Let $A\in Mat_n(\mathbb{C})$ and $T:Mat_n(\mathbb{R})\rightarrow Mat_n(\mathbb{R})$ defined by $T(X)=AX$ Calculate $\det(T)$

Question

Let $A\in Mat_n(\mathbb{C})$ and $T:Mat_n(\mathbb{R})\rightarrow Mat_n(\mathbb{R})$ defined by $T(X)=AX$ Calculate $\det(T)$

My work:

My definition: $\det(T)=\sum _{\sigma\in G_n}ε(\sigma)a_{1\sigma(1)}\cdots a_{n\sigma(n)}$

Let $B=\{A_1,\cdots,A_n\}$ a basis of $Mat_n(\mathbb{R})$ then

$\det_B(T(A_1),\cdots,T(A_n))=\alpha\det_B(A_1,\cdots,A_n)$ with $\alpha\in\mathbb{K}$

Moreover $\det_B(T(A_1),\cdots,T(A_n))=\det_B(A_1X,...,A_nX)$

Here i'm stuck. can someone help me?

Answer 1

Here's one eigenvalue-free approach: let $e_1,\dots,e_n$ denote the canonical basis of $\Bbb R^n$ (consisting of column-vectors). We compute the matrix of $T$ with respect to the basis $$ \mathcal B = (e_1^Te_1,e_2e_1^T,\dots,e_ne_1^T,\\ \qquad e_1^Te_2,e_2e_2^T,\dots,e_ne_2^T,\\ \vdots\\ \qquad e_1^Te_n,e_2e_n^T,\dots,e_ne_n^T) $$ To be the block-diagonal matrix $$ [T]_{\mathcal B} = \pmatrix{A&0& \cdots & 0\\ 0&A&\ddots & \vdots\\ \vdots & \ddots & \ddots & 0\\ 0 & \cdots & 0 & A } $$ Show that the determinant of this matrix is $\det(A)^n$.