Mixing
An example of a square matrix with the same eigenvectors but different eigenvalues
$begingroup$
Is there an example such that $A$ and $B$, three by three, that have the same eigenvectors, but different eigenvalues?
What would be the eigenvectors and eigenvalues if it exists because I'm stuck on this practice problem.
I know that if matrices $A$ and $B$ can be written such that $AB=BA$, they share the same eigenvectors, but what about their eigenvalues? precisely if they're squared matrices ($3times 3$ case)
linear-algebra eigenvalues-eigenvectors
edited Jan 29 at 3:33
J. W. Tanner
4,0611320
asked Jan 29 at 2:58
ximxim
516
$endgroup$
add a comment |
$begingroup$
Is there an example such that $A$ and $B$, three by three, that have the same eigenvectors, but different eigenvalues?
What would be the eigenvectors and eigenvalues if it exists because I'm stuck on this practice problem.
I know that if matrices $A$ and $B$ can be written such that $AB=BA$, they share the same eigenvectors, but what about their eigenvalues? precisely if they're squared matrices ($3times 3$ case)
linear-algebra eigenvalues-eigenvectors
edited Jan 29 at 3:33
J. W. Tanner
4,0611320
asked Jan 29 at 2:58
ximxim
516
$endgroup$
add a comment |
$begingroup$
Is there an example such that $A$ and $B$, three by three, that have the same eigenvectors, but different eigenvalues?
What would be the eigenvectors and eigenvalues if it exists because I'm stuck on this practice problem.
I know that if matrices $A$ and $B$ can be written such that $AB=BA$, they share the same eigenvectors, but what about their eigenvalues? precisely if they're squared matrices ($3times 3$ case)
linear-algebra eigenvalues-eigenvectors
edited Jan 29 at 3:33
J. W. Tanner
4,0611320
asked Jan 29 at 2:58
ximxim
516
$endgroup$
Is there an example such that $A$ and $B$, three by three, that have the same eigenvectors, but different eigenvalues?
What would be the eigenvectors and eigenvalues if it exists because I'm stuck on this practice problem.
I know that if matrices $A$ and $B$ can be written such that $AB=BA$, they share the same eigenvectors, but what about their eigenvalues? precisely if they're squared matrices ($3times 3$ case)
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
edited Jan 29 at 3:33
J. W. Tanner
4,0611320
asked Jan 29 at 2:58
ximxim
516
edited Jan 29 at 3:33
J. W. Tanner
4,0611320
asked Jan 29 at 2:58
ximxim
516
edited Jan 29 at 3:33
J. W. Tanner
4,0611320
edited Jan 29 at 3:33
J. W. Tanner
4,0611320
edited Jan 29 at 3:33
J. W. Tanner
4,0611320
4,0611320
asked Jan 29 at 2:58
ximxim
516
asked Jan 29 at 2:58
ximxim
516
asked Jan 29 at 2:58
ximxim
516
516
add a comment |
add a comment |
1 Answer
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$begingroup$
How about $I$ and $-I$? Then all $xneq0$ are eigenvectors. But the eigenvalues are $1$ and $-1$ respectively.
For a less trivial example, how about $begin{pmatrix}1&0&0\0&2&1\0&0&2end{pmatrix}$ and $begin{pmatrix}4&0&0\0&3&1\0&0&3end{pmatrix}$?
They share eigenvectors $(1,0,0)$ and $(0,1,0)$, but for different eigenvalues.
As to your question about commuting matrices, you could take any two diagonal matrices, with different entries on the respective diagonals.
answered Jan 29 at 3:01
Chris CusterChris Custer
14.2k3827
$endgroup$
1
$begingroup$
It only works for the identity matrices?
$endgroup$
– xim
Jan 29 at 3:04
$begingroup$
Hmm. I'll have to think about it. It looks like multiples of identity matrices will work. How about a permutation matrix?
$endgroup$
– Chris Custer
Jan 29 at 3:33
$begingroup$
It would work if you put different values on the diagonal. Maybe using Jordan normal form we can make examples.
$endgroup$
– Chris Custer
Jan 29 at 3:55
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
How about $I$ and $-I$? Then all $xneq0$ are eigenvectors. But the eigenvalues are $1$ and $-1$ respectively.
For a less trivial example, how about $begin{pmatrix}1&0&0\0&2&1\0&0&2end{pmatrix}$ and $begin{pmatrix}4&0&0\0&3&1\0&0&3end{pmatrix}$?
They share eigenvectors $(1,0,0)$ and $(0,1,0)$, but for different eigenvalues.
As to your question about commuting matrices, you could take any two diagonal matrices, with different entries on the respective diagonals.
answered Jan 29 at 3:01
Chris CusterChris Custer
14.2k3827
$endgroup$
1
$begingroup$
It only works for the identity matrices?
$endgroup$
– xim
Jan 29 at 3:04
$begingroup$
Hmm. I'll have to think about it. It looks like multiples of identity matrices will work. How about a permutation matrix?
