chen
Dec 16 '18 at 22:43
add a comment |
$begingroup$
Your thinking is too advanced. If $v$ is an eigenvector of $A$ with eigenvalue $1$, then $(I-A)v=0$, meaning $I-A$ has non-trivial kernel and is singular.
answered Dec 16 '18 at 22:39
ArthurArthur
118k7117200
$endgroup$
$begingroup$
I get it. This proof is nice and really consider the repeated eigenvalues, Jordan block..etc. Thanks!
$endgroup$
– sleeve chen
Dec 16 '18 at 22:43
add a comment |
$begingroup$
Your thinking is too advanced. If $v$ is an eigenvector of $A$ with eigenvalue $1$, then $(I-A)v=0$, meaning $I-A$ has non-trivial kernel and is singular.
answered Dec 16 '18 at 22:39
ArthurArthur
118k7117200
$endgroup$
Your thinking is too advanced. If $v$ is an eigenvector of $A$ with eigenvalue $1$, then $(I-A)v=0$, meaning $I-A$ has non-trivial kernel and is singular.
answered Dec 16 '18 at 22:39
ArthurArthur
118k7117200
answered Dec 16 '18 at 22:39
ArthurArthur
118k7117200
answered Dec 16 '18 at 22:39
ArthurArthur
118k7117200
answered Dec 16 '18 at 22:39
ArthurArthur
118k7117200
118k7117200
$begingroup$
I get it. This proof is nice and really consider the repeated eigenvalues, Jordan block..etc. Thanks!
$endgroup$
– sleeve chen
Dec 16 '18 at 22:43
add a comment |
$begingroup$
I get it. This proof is nice and really consider the repeated eigenvalues, Jordan block..etc. Thanks!
$endgroup$
– sleeve chen
Dec 16 '18 at 22:43
$begingroup$
I get it. This proof is nice and really consider the repeated eigenvalues, Jordan block..etc. Thanks!
$endgroup$
– sleeve chen
Dec 16 '18 at 22:43
$begingroup$
I get it. This proof is nice and really consider the repeated eigenvalues, Jordan block..etc. Thanks!
$endgroup$
– sleeve chen
Dec 16 '18 at 22:43
add a comment |
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