# Finding the trace & determinant of a linear operator which takes any square matrix as an input.

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.

• 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 $w$ – Omnomnomnom Oct 5 '18 at 2:09
• 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 $A$? – 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.

• 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. – Omnomnomnom Oct 5 '18 at 2:26
• 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? – John Oct 5 '18 at 2:29
• 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 – Omnomnomnom Oct 5 '18 at 2:32
• Thanks, but what I'm asking is whether the converse works i.e does every extension of the eigenbasis triangularize $A$? – John Oct 5 '18 at 2:37
• @John No, the converse does not hold – Omnomnomnom Oct 5 '18 at 2:53