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Let $\alpha$, $\beta$ be linear operators on a finitely-dimensional space $V$ over a field $F$. Let $\gamma=\beta\circ \alpha$ and $\delta=\alpha\circ\beta$. Suppose $\gamma$ is diagonalizable.

Prove that $\gamma^2$ and $\delta^2$ are conjugate, i.e., there exists a bijective linear operator $\phi$ on $V$ such that $\phi^{-1}\circ \gamma^2\circ \phi=\delta^2$.


(the professor said I should use the following lemma: Let $\alpha,\beta$ be linear operators on a finite dimensional vector space $V$ over field $F$. Let $\gamma=\alpha\circ\beta$ and $\delta=\beta\circ\alpha$. Prove that:

(1). $m_\delta(x)$ divides $xm_\gamma(x)$. Here $m_T(x)$ denotes the minimal polynomial of the linear operator $T$.

(2). If $\lambda\neq 0$, then $Dim(Ker(\gamma-\lambda \cdot Id))=Dim(Ker(\delta-\lambda \cdot Id))$.

Prof.'s hint:

As for your first question, we could start with the following

Lemma 1. $(\beta\circ\alpha)\circ(\beta\circ\alpha)^n = \beta\circ(\alpha\circ\beta)^n\circ\alpha$.

Proof. Exercise. [Hint. Verify the statement for $n=1$ and then induction on $n$. Notice that it's also true for $n=0$ :-D ]

Thanks to this lemma, we are able to prove

Lemma 2. $xm_\gamma (x)$ annihilates $\delta$.

From which statement (1) follows because the minimal polynomial of $\delta$ divides every polynomial that annihilates $\delta$.

Proof of lemma 2. Let $m_\gamma (x) = x^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0$. Then if we evaluate $xm_\gamma (x)$ on $\delta$, we get

\begin{eqnarray*} \delta \circ m_\gamma (\delta ) &= \beta\circ\alpha \circ \left((\beta\circ\alpha)^n + a_{n-1} (\beta\circ\alpha)^{n-1}+ \dots + a_1(\beta\circ\alpha) + a_0\mathrm{Id} \right) \\ &= \beta \circ \left((\alpha\circ\beta)^n + a_{n-1} (\alpha\circ\beta)^{n-1}+ \dots + a_1(\alpha\circ\beta) + a_0\mathrm{Id} \right)\circ\alpha \end{eqnarray*}

The last equality follows from lemma 1. But now, inside the parenthesis, we have $m_\gamma(\gamma)$, which is zero. QED

As for your second question, I make the following claim:

Proposition. Assume $\lambda \neq 0$ is an eigenvalue of $\gamma$. Then, restricted to $\mathrm{ker}\ (\gamma - \lambda\mathrm{Id})$ and $\mathrm{ker}\ (\delta -\lambda\mathrm{Id} )$, we have $\alpha\circ\beta = \lambda\mathrm{Id}$ and $\beta\circ\alpha = \lambda\mathrm{Id}$.

Hence, restricted to those subspaces, $\alpha$ and $\beta$ are isomorphisms. In particular, their dimensions are equal.

Proof of the proposition. Exercise. [Hint: verify that, if $v$ is an eigenvector of $\delta$ with eigenvalue $\lambda$, then so is $\alpha (v)$ for $\gamma$. Analogously, if $w$ is an eigenvector of $\gamma$ with eigenvalue $\lambda$, then so is $\beta (w)$ for $\delta$. Then, compute the compositions $\alpha \circ \beta$ and $\beta \circ\alpha$ restricted to those subspaces.])


I have followed Prof. Igor's instructions and proved that for each nonzero eigenvalue $\lambda$ of $\gamma^2$, the dimension of the eigenspace of $\delta^2$ is equal to the dimension of the eigenspace of $\delta^2$: $Dim(L_\lambda(\gamma^2))=Dim(L_\lambda(\delta^2))$. But this does not hold for $\lambda=0$. How to prove $\delta^2$ is diagonalizible hence $\gamma^2$ and $\delta^2$ will have same Jordan canonical form? It is possible that $\delta^2$ not diagonalizeble but all its nonzero eigenvalues are same as the nonzero eigenvalues of $\gamma^2$ and their corresponding eigenspaces are isomorphic to the eigenspaces of $\gamma^2$.

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  • $\begingroup$ isn't the category-theory tag a bit far-fetched? $\endgroup$
    – magma
    Dec 30, 2013 at 10:41

1 Answer 1

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First, ask yourself, why $\gamma^2?$ Your argument, for invertible $\beta$ works for $\gamma$ and $\delta,$ without any squares.

Let's try it anyhow. We know that $\alpha \gamma = \delta \alpha.$ Let $v$ be an eigenvector of $\gamma.$ Then $\alpha \gamma (v) = \lambda(v) \alpha(v),$ so $\delta\alpha (v) = \lambda(v) \alpha(v).$ So, if the action of $\alpha$ on the space spanned by the nonzero eigenvectors of $\gamma$ is nondegenerate, we are done (conjugate by $\alpha$ on the space of nonzero e.v.s, and extend to the kernel in some reasonably arbitrary way).

Now note that the action of $\alpha$ is always nondegenerate on the nonzero eigenvectors of $\gamma^2 = \beta \alpha \beta \alpha$ -- since the space spanned by the nonzero eigenvectors of $\gamma^2$ is the same as that spanned by nonzero eigenvector of $\gamma,$ and so if a vector got killed by $\alpha$ it would also be killed by $\gamma^2.$

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  • $\begingroup$ Dear Prof. Igor, I have proved for each nonzero eigenvalue $\lambda$ of $\gamma^2$, the dimension of the eigenspace of $\delta^2$ is equal to the dimension of the eigenspace of $\delta^2$: $Dim(L_\lambda(\gamma^2))=Dim(L_\lambda(\delta^2))$. But this does not hold for $\lambda=0$. How to prove $\delta^2$ is diagonalizible hence $\gamma^2$ and $\delta^2$ will have same Jordan canonical form? $\endgroup$
    – Shiquan
    Dec 30, 2013 at 4:46
  • $\begingroup$ I am confused on: So, if the action of α on the space spanned by the nonzero eigenvectors of γ is nondegenerate, we are done (conjugate by α on the space of nonzero e.v.s, and extend to the kernel in some reasonably arbitrary way. $\endgroup$
    – Shiquan
    Dec 30, 2013 at 4:59

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