If $m_1,m_2$ are the minimal polynomials of $ST$ and $TS$ prove $m_2(x)=x^im_1(x)$ where $i=-1,0$ or $1$
Let V be a finite dimensional vector space, and let $S,T:V rightarrow V$ be linear transformations. Let $m_1,m_2$ denote the minimal polynomials of $ST$ and $TS$ respectively how would you prove:
$m_2(x)=x^im_1(x)$ where $i=-1,0$ or $1$ and $a$ is an eigenvalue of $ST$ iff a is an eigenvalue of $TS$?
linear-algebra matrices eigenvalues-eigenvectors minimal-polynomials
edited Dec 9 at 8:00
Martin Sleziak
44.7k7115270
asked Oct 31 '11 at 18:27
Freeman
88082163
add a comment |
Let V be a finite dimensional vector space, and let $S,T:V rightarrow V$ be linear transformations. Let $m_1,m_2$ denote the minimal polynomials of $ST$ and $TS$ respectively how would you prove:
$m_2(x)=x^im_1(x)$ where $i=-1,0$ or $1$ and $a$ is an eigenvalue of $ST$ iff a is an eigenvalue of $TS$?
linear-algebra matrices eigenvalues-eigenvectors minimal-polynomials
edited Dec 9 at 8:00
Martin Sleziak
44.7k7115270
asked Oct 31 '11 at 18:27
Freeman
88082163
I think one of those $m_1$s is supposed to be an $m_2$.
– Arturo Magidin
Oct 31 '11 at 18:30
@ArturoMagidin: Thanks -typo!
– Freeman
Oct 31 '11 at 18:31
add a comment |
Let V be a finite dimensional vector space, and let $S,T:V rightarrow V$ be linear transformations. Let $m_1,m_2$ denote the minimal polynomials of $ST$ and $TS$ respectively how would you prove:
$m_2(x)=x^im_1(x)$ where $i=-1,0$ or $1$ and $a$ is an eigenvalue of $ST$ iff a is an eigenvalue of $TS$?
linear-algebra matrices eigenvalues-eigenvectors minimal-polynomials
edited Dec 9 at 8:00
Martin Sleziak
44.7k7115270
asked Oct 31 '11 at 18:27
Freeman
88082163
Let V be a finite dimensional vector space, and let $S,T:V rightarrow V$ be linear transformations. Let $m_1,m_2$ denote the minimal polynomials of $ST$ and $TS$ respectively how would you prove:
$m_2(x)=x^im_1(x)$ where $i=-1,0$ or $1$ and $a$ is an eigenvalue of $ST$ iff a is an eigenvalue of $TS$?
linear-algebra matrices eigenvalues-eigenvectors minimal-polynomials
linear-algebra matrices eigenvalues-eigenvectors minimal-polynomials
edited Dec 9 at 8:00
Martin Sleziak
44.7k7115270
asked Oct 31 '11 at 18:27
Freeman
88082163
edited Dec 9 at 8:00
Martin Sleziak
44.7k7115270
asked Oct 31 '11 at 18:27
Freeman
88082163
edited Dec 9 at 8:00
Martin Sleziak
44.7k7115270
edited Dec 9 at 8:00
Martin Sleziak
44.7k7115270
edited Dec 9 at 8:00
Martin Sleziak
44.7k7115270
44.7k7115270
asked Oct 31 '11 at 18:27
Freeman
88082163
asked Oct 31 '11 at 18:27
Freeman
88082163
asked Oct 31 '11 at 18:27
Freeman
88082163
88082163
I think one of those $m_1$s is supposed to be an $m_2$.
– Arturo Magidin
Oct 31 '11 at 18:30
@ArturoMagidin: Thanks -typo!
– Freeman
Oct 31 '11 at 18:31
add a comment |
I think one of those $m_1$s is supposed to be an $m_2$.
– Arturo Magidin
Oct 31 '11 at 18:30
@ArturoMagidin: Thanks -typo!
– Freeman
Oct 31 '11 at 18:31
I think one of those $m_1$s is supposed to be an $m_2$.
– Arturo Magidin
Oct 31 '11 at 18:30
I think one of those $m_1$s is supposed to be an $m_2$.
– Arturo Magidin
Oct 31 '11 at 18:30
@ArturoMagidin: Thanks -typo!
– Freeman
Oct 31 '11 at 18:31
@ArturoMagidin: Thanks -typo!
– Freeman
Oct 31 '11 at 18:31
add a comment |
1 Answer
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Suppose that $a$ is an eigenvalue of $ST$. If $mathbf{v}in E_a$, the eigenspace of $ST$ associated to $a$, then
$$TSBigl(T(mathbf{v})Bigr) = TBigl(ST(mathbf{v})Bigr) = T(amathbf{v}) = aT(mathbf{v}),$$
so either $T(mathbf{v})=mathbf{0}$, or else $a$ is also an eigenvalue of $TS$.
