Real-valued matrices and their eigenvalues
Why are the eigenvalues of the symmetric matrix $A^TB^TBA$ equal to the eigenvalues of the matrix $B^TBAA^T$ where $A in mathbb{R}^{3 times 3}$ and $B$ is either:
$B = begin{pmatrix} {1}, {0}, {0} end{pmatrix}$, $begin{pmatrix} {1}, {0}, {0}
\
{0}, {0}, {1}end{pmatrix}$ or $I_3$?
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 11 '18 at 15:18
M6126
154
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Why are the eigenvalues of the symmetric matrix $A^TB^TBA$ equal to the eigenvalues of the matrix $B^TBAA^T$ where $A in mathbb{R}^{3 times 3}$ and $B$ is either:
$B = begin{pmatrix} {1}, {0}, {0} end{pmatrix}$, $begin{pmatrix} {1}, {0}, {0}
\
{0}, {0}, {1}end{pmatrix}$ or $I_3$?
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 11 '18 at 15:18
M6126
154
add a comment |
Why are the eigenvalues of the symmetric matrix $A^TB^TBA$ equal to the eigenvalues of the matrix $B^TBAA^T$ where $A in mathbb{R}^{3 times 3}$ and $B$ is either:
$B = begin{pmatrix} {1}, {0}, {0} end{pmatrix}$, $begin{pmatrix} {1}, {0}, {0}
\
{0}, {0}, {1}end{pmatrix}$ or $I_3$?
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 11 '18 at 15:18
M6126
154
Why are the eigenvalues of the symmetric matrix $A^TB^TBA$ equal to the eigenvalues of the matrix $B^TBAA^T$ where $A in mathbb{R}^{3 times 3}$ and $B$ is either:
$B = begin{pmatrix} {1}, {0}, {0} end{pmatrix}$, $begin{pmatrix} {1}, {0}, {0}
\
{0}, {0}, {1}end{pmatrix}$ or $I_3$?
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 11 '18 at 15:18
M6126
154
asked Dec 11 '18 at 15:18
M6126
154
asked Dec 11 '18 at 15:18
M6126
154
asked Dec 11 '18 at 15:18
M6126
154
asked Dec 11 '18 at 15:18
M6126
154
154
add a comment |
add a comment |
1 Answer
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In fact, the nonzero eigenvalues of $CD$ and $DC$ are always the same for any matrices $C$ and $D$ that can be multiplied in both directions.
See e.g. here.
This is just the case $C = A^T$, $D = B^T B A$.
answered Dec 11 '18 at 16:27
Robert Israel
318k23208457
add a comment |
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1 Answer
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1 Answer
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active
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In fact, the nonzero eigenvalues of $CD$ and $DC$ are always the same for any matrices $C$ and $D$ that can be multiplied in both directions.
See e.g. here.
This is just the case $C = A^T$, $D = B^T B A$.
answered Dec 11 '18 at 16:27
Robert Israel
318k23208457
add a comment |
In fact, the nonzero eigenvalues of $CD$ and $DC$ are always the same for any matrices $C$ and $D$ that can be multiplied in both directions.
See e.g. here.
This is just the case $C = A^T$, $D = B^T B A$.
answered Dec 11 '18 at 16:27
Robert Israel
318k23208457
add a comment |
In fact, the nonzero eigenvalues of $CD$ and $DC$ are always the same for any matrices $C$ and $D$ that can be multiplied in both directions.
See e.g. here.
This is just the case $C = A^T$, $D = B^T B A$.
answered Dec 11 '18 at 16:27
Robert Israel
318k23208457
In fact, the nonzero eigenvalues of $CD$ and $DC$ are always the same for any matrices $C$ and $D$ that can be multiplied in both directions.
See e.g. here.
This is just the case $C = A^T$, $D = B^T B A$.
answered Dec 11 '18 at 16:27
Robert Israel
318k23208457
answered Dec 11 '18 at 16:27
Robert Israel
318k23208457
answered Dec 11 '18 at 16:27
Robert Israel
318k23208457
answered Dec 11 '18 at 16:27
Robert Israel
318k23208457
318k23208457
add a comment |
add a comment |
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