# How to approach a change of basis using matrices instead of vectors?

All the exercices we're doing in my linear algebra class use simple vectors when we're working on a basis, but every exam has a number that use a basis using $2 \times 2$ matrices. We've never covered that in class or examples so I have no idea how to approach it.

Example: Let $V$ be the vectorial space of the $2 \times 2$ square matrices, with the following basis:

$$\beta = \{ e_1 = \begin{pmatrix} 1 & 0\\ 0 & 0 \\ \end{pmatrix}, e_2 = \begin{pmatrix} 0 & 1\\ 0 & 0 \\ \end{pmatrix}, e_3 = \begin{pmatrix} 0 & 0\\ 1 & 0 \\ \end{pmatrix}, e_4 = \begin{pmatrix} 0 & 0\\ 0 & 1 \\ \end{pmatrix}\}$$

Let $M = \begin{pmatrix} 1 & 2\\ 3 & 4 \\ \end{pmatrix}$ and $T$ the linear transformation of $V$ defined by $T(A) = AM$ where $A ∈ V$

Find the matrix $[T]_{\beta\beta}$, representing $T$ in the basis $\beta$

Normally, if I was working with a basis made out of vectors instead of matrices, I would make an augmented matrix with the vectors and put in the original coordinates of the vector and find the new coordinates using the Gauss Jordan method. But I can't make a matrix with other matrices, can't I? I'm at a complete loss.

What I've been able to work out so far:

$$x = [M]_{\beta\beta}$$ $$x = \alpha_1 \begin{pmatrix} 1 & 0\\ 0 & 0 \\ \end{pmatrix} + \alpha_2 \begin{pmatrix} 0 & 1\\ 0 & 0 \\ \end{pmatrix} + \alpha_3 \begin{pmatrix} 0 & 0\\ 1 & 0 \\ \end{pmatrix} + \alpha_4 \begin{pmatrix} 0 & 0\\ 0 & 1 \\ \end{pmatrix} +$$

$$x = \begin{pmatrix} \alpha_1 & \alpha_2\\ \alpha_3 & \alpha_4 \\ \end{pmatrix}(e_1 + e_2 + e_3 + e_4)$$

Hint:

A matrix $$A=\begin{bmatrix}a&b\\ c&d \end{bmatrix}$$ in terms of the basis $\beta$ is expressed as: $$A=ae_1+be_2+ce_3+de_4$$

and we can write this as a ''vector'' of the components: $$A=\begin{bmatrix}a\\b\\ c\\d \end{bmatrix}$$

With this notation we have: $$T(A)=[T]_{\beta \beta}\begin{bmatrix}a\\b\\ c\\d \end{bmatrix}=AM=\begin{bmatrix}a&b\\ c&d \end{bmatrix}\begin{bmatrix}1&2\\ 3&4 \end{bmatrix}= \begin{bmatrix}a+3b&2a+4b\\ c+3d&2c+4d \end{bmatrix}= \begin{bmatrix}a+3b\\2a+4b\\ c+3d\\2c+4d \end{bmatrix}$$

So you can use the usual ''machinery of vectors'' that you know to find the matrix $[T]_{\beta \beta}$.

The matrix that represents the transformation $T$ in the basis $\beta$ is: $$[T]_{\beta \beta}= \begin{bmatrix} 1&3&0&0\\ 2&4&0&0\\ 0&0&1&3\\ 0&0&2&4 \end{bmatrix}$$

as you can easily verify.

• But how is the original basis B taken into account? And can you elaborate on "the usual machinery of vectors"? Usually the vectors coordinates are real numbers, whereas here we have several variables. – Dat8StringGuy Apr 23 '17 at 13:23
• The original basis are the vectors $e_1=[1,0,0,0]^T$, $e_2=[0,1,0,0]^T$, $e_3=[0,0,1,0]^T$ ,$e_4=[0,0,0,1]^T$ by means of wich is expressed the matrix $A$ as a linear combination. – Emilio Novati Apr 23 '17 at 13:29
• I'm not sure to understand the second question. I suppose that you know the ususal way to represents a linear transformation by means of a matrix, but I add something to may answer. – Emilio Novati Apr 23 '17 at 13:31