So you have two cases: if $T(mathbf{v})=mathbf{0}$ for every $mathbf{v}in E_a$, then $T$ is not invertible; what can you conclude $a$ and about $TS$ in that case?
The other case is that there exists $mathbf{v}in E_a$ with $T(mathbf{v})neqmathbf{0}$. What can you conclude about $TS$ in that case?
Can you apply the same argument starting with $TS$ instead of with $ST$?
For the minimal polynomials, notice that for any $ngt 0$,
$$T(ST)^nS = (TS)^{n+1}.$$
If $m_1(x) = x^k + a_{k-1}x^{k-1}+cdots + a_1x + a_0$ is the minimal polynomial of $ST$, then
$$begin{align*}
(ST)^k + a_{k-1}(ST)^{k-1}+cdots + a_1(ST) + a_0I &= 0\
TBigl((ST)^k + a_{k-1}(ST)^{k-1}+cdots + a_1(ST) + a_0IBigr)S &= 0\
T(ST)^kS + a_{k-1}T(ST)^{k-1}S + cdots + a_1T(ST)S + a_0TS &= 0\
(TS)^{k+1} + a_{k-1}(TS)^k + cdots + a_1(TS)^2 + a_0(TS) & = 0,
end{align*}$$
so $TS$ satisfies $xm_1(x)$. Hence $m_2(x)$ divides $xm_1(x)$. By a symmetric argument, $m_2(x)$ divides $xm_1(x)$.
If $0$ is not an eigenvalue of $ST$ nor of $TS$, then $x$ cannot be a factor of the minimal polynomials; what can you conclude then?
If $0$ is an eigenvalue of $ST$ and of $TS$, then you cannot simply "take them out"; but it does tell you that every irreducible factor, except perhaps for the irreducible factor $x$, must be the same (and raised to the same degree) in $m_1(x)$ and in $m_2(x)$. Now you just need to deal with $x$, and show that the degree to which it shows in $m_1(x)$ and in $m_2(x)$ can differ by at most $1$, which will give you the result you want.
answered Oct 31 '11 at 19:03
Arturo Magidin
260k32584904
Thank you Arturo, this is an incredibly helpful response, just sad I can't vote it up more than once ;)
– Freeman
Oct 31 '11 at 19:39
add a comment |
Your Answer
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1 Answer
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1 Answer
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Suppose that $a$ is an eigenvalue of $ST$. If $mathbf{v}in E_a$, the eigenspace of $ST$ associated to $a$, then
$$TSBigl(T(mathbf{v})Bigr) = TBigl(ST(mathbf{v})Bigr) = T(amathbf{v}) = aT(mathbf{v}),$$
so either $T(mathbf{v})=mathbf{0}$, or else $a$ is also an eigenvalue of $TS$.
So you have two cases: if $T(mathbf{v})=mathbf{0}$ for every $mathbf{v}in E_a$, then $T$ is not invertible; what can you conclude $a$ and about $TS$ in that case?
The other case is that there exists $mathbf{v}in E_a$ with $T(mathbf{v})neqmathbf{0}$. What can you conclude about $TS$ in that case?
Can you apply the same argument starting with $TS$ instead of with $ST$?
For the minimal polynomials, notice that for any $ngt 0$,
$$T(ST)^nS = (TS)^{n+1}.$$
If $m_1(x) = x^k + a_{k-1}x^{k-1}+cdots + a_1x + a_0$ is the minimal polynomial of $ST$, then
$$begin{align*}
(ST)^k + a_{k-1}(ST)^{k-1}+cdots + a_1(ST) + a_0I &= 0\
TBigl((ST)^k + a_{k-1}(ST)^{k-1}+cdots + a_1(ST) + a_0IBigr)S &= 0\
T(ST)^kS + a_{k-1}T(ST)^{k-1}S + cdots + a_1T(ST)S + a_0TS &= 0\
(TS)^{k+1} + a_{k-1}(TS)^k + cdots + a_1(TS)^2 + a_0(TS) & = 0,
end{align*}$$
so $TS$ satisfies $xm_1(x)$. Hence $m_2(x)$ divides $xm_1(x)$. By a symmetric argument, $m_2(x)$ divides $xm_1(x)$.
If $0$ is not an eigenvalue of $ST$ nor of $TS$, then $x$ cannot be a factor of the minimal polynomials; what can you conclude then?
