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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.

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  • $\begingroup$ 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. $\endgroup$ – Harald Hanche-Olsen Jan 26 '14 at 10:26
  • $\begingroup$ 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? $\endgroup$ – Maria Jan 26 '14 at 10:29
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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 $$\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 2 & 0 & 1 & -1 \end{array} \right)$$

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Note that, if $A=\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)$, then $T(A)=T\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)=\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)^{t}=\left( \begin{array}{cc} a & c \\ b & d \\ \end{array} \right)$

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  • $\begingroup$ How do I go from here to matrix M? I have four 2x2 basis matrices and in the end I will have a 4x4 matrix after applying the transformation. $\endgroup$ – Maria Jan 26 '14 at 11:04
  • $\begingroup$ The transformation takes 2x2 matrices on 2x2 matrices. The transformation matrix is 4x4. $\endgroup$ – VVCM Jan 26 '14 at 11:11
  • $\begingroup$ If I transpose the 2x2 matrices on B, I get B' with the corresponding transposed matrices. My difficulty is in going from there to calculate M. $\endgroup$ – Maria Jan 26 '14 at 11:15
  • $\begingroup$ Let B = {u, v, w, z}. Calculate T (u) and then write T (u) = au + bv + cw + dz. Determine a, b​​, c and d. The vector (a, b, c, d) is the first column of M. Repeat for v, to find the second column of M. Etc. $\endgroup$ – VVCM Jan 26 '14 at 11:24

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