Discussion
Let \(A = \begin{pmatrix} 0 & 2 \\ 2 & 0 \end{pmatrix}\) and \(T : M_{2}(\mathbb{C}) \to M_{2}(\mathbb{C})\) be the linear transformation given by \(T(B) = AB\). The characteristic polynomial of \(T\) is
Answer: A
The characteristic polynomial is \(X^{4} - 8X^{2} + 16\).
Left multiplication by \(A\) acts on each column of \(B\) separately: if \(B = [b_{1}\ b_{2}]\), then \(AB = [Ab_{1}\ Ab_{2}]\). So \(T\) is \(A\) acting on two independent copies of \(\mathbb{C}^{2}\), and its characteristic polynomial is the square of that of \(A\).
\[ \begin{aligned} \det(XI - A) &= X^{2} - 4 \\ \chi_{T}(X) &= (X^{2} - 4)^{2} = X^{4} - 8X^{2} + 16 \end{aligned} \]
\(X^{2} - 4\) is the polynomial of \(A\) itself, but \(T\) acts on a 4-dimensional space so its polynomial has degree 4. \(X^{4} - 16\) has eigenvalues \(\pm 2, \pm 2i\), but \(T\)'s eigenvalues are only \(\pm 2\), each twice.
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