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 $1$












2












$begingroup$


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







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  • $begingroup$
    I think one of those $m_1$s is supposed to be an $m_2$.
    $endgroup$
    – Arturo Magidin
    Oct 31 '11 at 18:30










  • $begingroup$
    @ArturoMagidin: Thanks -typo!
    $endgroup$
    – Freeman
    Oct 31 '11 at 18:31
















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think one of those $m_1$s is supposed to be an $m_2$.
    $endgroup$
    – Arturo Magidin
    Oct 31 '11 at 18:30










  • $begingroup$
    @ArturoMagidin: Thanks -typo!
    $endgroup$
    – Freeman
    Oct 31 '11 at 18:31














2












2








2


3



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 8:00









Martin Sleziak

44.7k9117272




44.7k9117272










asked Oct 31 '11 at 18:27









FreemanFreeman

89092163




89092163












  • $begingroup$
    I think one of those $m_1$s is supposed to be an $m_2$.
    $endgroup$
    – Arturo Magidin
    Oct 31 '11 at 18:30










  • $begingroup$
    @ArturoMagidin: Thanks -typo!
    $endgroup$
    – Freeman
    Oct 31 '11 at 18:31


















  • $begingroup$
    I think one of those $m_1$s is supposed to be an $m_2$.
    $endgroup$
    – Arturo Magidin
    Oct 31 '11 at 18:30










  • $begingroup$
    @ArturoMagidin: Thanks -typo!
    $endgroup$
    – Freeman
    Oct 31 '11 at 18:31
















$begingroup$
I think one of those $m_1$s is supposed to be an $m_2$.
$endgroup$
– Arturo Magidin
Oct 31 '11 at 18:30




$begingroup$
I think one of those $m_1$s is supposed to be an $m_2$.
$endgroup$
– Arturo Magidin
Oct 31 '11 at 18:30












$begingroup$
@ArturoMagidin: Thanks -typo!
$endgroup$
– Freeman
Oct 31 '11 at 18:31




$begingroup$
@ArturoMagidin: Thanks -typo!
$endgroup$
– Freeman
Oct 31 '11 at 18:31










1 Answer
1






active

oldest

votes


















3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you Arturo, this is an incredibly helpful response, just sad I can't vote it up more than once ;)
    $endgroup$
    – Freeman
    Oct 31 '11 at 19:39











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1 Answer
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1 Answer
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active

oldest

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active

oldest

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active

oldest

votes









3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you Arturo, this is an incredibly helpful response, just sad I can't vote it up more than once ;)
    $endgroup$
    – Freeman
    Oct 31 '11 at 19:39
















3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you Arturo, this is an incredibly helpful response, just sad I can't vote it up more than once ;)
    $endgroup$
    – Freeman
    Oct 31 '11 at 19:39














3












3








3





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 31 '11 at 19:03









Arturo MagidinArturo Magidin

262k34586908




262k34586908












  • $begingroup$
    Thank you Arturo, this is an incredibly helpful response, just sad I can't vote it up more than once ;)
    $endgroup$
    – Freeman
    Oct 31 '11 at 19:39


















  • $begingroup$
    Thank you Arturo, this is an incredibly helpful response, just sad I can't vote it up more than once ;)
    $endgroup$
    – Freeman
    Oct 31 '11 at 19:39
















$begingroup$
Thank you Arturo, this is an incredibly helpful response, just sad I can't vote it up more than once ;)
$endgroup$
– Freeman
Oct 31 '11 at 19:39




$begingroup$
Thank you Arturo, this is an incredibly helpful response, just sad I can't vote it up more than once ;)
$endgroup$
– Freeman
Oct 31 '11 at 19:39


















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