We define $T(X) = AX$ with $A$ & $X$ square matrices of size $n$ with complex entries.

First, write down all the eigenvalues (with their respective algebraic multiplicities) of $T$. Use this to compute the trace & determinant of $A$.

I have proven that $A$ and $T$ have the same eigenvalues. My guess is that each eigenvalue of $A$ occurs with algebraic multiplicity $n$ times in order to give us a total sum of $n^2$, the dimension of the space. This is the part I'd like some help proving.

Possible idea : Compute $T$ for a basis which puts it into upper triangular form and then read off the diagonal entries for the eigenvalues.

Any help would be great.

  • $\begingroup$ One approach is to note that if $v$ is an eigenvector of $A$, then $vw^T$ is an eigenvector of $T$ for any non-zero vector $w$ $\endgroup$ – Omnomnomnom Oct 5 '18 at 2:09
  • $\begingroup$ But then what basis would we take to make $T$ upper triangular? I was thinking we complete an eigenbasis of $A$ to a basis for all column vectors and take another basis $w_i$ of all column vectors so that $v_i w_j^T$ is a basis for the set of all matrices. But how do you compute $T$ on $v_i w_j^T$ when $v_i$ is not an eigenvector for $A$? $\endgroup$ – John Oct 5 '18 at 2:15

Suppose that the basis $\mathcal B = \{v_1,\dots,v_n\}$ triangularizes $A$ (so that $[A]_{\mathcal B}$ is upper triangular), and take any basis $\{w_1,\dots,w_n\}$ of $\Bbb R^n$. Let $\mathcal B^*$ denote the basis of $\Bbb C^{n \times n}$ given by $$ \mathcal B^* = \{v_i w_j^T : 1 \leq i,j \leq n\} $$ where the tuples $(j,i)$ are taken in lexicographical order. The matrix of $T$ relative to $\mathcal B^*$ is given by the block-diagonal matrix $$ [T]_{\mathcal B^*} = \pmatrix{[A]_{\mathcal B} \\ & \ddots & \\ && [A]_{\mathcal B}} = I \otimes [A]_{\mathcal B} $$ where $I$ denotes the identity matrix and $\otimes$ denotes the Kroneker product.

  • $\begingroup$ If we order the tuples $(i,j)$ in lexicographical order, then we instead end up with the matrix $$ [A]_{\mathcal B} \otimes I $$ which is also upper triangular, but less convenient to write in its block form. $\endgroup$ – Omnomnomnom Oct 5 '18 at 2:26
  • $\begingroup$ So a basis of the form $v_i w_j^T$ (over all $i$ and $j$) where $v_i$ (over all $i$ again) are a basis formed by extending the eigenbasis of $A$ doesn't work? $\endgroup$ – John Oct 5 '18 at 2:29
  • $\begingroup$ It is of course possible to extend the eigenbasis into a basis that triangularizes $A$ (e.g. the Jordan basis). Any basis that triangularizes $A$ is an extension of the eigenbasis $\endgroup$ – Omnomnomnom Oct 5 '18 at 2:32
  • $\begingroup$ Thanks, but what I'm asking is whether the converse works i.e does every extension of the eigenbasis triangularize $A$? $\endgroup$ – John Oct 5 '18 at 2:37
  • 1
    $\begingroup$ @John No, the converse does not hold $\endgroup$ – Omnomnomnom Oct 5 '18 at 2:53

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