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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?

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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
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1 Answer

up vote 1 down vote accepted

Suppose that $a$ is an eigenvalue of $ST$. If $\mathbf{v}\in E_a$, the eigenspace of $ST$ associated to $a$, then
$$TS\Bigl(T(\mathbf{v})\Bigr) = T\Bigl(ST(\mathbf{v})\Bigr) = T(a\mathbf{v}) = aT(\mathbf{v}),$$ 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 $T(\mathbf{v})\neq\mathbf{0}$. 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 $n\gt 0$, $$T(ST)^nS = (TS)^{n+1}.$$

If $m_1(x) = x^k + a_{k-1}x^{k-1}+\cdots + a_1x + a_0$ is the minimal polynomial of $ST$, then $$\begin{align*} (ST)^k + a_{k-1}(ST)^{k-1}+\cdots + a_1(ST) + a_0I &= 0\\ T\Bigl((ST)^k + a_{k-1}(ST)^{k-1}+\cdots + a_1(ST) + a_0I\Bigr)S &= 0\\ T(ST)^kS + a_{k-1}T(ST)^{k-1}S + \cdots + a_1T(ST)S + a_0TS &= 0\\ (TS)^{k+1} + a_{k-1}(TS)^k + \cdots + a_1(TS)^2 + a_0(TS) & = 0, \end{align*}$$ 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.

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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
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