Representing a linear transformation as a matrix in terms of a given basis

Question

I am new to linear algebra, I need help in understanding how to represent a linear transformation into standard basis of a matrix

Consider $M_{2}(\mathbb{R}),$ the vector space of all $2 \times 2$ real matrices. Let $$ A=\left(\begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array}\right) $$ and if we define $\mathcal{A}(B)=A B$ for any $B \in M_{2}(\mathbb{R})$. Show that $\mathcal{A}$ is a linear transformation on $M_{2}(\mathbb{R})$ and find the matrix of $\mathcal{A}$ under the basis $E_{i j}, i, j=1,2$

I can show the linearity part by considering the action of this linear transformation on matrix $B+ \lambda C$, in fact I know this will be true for any matrix $A$.
But how to represent this in terms of given basis.

Note here I have taken basis $E_{i j}$ be the $2 \times 2$ matrix with $(i, j)^{\text {th }}$ entry 1 and other entries 0.

Answer 1

You have a linear transformation $\mathcal{A}\colon M_{2\times 2}(\mathbb{R})\to M_{2\times 2}(\mathbb{R})$ and you would like to know its matrix in terms of a standard basis.

What you'd like to know is the coefficients $a_{ij}^{kl}$ when you write $\mathcal{A}E_{11} = a_{11}^{11}E_{11}+a_{12}^{11}E_{12}+a_{21}^{11}E_{21}+a_{22}^{11}E_{22}$ for example. Once you have these coefficients, you arrange them into an appropriate $4\times 4$ matrix form $T$ so that $\mathcal{A}(B) = T\begin{bmatrix} b_{11}&b_{12}&b_{21}&b_{22}\end{bmatrix}^T$ where $b_{ij}$ are the coefficients of $B$ with respect to the $E_{ij}$ basis.

You may also want to look into Representing a Linear Transformation as a Matrix

Answer 2

Let's look at how a 'common' matrix's coefficients are found. Suppose we want the matrix $R$ of a rotation by $90^\circ$ in $\mathbb{R}^2$. First see what action $R$ has on each basis element, then put these as columns of the matrix, keeping the same order. Thus since $R\mathbf{i}=\mathbf{j}$ and $R\mathbf{j}=-\mathbf{i}$, the matrix of $R$ would be $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$.

$M_2(\mathbb{R})$ has dimension 4, whose basis can be taken to be $E_{11}$, $E_{12}$, $E_{21}$, $E_{22}$. To find the matrix of $\mathcal{A}$, it is enough to find its action on the basis elements; the output $\mathcal{A}E_{ij}$ will be a column of the matrix.

$$\mathcal{A}(E_{11})=\begin{pmatrix}1&-1\\-1&1\end{pmatrix}\begin{pmatrix}1&0\\0&0\end{pmatrix}=\begin{pmatrix}1&0\\-1&0\end{pmatrix}=\begin{pmatrix}1&0\\0&0\end{pmatrix}-\begin{pmatrix}0&0\\1&0\end{pmatrix}=E_{11}-E_{21}$$ Similarly, $$\mathcal{A}(E_{12})=E_{12}-E_{22},\qquad\mathcal{A}(E_{21})=-E_{11}+E_{21},\qquad\mathcal{A}(E_{22})=-E_{12}+E_{22}$$ Hence the matrix of $\mathcal{A}$ with this basis in the given order is $$\begin{pmatrix}1&0&-1&0\\0&1&0&-1\\-1&0&1&0\\0&-1&0&1 \end{pmatrix}$$

For example, $\mathcal{A}(E_{12})$ corresponds to $$\begin{pmatrix}1&0&-1&0\\0&1&0&-1\\-1&0&1&0\\0&-1&0&1 \end{pmatrix}\begin{pmatrix}0\\1\\0\\0\end{pmatrix}=\begin{pmatrix}0\\1\\0\\-1\end{pmatrix}$$ which corresponds to $E_{12}-E_{22}$.