A conceptual Question on Diagonalization of Matrix
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Let S belong to F(m x n) and T belong to (n x m). Let ST be invertible. Prove that ST is diagonalizable if and only if TS is diagonalizable for: a) when n = m b) when n > m c) Can ST be invertible when n < m?
I tried a lot in different ways but couldn't get this done. Can anyone help me on this? Thanks in advance. :)
linear-algebra diagonalization
asked Dec 7 '18 at 1:15
Ashis JanaAshis Jana
22
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add a comment |
$begingroup$
Let S belong to F(m x n) and T belong to (n x m). Let ST be invertible. Prove that ST is diagonalizable if and only if TS is diagonalizable for: a) when n = m b) when n > m c) Can ST be invertible when n < m?
I tried a lot in different ways but couldn't get this done. Can anyone help me on this? Thanks in advance. :)
linear-algebra diagonalization
asked Dec 7 '18 at 1:15
Ashis JanaAshis Jana
22
$endgroup$
$begingroup$
I recommend for c) you interpret your matrices as morhpisms and use dimensions
$endgroup$
– Enkidu
Dec 7 '18 at 9:09
add a comment |
$begingroup$
Let S belong to F(m x n) and T belong to (n x m). Let ST be invertible. Prove that ST is diagonalizable if and only if TS is diagonalizable for: a) when n = m b) when n > m c) Can ST be invertible when n < m?
I tried a lot in different ways but couldn't get this done. Can anyone help me on this? Thanks in advance. :)
linear-algebra diagonalization
asked Dec 7 '18 at 1:15
Ashis JanaAshis Jana
22
$endgroup$
Let S belong to F(m x n) and T belong to (n x m). Let ST be invertible. Prove that ST is diagonalizable if and only if TS is diagonalizable for: a) when n = m b) when n > m c) Can ST be invertible when n < m?
I tried a lot in different ways but couldn't get this done. Can anyone help me on this? Thanks in advance. :)
linear-algebra diagonalization
linear-algebra diagonalization
asked Dec 7 '18 at 1:15
Ashis JanaAshis Jana
22
asked Dec 7 '18 at 1:15
Ashis JanaAshis Jana
22
asked Dec 7 '18 at 1:15
Ashis JanaAshis Jana
22
asked Dec 7 '18 at 1:15
Ashis JanaAshis Jana
22
asked Dec 7 '18 at 1:15
Ashis JanaAshis Jana
22
22
$begingroup$
I recommend for c) you interpret your matrices as morhpisms and use dimensions
$endgroup$
– Enkidu
Dec 7 '18 at 9:09
add a comment |
$begingroup$
I recommend for c) you interpret your matrices as morhpisms and use dimensions
$endgroup$
– Enkidu
Dec 7 '18 at 9:09
$begingroup$
I recommend for c) you interpret your matrices as morhpisms and use dimensions
$endgroup$
– Enkidu
Dec 7 '18 at 9:09
$begingroup$
I recommend for c) you interpret your matrices as morhpisms and use dimensions
$endgroup$
– Enkidu
Dec 7 '18 at 9:09
add a comment |
1 Answer
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Hints for (b). Denote the minimal polynomials of $ST$ and $TS$ by $m_{ST}$ and $m_{TS}$ respectively.
Suppose $ST$ is diagonalisable. Since $ST$ is invertible, $0$ is not a root of $m_{ST}$. Hence $p(x)=x,m_{ST}(x)$ is a product of distinct linear factors. Now, show that $p(TS)=0$ and hence $m_{TS}|p$.
Conversely, suppose $TS$ is diagonalisable, so that $m_{TS}$ is a product of distinct linear factors. By considering $S,m_{TS}(TS),T$ and by using the assumption that $ST$ is invertible, show that $m_{TS}(ST)=0$ and hence $m_{ST}|m_{TS}$.
answered Dec 7 '18 at 10:39
user1551user1551
72.4k566127
$endgroup$
$begingroup$
Thanks a lot. :)
$endgroup$
– Ashis Jana
Dec 8 '18 at 0:19
add a comment |
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1 Answer
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$begingroup$
Hints for (b). Denote the minimal polynomials of $ST$ and $TS$ by $m_{ST}$ and $m_{TS}$ respectively.
