Let $A$ be an invertible, diagonalizable matrix and let $V$ be the space of $n \times n$ matrices. Define $T\colon V\to V$ be $T(X)=A^{-1}XA$. Find the eigenvalues, minimal, and characteristic polynomial of $T$.

I think I have what the eigenvalues could be. If we let $A=QDQ^{-1}$, where $D$ is diagonal with entries $\lambda_1,\dots,\lambda_n$ , then $T(X)=\lambda X$ is equivalent to $D^{-1}ZD=\lambda Z$, where $Z=Q^{-1}XQ$. Hence, comparing entries on the left and right, I get that $\lambda z_{ii}=z_{ii}$ for each $i$, and if $i\not=j$, $\lambda z_{ij}=\frac{\lambda_j}{\lambda_i}z_{ij}$. So the possible eigenvalues for $T$ are $\lambda=1$ or $\lambda=\frac{\lambda_j}{\lambda_i}$, $i\not=j$.

From here though, I don't see how to find the minimal or characteristic polynomial of $T$.

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    $\begingroup$ As you can exhibit $n^2$ eigenvectors, $T$ is diagonalizable, and the product of all linear factors $(x-\lambda)$ is the characteristic polynomial, whereas the minimal polynomial is obtained by dropping repeated factors from this $\endgroup$ – Hagen von Eitzen Apr 27 '18 at 21:41
  • $\begingroup$ @HagenvonEitzen Could you explain why there $n^2$ independent eigenvectors? It seems $T$ can have at most $n^2-n+1$ distinct eigenvalues (if $\lambda_j/\lambda_i\not=1$ for all $i\not=j)$. But I don't see how you get $n^2$ eigenvectors? $\endgroup$ – Tom Chalmer Apr 27 '18 at 22:16
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    $\begingroup$ In the case when $A$ is diagonal is it is easy to write down $n^2$ independent eigenvectors. $\endgroup$ – Derek Holt Apr 29 '18 at 22:00
  • $\begingroup$ @DerekHolt since $\lambda z_{ij}=\frac{\lambda_j}{\lambda_i}z_{ij}$ if $i\not=j$, and $\lambda z_{ii}=z_ii$, would the $n^2$ eigenvectors just be the standard basis vectors for the nxn matrices? That is, $E_{ij}$ being the matrix that is $1$ in the ij place and $0$ everywhere else? $\endgroup$ – Tom Chalmer Apr 29 '18 at 22:05
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    $\begingroup$ Yes that's what I had in mind! You can then conjugate them by $Q$ to get the eigenvectors for the original $A$. $\endgroup$ – Derek Holt Apr 30 '18 at 7:31

Indeed, those are the eigenvalues. In fact, $T$ is equivalent to the linear map $(A^T\otimes A^{-1})\mbox{vec}(X)$. Thus, the Jordan form consists of $1$-by-$1$ blocks (as $A$ is diagonalizable) and the eigenvalues are $\lambda_i\lambda_j^{-1}$ for all $i,j=1,\ldots,n$. Hence, the characteristic polynomial is $p(z)=(z-1)^n\prod_{i\neq j}(z-\lambda_i\lambda_j^{-1})$. To get the minimal polynomial, remove the repeating factors in $p(z)$.

  • $\begingroup$ I'm sorry, i'm not super familiar with the tensor notation. Could you explain what that part means? $\endgroup$ – Tom Chalmer Apr 27 '18 at 22:18

A possible way to show that those are the eigenvalues is to exibit the matrix of T.

It is useful to first note that the function $T(X)=AXA^{-1}$ is equivalent to $S(X)=DAD^{-1}$, in fact $$T(X)= Q D Q^{-1} X Q D^{-1} Q^{-1} = \beta_{Q} \circ S \circ \beta_{Q^{-1}}(X) $$ where $\beta_Q(X) = Q X Q^{-1}$ is an invertible linear map with determinant one.

We can now study $S$.

Let $\mathcal{E}$ be the canonical basis of $V$, space of the $n \times n$ matrices, ordered by columns.

The matrices of $L_{D^{-1}}(X) = D^{-1}X$ and $R_{D}(X)=XD$ in the basis $\mathcal{E}$ are easily shown by direct computation to be diagonal. Denoting with $l_k$ the $k$-th element on the diagonal of the first matrix and $r_k$ the $k$-th element on the diagonal of the second we have $l_{ni+j} = \frac{1}{\lambda _j}$ and $r_{ni+j}= \lambda_i$ for each $i,j \in \{1,..,n\}$ .

The matrix of $S$ is their product, wich is diagonal with entries of the form $s _{ni+j} = \frac{\lambda_i}{\lambda_j}$. The charateristic polinomial is then as follows: $$ p_S(x)= \prod_{\substack{1 \leq i \leq n \\ 1 \leq j \leq n }} (x - \frac{\lambda_i}{\lambda_j})$$ The minimal is the same with repeating factors removed.

  • $\begingroup$ The OP asks for a detailed proof of $spectrum(T)=(\lambda_i/\lambda_j)$; you prove only that each eigenvalue is in this form, which is clear. $\endgroup$ – loup blanc May 4 '18 at 22:10
  • $\begingroup$ @loupblanc I added the exact form of the matrices that fully determines the spectrum. Let me know if it clarifies and if you have any suggestion to improve the exposition. $\endgroup$ – Dario May 5 '18 at 14:05
  • $\begingroup$ OK. Now your proof is complete. $\endgroup$ – loup blanc May 6 '18 at 10:34

The @chhro 's answer is correct (I upvoted it). We can add that follows:

$A,A^{-1},A^T$ are diagonalizable. Let $spectrum(A)=(\lambda_i)$ and let $(u_i)$ be an associated eigen-basis of $A$, $(v_i)$ be an associated eigen-basis of $A^T$. Then $A^{-1}u_i=\dfrac{1}{\lambda_i}u_i$ and $v_j^TA=\lambda_jv_j^T$.

Note that $\mathcal{B}=(u_iv_j^T)_{i,j}$ has $n^2$ elements and is a basis of $M_n$. Moreover, $T(u_iv_j^T)=\dfrac{\lambda_j}{\lambda_i}u_iv_j^T$ and therefore $\mathcal{B}$ is a basis of eigenvectors of $T$ and $spectrum(T)=(\lambda_j/\lambda_i)_{i,j}$.

Conclusion: $T$ is diagonalizable, its characteristic polynomial is $p(x)=\Pi_{i,j}(x-\lambda_i/\lambda_j)$ and its minimal polynomial is constituted by the product of distinct terms of $p(x)$.

EDIT to @Tom Chalmer . The OP offers a bounty for a detailed answer to his above question. He obtains three answers, two absolutely complete, including that of @Dario . Not only does he not give the promised bounty but he does not upvote any answer and he did not deign to write any comment for two of the answers.

I find this attitude very contemptuous and therefore absolutely inadmissible.


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