Change of basis linear transformation Let $S : M_{2,2} → M_{2,2}$ be the linear transformation defined by $S(A) = A + A^T$
.
Consider the bases for $M_{2,2}$ given by:
$$B = \left\{\begin{bmatrix} 1&0\\0&0 \end{bmatrix}, \begin{bmatrix} 0&1\\0&0 \end{bmatrix}, \begin{bmatrix} 0&0\\1&0 \end{bmatrix}, \begin{bmatrix} 0&0\\0&1 \end{bmatrix} \right\}$$
and
$$c = \left\{\begin{bmatrix} 1&0\\0&0 \end{bmatrix}, \begin{bmatrix} 2&-1\\0&0 \end{bmatrix}, \begin{bmatrix} 0&0\\1&1 \end{bmatrix}, \begin{bmatrix} -3&0\\0&2 \end{bmatrix} \right\}$$
(a) Determine the transition matrix $P_{B,C}$ from C to B.
(b) Determine the transition matrix $P_{C,B}$ from B to C.
This question is very different from the types I have encountered thus far so please help me solve it, I have been struggling for a few hours now.

We have that by matrix representation of the same vector $v$ with respect to the two basis


*

*$v=M_Bv_B$

*$v=M_Cv_C$
with
$$M_B=\left(\begin{array} &1 & 0 & 0 & 0 \\ 0 & 1 &0 &0 \\ 0& 0 &1&0\\ 0 & 0 &0 & 1 \end{array} \right), \quad M_C=\left(\begin{array} &1 & 2 & 0 & -3 \\ 0 & -1 &0 &0 \\ 0& 0 &1&0\\ 0 & 0 &1 & 2 \end{array} \right)$$
therefore
$$M_Bv_B=M_Cv_C \implies v_B=M_B^{-1}M_Cv_C \quad \land \quad v_C=M_C^{-1}M_Bv_B$$
that is $P_{C,B}=M_C^{-1}M_B$ and $P_{B,C}=M_B^{-1}M_C$.
