Mixing
How do eigenvalues of a matrix X change if we linear transform the matrix X?
$begingroup$
I have a matrix $X$ which has eigenvalues $U$.
Now create a new matrix $Y = AX$ where $A$ is a nonsingular matrix.
How do the eigenvectors and eigenvalues of $Y$ change in relation to the eigenvectors and eigenvalues of $X$ and the matrix $A$?
eigenvalues-eigenvectors transformation
edited Jan 15 at 15:07
Adam Higgins
611113
asked Jan 15 at 14:38
sivasiva
263
$endgroup$
add a comment |
$begingroup$
I have a matrix $X$ which has eigenvalues $U$.
Now create a new matrix $Y = AX$ where $A$ is a nonsingular matrix.
How do the eigenvectors and eigenvalues of $Y$ change in relation to the eigenvectors and eigenvalues of $X$ and the matrix $A$?
eigenvalues-eigenvectors transformation
edited Jan 15 at 15:07
Adam Higgins
611113
asked Jan 15 at 14:38
sivasiva
263
$endgroup$
$begingroup$
$Y=AX$ really does not imply any simple relation between eigenvectors/values of $X$ and of $Y$.
$endgroup$
– David C. Ullrich
Jan 15 at 16:00
add a comment |
$begingroup$
I have a matrix $X$ which has eigenvalues $U$.
Now create a new matrix $Y = AX$ where $A$ is a nonsingular matrix.
How do the eigenvectors and eigenvalues of $Y$ change in relation to the eigenvectors and eigenvalues of $X$ and the matrix $A$?
eigenvalues-eigenvectors transformation
edited Jan 15 at 15:07
Adam Higgins
611113
asked Jan 15 at 14:38
sivasiva
263
$endgroup$
I have a matrix $X$ which has eigenvalues $U$.
Now create a new matrix $Y = AX$ where $A$ is a nonsingular matrix.
How do the eigenvectors and eigenvalues of $Y$ change in relation to the eigenvectors and eigenvalues of $X$ and the matrix $A$?
eigenvalues-eigenvectors transformation
eigenvalues-eigenvectors transformation
edited Jan 15 at 15:07
Adam Higgins
611113
asked Jan 15 at 14:38
sivasiva
263
edited Jan 15 at 15:07
Adam Higgins
611113
asked Jan 15 at 14:38
sivasiva
263
edited Jan 15 at 15:07
Adam Higgins
611113
edited Jan 15 at 15:07
Adam Higgins
611113
edited Jan 15 at 15:07
Adam Higgins
611113
611113
asked Jan 15 at 14:38
sivasiva
263
asked Jan 15 at 14:38
sivasiva
263
asked Jan 15 at 14:38
sivasiva
263
263
$begingroup$
$Y=AX$ really does not imply any simple relation between eigenvectors/values of $X$ and of $Y$.
$endgroup$
– David C. Ullrich
Jan 15 at 16:00
add a comment |
$begingroup$
$Y=AX$ really does not imply any simple relation between eigenvectors/values of $X$ and of $Y$.
$endgroup$
– David C. Ullrich
Jan 15 at 16:00
$begingroup$
$Y=AX$ really does not imply any simple relation between eigenvectors/values of $X$ and of $Y$.
$endgroup$
– David C. Ullrich
Jan 15 at 16:00
$begingroup$
$Y=AX$ really does not imply any simple relation between eigenvectors/values of $X$ and of $Y$.
$endgroup$
– David C. Ullrich
Jan 15 at 16:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If the eigenvector equation for $X$ is
$$
Xv=lambda v
$$
then
$$
Yv=AXv=Alambda v=lambda Av
$$
so it comes down to what effect $A$ has on $v$ which will vary according to $A$.
In special cases, such as $A$ being a constant diagonal matrix with value $k$ on the diagonal then the above equation continues
$$
lambda Av=lambda k Iv = lambda kv
$$
In such a special case the eigenvectors of $X$ and $A$ are the same and the eigenvalues are multiplied by $k$.
answered Jan 15 at 17:02
PM.PM.
3,3532925
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If the eigenvector equation for $X$ is
$$
Xv=lambda v
$$
then
$$
Yv=AXv=Alambda v=lambda Av
$$
so it comes down to what effect $A$ has on $v$ which will vary according to $A$.
In special cases, such as $A$ being a constant diagonal matrix with value $k$ on the diagonal then the above equation continues
$$
lambda Av=lambda k Iv = lambda kv
$$
In such a special case the eigenvectors of $X$ and $A$ are the same and the eigenvalues are multiplied by $k$.
answered Jan 15 at 17:02
PM.PM.
3,3532925
$endgroup$
add a comment |
$begingroup$
If the eigenvector equation for $X$ is
$$
Xv=lambda v
$$
then
$$
Yv=AXv=Alambda v=lambda Av
$$
so it comes down to what effect $A$ has on $v$ which will vary according to $A$.
In special cases, such as $A$ being a constant diagonal matrix with value $k$ on the diagonal then the above equation continues
$$
lambda Av=lambda k Iv = lambda kv
$$
In such a special case the eigenvectors of $X$ and $A$ are the same and the eigenvalues are multiplied by $k$.
answered Jan 15 at 17:02
PM.PM.
3,3532925
$endgroup$
add a comment |
$begingroup$
If the eigenvector equation for $X$ is
$$
Xv=lambda v
$$
then
$$
Yv=AXv=Alambda v=lambda Av
$$
so it comes down to what effect $A$ has on $v$ which will vary according to $A$.
In special cases, such as $A$ being a constant diagonal matrix with value $k$ on the diagonal then the above equation continues
$$
lambda Av=lambda k Iv = lambda kv
$$
In such a special case the eigenvectors of $X$ and $A$ are the same and the eigenvalues are multiplied by $k$.
answered Jan 15 at 17:02
PM.PM.
3,3532925
$endgroup$
If the eigenvector equation for $X$ is
$$
Xv=lambda v
$$
then
$$
Yv=AXv=Alambda v=lambda Av
$$
so it comes down to what effect $A$ has on $v$ which will vary according to $A$.
In special cases, such as $A$ being a constant diagonal matrix with value $k$ on the diagonal then the above equation continues
$$
lambda Av=lambda k Iv = lambda kv
$$
In such a special case the eigenvectors of $X$ and $A$ are the same and the eigenvalues are multiplied by $k$.
answered Jan 15 at 17:02
PM.PM.
3,3532925
answered Jan 15 at 17:02
PM.PM.
3,3532925
answered Jan 15 at 17:02
PM.PM.
3,3532925
answered Jan 15 at 17:02
PM.PM.
3,3532925
3,3532925
add a comment |
add a comment |
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$begingroup$
$Y=AX$ really does not imply any simple relation between eigenvectors/values of $X$ and of $Y$.
$endgroup$
– David C. Ullrich
Jan 15 at 16:00