# Linear transformation and matrix basis

Consider the linear transformation $T:M_{2x2}(\mathbb{R}) \rightarrow M_{2x2}(\mathbb{R})$ defined by $T(A)= A^T$. Consider the basis $B$ of $M_{2x2}(\mathbb{R})$, defined by:

$$B=\left(\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix},\begin{bmatrix}0&1\\1&0\end{bmatrix},\begin{bmatrix}1&0\\1&0\end{bmatrix}\right)$$

What is the matrix $M=(T; B, B)$ that represents $T$ in this basis? I 've been around this exercise all weekend and seem to be going around in circles. Woud really appreciate some help on this. Thanks.

• What have you tried? Like applying $T$ to each of your basis vectors in turn, and expressing the result as a linear combination of the basis vectors? It will be useful to name the basis vectors, like $e_1$, $e_2$, $e_3$, $e_4$ or whatever naming scheme you fancy. – Harald Hanche-Olsen Jan 26 '14 at 10:26
• What I find difficult is that my vectors are all 2x2 matrices and I don't know where to go from there. Shuld I transpose these vectors? – Maria Jan 26 '14 at 10:29

Write you basis as $b_1, \cdots, b_4$. Then $T(b_1)=b_1=1\cdot b_1 +0\cdot b_2+0\cdot b_3+0\cdot b_4$. Similarly, $T(b_2)=b_2, T(b_3)=b_3$. $T(b_4)=2\cdot b_1+0\cdot b_2+1\cdot b_3+(-1)\cdot b_4$. Hence the matrix for T is $$\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 2 & 0 & 1 & -1 \end{array} \right)$$
Note that, if $A=\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)$, then $T(A)=T\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)=\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)^{t}=\left( \begin{array}{cc} a & c \\ b & d \\ \end{array} \right)$