1
$\begingroup$

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 $f$?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – qiubit Dec 29 '14 at 23:58
  • $\begingroup$ 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. $\endgroup$ – Bernard Dec 30 '14 at 0:15
1
$\begingroup$

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 $(\alpha_{i,j})$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.