Mixing
Diagonalizability in relation to characteristic polynomial and row equivalence
$begingroup$
I am new to linear algebra, and am unsure re the following question:
True or False?
Let A and B be matrices of n x n.
If A and B are diagonalizable and they have the same characteristic polynomial, then A and B are similar.
If A and B are row equivalent and A is diagonalizable, then B is diagonalizable.
My intuitive answer is "false" to 1, and "true" to 2.
However, I am not sure, and either way, I would ideally like to be able to prove it...
Many thanks!
linear-algebra diagonalization
asked Jan 19 at 13:40
daltadalta
1238
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add a comment |
$begingroup$
I am new to linear algebra, and am unsure re the following question:
True or False?
Let A and B be matrices of n x n.
If A and B are diagonalizable and they have the same characteristic polynomial, then A and B are similar.
If A and B are row equivalent and A is diagonalizable, then B is diagonalizable.
My intuitive answer is "false" to 1, and "true" to 2.
However, I am not sure, and either way, I would ideally like to be able to prove it...
Many thanks!
linear-algebra diagonalization
asked Jan 19 at 13:40
daltadalta
1238
$endgroup$
add a comment |
$begingroup$
I am new to linear algebra, and am unsure re the following question:
True or False?
Let A and B be matrices of n x n.
If A and B are diagonalizable and they have the same characteristic polynomial, then A and B are similar.
If A and B are row equivalent and A is diagonalizable, then B is diagonalizable.
My intuitive answer is "false" to 1, and "true" to 2.
However, I am not sure, and either way, I would ideally like to be able to prove it...
Many thanks!
linear-algebra diagonalization
asked Jan 19 at 13:40
daltadalta
1238
$endgroup$
I am new to linear algebra, and am unsure re the following question:
True or False?
Let A and B be matrices of n x n.
If A and B are diagonalizable and they have the same characteristic polynomial, then A and B are similar.
If A and B are row equivalent and A is diagonalizable, then B is diagonalizable.
My intuitive answer is "false" to 1, and "true" to 2.
However, I am not sure, and either way, I would ideally like to be able to prove it...
Many thanks!
linear-algebra diagonalization
linear-algebra diagonalization
asked Jan 19 at 13:40
daltadalta
1238
asked Jan 19 at 13:40
daltadalta
1238
asked Jan 19 at 13:40
daltadalta
1238
asked Jan 19 at 13:40
daltadalta
1238
asked Jan 19 at 13:40
daltadalta
1238
1238
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Wrong on both counts.
For (1): If $A$ is diagonalizable then $A$ is similar to $D$, where $D$ is a diagonal matrix that has the eigenvalues of $A$ on the diagonal. If $A$ and $B$ have the same characteristic polynomial then they have the same eigenvalues, so if $B$ is diagonalizable it's similar to the same $D$ as we used for $A$. So $A$ and $B$ are similar.
For (2): $begin{bmatrix}1&1\0&1end{bmatrix}$ is row-equivalent to $begin{bmatrix}1&0\0&1end{bmatrix}$.
answered Jan 19 at 14:50
David C. UllrichDavid C. Ullrich
61k43994
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add a comment |
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1 Answer
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$begingroup$
Wrong on both counts.
For (1): If $A$ is diagonalizable then $A$ is similar to $D$, where $D$ is a diagonal matrix that has the eigenvalues of $A$ on the diagonal. If $A$ and $B$ have the same characteristic polynomial then they have the same eigenvalues, so if $B$ is diagonalizable it's similar to the same $D$ as we used for $A$. So $A$ and $B$ are similar.
For (2): $begin{bmatrix}1&1\0&1end{bmatrix}$ is row-equivalent to $begin{bmatrix}1&0\0&1end{bmatrix}$.
answered Jan 19 at 14:50
David C. UllrichDavid C. Ullrich
61k43994
$endgroup$
add a comment |
$begingroup$
Wrong on both counts.
For (1): If $A$ is diagonalizable then $A$ is similar to $D$, where $D$ is a diagonal matrix that has the eigenvalues of $A$ on the diagonal. If $A$ and $B$ have the same characteristic polynomial then they have the same eigenvalues, so if $B$ is diagonalizable it's similar to the same $D$ as we used for $A$. So $A$ and $B$ are similar.
For (2): $begin{bmatrix}1&1\0&1end{bmatrix}$ is row-equivalent to $begin{bmatrix}1&0\0&1end{bmatrix}$.
answered Jan 19 at 14:50
David C. UllrichDavid C. Ullrich
61k43994
$endgroup$
add a comment |
$begingroup$
Wrong on both counts.
For (1): If $A$ is diagonalizable then $A$ is similar to $D$, where $D$ is a diagonal matrix that has the eigenvalues of $A$ on the diagonal. If $A$ and $B$ have the same characteristic polynomial then they have the same eigenvalues, so if $B$ is diagonalizable it's similar to the same $D$ as we used for $A$. So $A$ and $B$ are similar.
For (2): $begin{bmatrix}1&1\0&1end{bmatrix}$ is row-equivalent to $begin{bmatrix}1&0\0&1end{bmatrix}$.
answered Jan 19 at 14:50
David C. UllrichDavid C. Ullrich
61k43994
$endgroup$
Wrong on both counts.
For (1): If $A$ is diagonalizable then $A$ is similar to $D$, where $D$ is a diagonal matrix that has the eigenvalues of $A$ on the diagonal. If $A$ and $B$ have the same characteristic polynomial then they have the same eigenvalues, so if $B$ is diagonalizable it's similar to the same $D$ as we used for $A$. So $A$ and $B$ are similar.
For (2): $begin{bmatrix}1&1\0&1end{bmatrix}$ is row-equivalent to $begin{bmatrix}1&0\0&1end{bmatrix}$.
answered Jan 19 at 14:50
David C. UllrichDavid C. Ullrich
61k43994
answered Jan 19 at 14:50
David C. UllrichDavid C. Ullrich
61k43994
answered Jan 19 at 14:50
David C. UllrichDavid C. Ullrich
61k43994
answered Jan 19 at 14:50
David C. UllrichDavid C. Ullrich
61k43994
61k43994
add a comment |
add a comment |
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