# What does it mean to find a matrix of linear transformation in given basis?

I have a following problem to do:

A linear transformation $f: \Bbb{R}^3 \rightarrow \Bbb{R}^2$ is defined with a formula:

$$f(\mathbf{x}) =( \begin{smallmatrix} x_1+x_2\\ x_2+x_3 \end{smallmatrix})$$

Find $ker (f), im (f)$ and the matrix of $f$ in bases

($\mathbf{e_1},\mathbf{e_1+e_2},\mathbf{e_1+e_2+e_3}), (\mathbf{e_1+e_2}, \mathbf{e_1-e_2}).$

I managed to do the 1st part of this problem - it's easy to see that $ker(f) = span \{(1,-1,1)^T\}$ and $im(f) = \Bbb{R}^2$ but I can't understand what should be done in the second part. What does it mean to find a matrix of $f$?

Any linear transformation between two finite dimensional vector spaces can be represented as a matrix. Suppose that you have two vector spaces $V_1$ and $V_2$, where the first has basis $e_1, e_2, e_3$, and the second has basis $f_1, f_2$.

The identification goes as follows. You identify $e_1 = \left(\begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix}\right)$, $e_2 = \left(\begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix}\right)$, $e_3 = \left(\begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix}\right)$, and $f_1 = \left(\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right)$, $f_2 = \left(\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right)$. This gives you a map from $V_1$ to $\mathbb{R}^3$, and from $V_2$ to $\mathbb{R}^2$, which are now very familiar spaces.

Then, if you have a linear transformation $L: V_1 \rightarrow V_2$, then you can find a matrix $M_L$ such that when you multiply vectors in $\mathbb{R}^3$, it gives you vectors in $\mathbb{R}^2$ in exactly the way that applying the transformation $L$ would.

So, suppose that $L(e_1) = f_1 + f_2$, $L(e_2) = f_1$, and $L(e_3) = f_2$. By our identification, we are saying that $L$ maps in the following way: $\left(\begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix}\right) \mapsto \left(\begin{smallmatrix} 1 \\ 1 \end{smallmatrix}\right)$, $\left(\begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix}\right) \mapsto \left(\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right)$, and $\left(\begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix}\right) \mapsto \left(\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right)$.

There is a unique matrix that does this, and that is $M_L = \left(\begin{smallmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \end{smallmatrix}\right)$. Your problem is similar.

• According to this explanation we only need one basis to determine the matrix, right? So does that mean that if I have 2 different bases given that's like 2 independent excercises I have to do? – qiubit Dec 29 '14 at 23:58
• You need a basis for $V_1$ and a basis for $V_2$. Now, if you change basis in these spaces, the matrix of a linear map changes too, but there are formulae for what happens to the matrix. – Bernard Dec 30 '14 at 0:15

The following is one way to define the matrix that represents a linear transformation with respect to specific bases. There are other ways to think of it, but eventually they all lead to the same matrix.

Let $U,V,$ be finitely generated vector fields over the field $F$. Let $u_1,\ldots,u_n,$ be a basis of $U$, $v_1,\ldots,v_m,$ a basis of $V$, and let $T:U\to V$ be a linear transformation. For $j=1,\ldots,n,$ let $\alpha_{1,j},\ldots\alpha_{m,j}\in F$ such that $$\sum_{i=1}^m\alpha_{i,j}\cdot v_i=T(u_j).$$Note that the above equation defines the $\alpha$'s well since $v_1,\ldots,v_m,$ is a basis of $V$.

The matrix that represents $T$ with respect to the above bases is the $m\times n$ matrix whose entries are $(\alpha_{i,j})$.