If $0$ is an eigenvalue of $ST$ and of $TS$, then you cannot simply "take them out"; but it does tell you that every irreducible factor, except perhaps for the irreducible factor $x$, must be the same (and raised to the same degree) in $m_1(x)$ and in $m_2(x)$. Now you just need to deal with $x$, and show that the degree to which it shows in $m_1(x)$ and in $m_2(x)$ can differ by at most $1$, which will give you the result you want.
answered Oct 31 '11 at 19:03
Arturo Magidin
260k32584904
Thank you Arturo, this is an incredibly helpful response, just sad I can't vote it up more than once ;)
– Freeman
Oct 31 '11 at 19:39
add a comment |
Suppose that $a$ is an eigenvalue of $ST$. If $mathbf{v}in E_a$, the eigenspace of $ST$ associated to $a$, then
$$TSBigl(T(mathbf{v})Bigr) = TBigl(ST(mathbf{v})Bigr) = T(amathbf{v}) = aT(mathbf{v}),$$
so either $T(mathbf{v})=mathbf{0}$, or else $a$ is also an eigenvalue of $TS$.
So you have two cases: if $T(mathbf{v})=mathbf{0}$ for every $mathbf{v}in E_a$, then $T$ is not invertible; what can you conclude $a$ and about $TS$ in that case?
The other case is that there exists $mathbf{v}in E_a$ with $T(mathbf{v})neqmathbf{0}$. What can you conclude about $TS$ in that case?
Can you apply the same argument starting with $TS$ instead of with $ST$?
For the minimal polynomials, notice that for any $ngt 0$,
$$T(ST)^nS = (TS)^{n+1}.$$
If $m_1(x) = x^k + a_{k-1}x^{k-1}+cdots + a_1x + a_0$ is the minimal polynomial of $ST$, then
$$begin{align*}
(ST)^k + a_{k-1}(ST)^{k-1}+cdots + a_1(ST) + a_0I &= 0\
TBigl((ST)^k + a_{k-1}(ST)^{k-1}+cdots + a_1(ST) + a_0IBigr)S &= 0\
T(ST)^kS + a_{k-1}T(ST)^{k-1}S + cdots + a_1T(ST)S + a_0TS &= 0\
(TS)^{k+1} + a_{k-1}(TS)^k + cdots + a_1(TS)^2 + a_0(TS) & = 0,
end{align*}$$
so $TS$ satisfies $xm_1(x)$. Hence $m_2(x)$ divides $xm_1(x)$. By a symmetric argument, $m_2(x)$ divides $xm_1(x)$.
If $0$ is not an eigenvalue of $ST$ nor of $TS$, then $x$ cannot be a factor of the minimal polynomials; what can you conclude then?
If $0$ is an eigenvalue of $ST$ and of $TS$, then you cannot simply "take them out"; but it does tell you that every irreducible factor, except perhaps for the irreducible factor $x$, must be the same (and raised to the same degree) in $m_1(x)$ and in $m_2(x)$. Now you just need to deal with $x$, and show that the degree to which it shows in $m_1(x)$ and in $m_2(x)$ can differ by at most $1$, which will give you the result you want.
answered Oct 31 '11 at 19:03
Arturo Magidin
260k32584904
Thank you Arturo, this is an incredibly helpful response, just sad I can't vote it up more than once ;)
– Freeman
Oct 31 '11 at 19:39
add a comment |
Suppose that $a$ is an eigenvalue of $ST$. If $mathbf{v}in E_a$, the eigenspace of $ST$ associated to $a$, then
$$TSBigl(T(mathbf{v})Bigr) = TBigl(ST(mathbf{v})Bigr) = T(amathbf{v}) = aT(mathbf{v}),$$
so either $T(mathbf{v})=mathbf{0}$, or else $a$ is also an eigenvalue of $TS$.
So you have two cases: if $T(mathbf{v})=mathbf{0}$ for every $mathbf{v}in E_a$, then $T$ is not invertible; what can you conclude $a$ and about $TS$ in that case?
The other case is that there exists $mathbf{v}in E_a$ with $T(mathbf{v})neqmathbf{0}$. What can you conclude about $TS$ in that case?
Can you apply the same argument starting with $TS$ instead of with $ST$?
For the minimal polynomials, notice that for any $ngt 0$,
$$T(ST)^nS = (TS)^{n+1}.$$
If $m_1(x) = x^k + a_{k-1}x^{k-1}+cdots + a_1x + a_0$ is the minimal polynomial of $ST$, then
$$begin{align*}
(ST)^k + a_{k-1}(ST)^{k-1}+cdots + a_1(ST) + a_0I &= 0\
TBigl((ST)^k + a_{k-1}(ST)^{k-1}+cdots + a_1(ST) + a_0IBigr)S &= 0\
T(ST)^kS + a_{k-1}T(ST)^{k-1}S + cdots + a_1T(ST)S + a_0TS &= 0\
(TS)^{k+1} + a_{k-1}(TS)^k + cdots + a_1(TS)^2 + a_0(TS) & = 0,
end{align*}$$
so $TS$ satisfies $xm_1(x)$. Hence $m_2(x)$ divides $xm_1(x)$. By a symmetric argument, $m_2(x)$ divides $xm_1(x)$.