$endgroup$
– Chris Custer
Jan 29 at 3:33
$begingroup$
It would work if you put different values on the diagonal. Maybe using Jordan normal form we can make examples.
$endgroup$
– Chris Custer
Jan 29 at 3:55
add a comment |
$begingroup$
How about $I$ and $-I$? Then all $xneq0$ are eigenvectors. But the eigenvalues are $1$ and $-1$ respectively.
For a less trivial example, how about $begin{pmatrix}1&0&0\0&2&1\0&0&2end{pmatrix}$ and $begin{pmatrix}4&0&0\0&3&1\0&0&3end{pmatrix}$?
They share eigenvectors $(1,0,0)$ and $(0,1,0)$, but for different eigenvalues.
As to your question about commuting matrices, you could take any two diagonal matrices, with different entries on the respective diagonals.
answered Jan 29 at 3:01
Chris CusterChris Custer
14.2k3827
$endgroup$
1
$begingroup$
It only works for the identity matrices?
$endgroup$
– xim
Jan 29 at 3:04
$begingroup$
Hmm. I'll have to think about it. It looks like multiples of identity matrices will work. How about a permutation matrix?
$endgroup$
– Chris Custer
Jan 29 at 3:33
$begingroup$
It would work if you put different values on the diagonal. Maybe using Jordan normal form we can make examples.
$endgroup$
– Chris Custer
Jan 29 at 3:55
add a comment |
$begingroup$
How about $I$ and $-I$? Then all $xneq0$ are eigenvectors. But the eigenvalues are $1$ and $-1$ respectively.
For a less trivial example, how about $begin{pmatrix}1&0&0\0&2&1\0&0&2end{pmatrix}$ and $begin{pmatrix}4&0&0\0&3&1\0&0&3end{pmatrix}$?
They share eigenvectors $(1,0,0)$ and $(0,1,0)$, but for different eigenvalues.
As to your question about commuting matrices, you could take any two diagonal matrices, with different entries on the respective diagonals.
answered Jan 29 at 3:01
Chris CusterChris Custer
14.2k3827
$endgroup$
How about $I$ and $-I$? Then all $xneq0$ are eigenvectors. But the eigenvalues are $1$ and $-1$ respectively.
For a less trivial example, how about $begin{pmatrix}1&0&0\0&2&1\0&0&2end{pmatrix}$ and $begin{pmatrix}4&0&0\0&3&1\0&0&3end{pmatrix}$?
They share eigenvectors $(1,0,0)$ and $(0,1,0)$, but for different eigenvalues.
As to your question about commuting matrices, you could take any two diagonal matrices, with different entries on the respective diagonals.
answered Jan 29 at 3:01
Chris CusterChris Custer
14.2k3827
edited Jan 29 at 4:34
answered Jan 29 at 3:01
Chris CusterChris Custer
14.2k3827
answered Jan 29 at 3:01
Chris CusterChris Custer
14.2k3827
answered Jan 29 at 3:01
Chris CusterChris Custer
14.2k3827
14.2k3827
1
$begingroup$
It only works for the identity matrices?
$endgroup$
– xim
Jan 29 at 3:04
$begingroup$
Hmm. I'll have to think about it. It looks like multiples of identity matrices will work. How about a permutation matrix?
$endgroup$
– Chris Custer
Jan 29 at 3:33
$begingroup$
It would work if you put different values on the diagonal. Maybe using Jordan normal form we can make examples.
$endgroup$
– Chris Custer
Jan 29 at 3:55
add a comment |
1
$begingroup$
It only works for the identity matrices?
$endgroup$
– xim
Jan 29 at 3:04
$begingroup$
Hmm. I'll have to think about it. It looks like multiples of identity matrices will work. How about a permutation matrix?
$endgroup$
– Chris Custer
Jan 29 at 3:33
$begingroup$
It would work if you put different values on the diagonal. Maybe using Jordan normal form we can make examples.
$endgroup$
– Chris Custer
Jan 29 at 3:55
1
1
$begingroup$
It only works for the identity matrices?
$endgroup$
– xim
Jan 29 at 3:04
$begingroup$
It only works for the identity matrices?
$endgroup$
– xim
Jan 29 at 3:04
$begingroup$
Hmm. I'll have to think about it. It looks like multiples of identity matrices will work. How about a permutation matrix?
$endgroup$
– Chris Custer
Jan 29 at 3:33
$begingroup$
Hmm. I'll have to think about it. It looks like multiples of identity matrices will work. How about a permutation matrix?
$endgroup$
– Chris Custer
Jan 29 at 3:33
$begingroup$
It would work if you put different values on the diagonal. Maybe using Jordan normal form we can make examples.
$endgroup$
– Chris Custer
Jan 29 at 3:55
$begingroup$
It would work if you put different values on the diagonal. Maybe using Jordan normal form we can make examples.
$endgroup$
– Chris Custer
Jan 29 at 3:55
add a comment |
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