Suppose $ST$ is diagonalisable. Since $ST$ is invertible, $0$ is not a root of $m_{ST}$. Hence $p(x)=x,m_{ST}(x)$ is a product of distinct linear factors. Now, show that $p(TS)=0$ and hence $m_{TS}|p$.
Conversely, suppose $TS$ is diagonalisable, so that $m_{TS}$ is a product of distinct linear factors. By considering $S,m_{TS}(TS),T$ and by using the assumption that $ST$ is invertible, show that $m_{TS}(ST)=0$ and hence $m_{ST}|m_{TS}$.
answered Dec 7 '18 at 10:39
user1551user1551
72.4k566127
$endgroup$
$begingroup$
Thanks a lot. :)
$endgroup$
– Ashis Jana
Dec 8 '18 at 0:19
add a comment |
$begingroup$
Hints for (b). Denote the minimal polynomials of $ST$ and $TS$ by $m_{ST}$ and $m_{TS}$ respectively.
Suppose $ST$ is diagonalisable. Since $ST$ is invertible, $0$ is not a root of $m_{ST}$. Hence $p(x)=x,m_{ST}(x)$ is a product of distinct linear factors. Now, show that $p(TS)=0$ and hence $m_{TS}|p$.
Conversely, suppose $TS$ is diagonalisable, so that $m_{TS}$ is a product of distinct linear factors. By considering $S,m_{TS}(TS),T$ and by using the assumption that $ST$ is invertible, show that $m_{TS}(ST)=0$ and hence $m_{ST}|m_{TS}$.
answered Dec 7 '18 at 10:39
user1551user1551
72.4k566127
$endgroup$
$begingroup$
Thanks a lot. :)
$endgroup$
– Ashis Jana
Dec 8 '18 at 0:19
add a comment |
$begingroup$
Hints for (b). Denote the minimal polynomials of $ST$ and $TS$ by $m_{ST}$ and $m_{TS}$ respectively.
Suppose $ST$ is diagonalisable. Since $ST$ is invertible, $0$ is not a root of $m_{ST}$. Hence $p(x)=x,m_{ST}(x)$ is a product of distinct linear factors. Now, show that $p(TS)=0$ and hence $m_{TS}|p$.
Conversely, suppose $TS$ is diagonalisable, so that $m_{TS}$ is a product of distinct linear factors. By considering $S,m_{TS}(TS),T$ and by using the assumption that $ST$ is invertible, show that $m_{TS}(ST)=0$ and hence $m_{ST}|m_{TS}$.
answered Dec 7 '18 at 10:39
user1551user1551
72.4k566127
$endgroup$
Hints for (b). Denote the minimal polynomials of $ST$ and $TS$ by $m_{ST}$ and $m_{TS}$ respectively.
Suppose $ST$ is diagonalisable. Since $ST$ is invertible, $0$ is not a root of $m_{ST}$. Hence $p(x)=x,m_{ST}(x)$ is a product of distinct linear factors. Now, show that $p(TS)=0$ and hence $m_{TS}|p$.
Conversely, suppose $TS$ is diagonalisable, so that $m_{TS}$ is a product of distinct linear factors. By considering $S,m_{TS}(TS),T$ and by using the assumption that $ST$ is invertible, show that $m_{TS}(ST)=0$ and hence $m_{ST}|m_{TS}$.
answered Dec 7 '18 at 10:39
user1551user1551
72.4k566127
answered Dec 7 '18 at 10:39
user1551user1551
72.4k566127
answered Dec 7 '18 at 10:39
user1551user1551
72.4k566127
answered Dec 7 '18 at 10:39
user1551user1551
72.4k566127
72.4k566127
$begingroup$
Thanks a lot. :)
$endgroup$
– Ashis Jana
Dec 8 '18 at 0:19
add a comment |
$begingroup$
Thanks a lot. :)
$endgroup$
– Ashis Jana
Dec 8 '18 at 0:19
$begingroup$
Thanks a lot. :)
$endgroup$
– Ashis Jana
Dec 8 '18 at 0:19
$begingroup$
Thanks a lot. :)
$endgroup$
– Ashis Jana
Dec 8 '18 at 0:19
add a comment |
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$begingroup$
I recommend for c) you interpret your matrices as morhpisms and use dimensions
$endgroup$
– Enkidu
Dec 7 '18 at 9:09