If $0$ is not an eigenvalue of $ST$ nor of $TS$, then $x$ cannot be a factor of the minimal polynomials; what can you conclude then?
If $0$ is an eigenvalue of $ST$ and of $TS$, then you cannot simply "take them out"; but it does tell you that every irreducible factor, except perhaps for the irreducible factor $x$, must be the same (and raised to the same degree) in $m_1(x)$ and in $m_2(x)$. Now you just need to deal with $x$, and show that the degree to which it shows in $m_1(x)$ and in $m_2(x)$ can differ by at most $1$, which will give you the result you want.
answered Oct 31 '11 at 19:03
Arturo Magidin
260k32584904
Suppose that $a$ is an eigenvalue of $ST$. If $mathbf{v}in E_a$, the eigenspace of $ST$ associated to $a$, then
$$TSBigl(T(mathbf{v})Bigr) = TBigl(ST(mathbf{v})Bigr) = T(amathbf{v}) = aT(mathbf{v}),$$
so either $T(mathbf{v})=mathbf{0}$, or else $a$ is also an eigenvalue of $TS$.
So you have two cases: if $T(mathbf{v})=mathbf{0}$ for every $mathbf{v}in E_a$, then $T$ is not invertible; what can you conclude $a$ and about $TS$ in that case?
The other case is that there exists $mathbf{v}in E_a$ with $T(mathbf{v})neqmathbf{0}$. What can you conclude about $TS$ in that case?
Can you apply the same argument starting with $TS$ instead of with $ST$?
For the minimal polynomials, notice that for any $ngt 0$,
$$T(ST)^nS = (TS)^{n+1}.$$
If $m_1(x) = x^k + a_{k-1}x^{k-1}+cdots + a_1x + a_0$ is the minimal polynomial of $ST$, then
$$begin{align*}
(ST)^k + a_{k-1}(ST)^{k-1}+cdots + a_1(ST) + a_0I &= 0\
TBigl((ST)^k + a_{k-1}(ST)^{k-1}+cdots + a_1(ST) + a_0IBigr)S &= 0\
T(ST)^kS + a_{k-1}T(ST)^{k-1}S + cdots + a_1T(ST)S + a_0TS &= 0\
(TS)^{k+1} + a_{k-1}(TS)^k + cdots + a_1(TS)^2 + a_0(TS) & = 0,
end{align*}$$
so $TS$ satisfies $xm_1(x)$. Hence $m_2(x)$ divides $xm_1(x)$. By a symmetric argument, $m_2(x)$ divides $xm_1(x)$.
If $0$ is not an eigenvalue of $ST$ nor of $TS$, then $x$ cannot be a factor of the minimal polynomials; what can you conclude then?
If $0$ is an eigenvalue of $ST$ and of $TS$, then you cannot simply "take them out"; but it does tell you that every irreducible factor, except perhaps for the irreducible factor $x$, must be the same (and raised to the same degree) in $m_1(x)$ and in $m_2(x)$. Now you just need to deal with $x$, and show that the degree to which it shows in $m_1(x)$ and in $m_2(x)$ can differ by at most $1$, which will give you the result you want.
answered Oct 31 '11 at 19:03
Arturo Magidin
260k32584904
answered Oct 31 '11 at 19:03
Arturo Magidin
260k32584904
answered Oct 31 '11 at 19:03
Arturo Magidin
260k32584904
answered Oct 31 '11 at 19:03
Arturo Magidin
260k32584904
260k32584904
Thank you Arturo, this is an incredibly helpful response, just sad I can't vote it up more than once ;)
– Freeman
Oct 31 '11 at 19:39
add a comment |
Thank you Arturo, this is an incredibly helpful response, just sad I can't vote it up more than once ;)
– Freeman
Oct 31 '11 at 19:39
Thank you Arturo, this is an incredibly helpful response, just sad I can't vote it up more than once ;)
– Freeman
Oct 31 '11 at 19:39
Thank you Arturo, this is an incredibly helpful response, just sad I can't vote it up more than once ;)
– Freeman
Oct 31 '11 at 19:39
add a comment |
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I think one of those $m_1$s is supposed to be an $m_2$.
– Arturo Magidin
Oct 31 '11 at 18:30
@ArturoMagidin: Thanks -typo!
– Freeman
Oct 31 '11 at 18:31