4.3.1

Let V be the vector space of real 2x2 symmetric matrices, let $$ A = \begin{bmatrix} 2 & 1 \\ 0 & 1 \\ \end{bmatrix}$$, and let L be a linear operator on V such that $$L(X) = AXA^T$$, where $$A^T $$ denotes the tranpose of A.

We have a basis $$(M_1,M_2,M_3) $$ for V, where $$M_1 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ \end{bmatrix}, M_2 = \begin{bmatrix} 0&1\\ 1&0\\ \end{bmatrix}, M_3 = \begin{bmatrix} 0&0\\ 0&1\\ \end{bmatrix}.$$

Then, we can explicitly compute the action of L on the elements of this basis: $$L(M_1) = \begin{bmatrix} 4&0\\ 0&0\\ \end{bmatrix}, L(M_2) = \begin{bmatrix} 4&2\\ 2&0\\ \end{bmatrix}, L(M_3) = \begin{bmatrix} 1&1\\ 1&1\\ \end{bmatrix} .$$

We can rewrite these equations as $$L(M_1) = 4M_1, \ L(M_2) = 4M_1 + 2M_2, \  L(M_3) = M_1 + M_2 + M_3 $$.

Thus we have $$L(M_1,M_2,M_3) = (M_1,M_2,M_3) \begin{bmatrix} 4&4&1\\ 0&2&1\\ 0&0&1\\ \end{bmatrix}$$, giving us the matrix of L with respect to the basis $$(M_1,M_2,M_3). $$