Find the matrices of the linear mappings in the following bases

Question

Find the matrices of the linear mappings in the following bases

$F(x,y,z) = (x-y+z, x+y-z)$

$\{(1,1,1), (1,0,0), (0,0,1)\}$ and $\{(1,2), (3,3)\}$

I don't really know how to start this task. It is a bit confusing for me as there is 2nd to 3d. Thanks for help

Answer 1

By definition, the matrix associated to a linear map between vector spaces with two given bases is the matrix that has for columns the images of the vectors of the first basis expressed in terms of the second basis. Hence, a linear map from an $n$ dimensional vector space to an $m$ dimensional vector space will have a matrix with $n$ columns (one for each basis element in the source vector space) and $m$ rows (one for each coefficient needed to express the image of a vector in terms of the basis of the target vector space).

So in your case you will get a $2\times 3$ matrix whose first column is the vector $F(1,1,1)$ expressed in terms of the basis $\{ (1,2),(3,3) \}$, and so on.

Answer 2

Here's how to find the matrix representation in a more general setting. Suppose we have a linear transformation $T: V\to W$ where the domain and codomain are vector spaces. Suppose $V$ has a basis $\mathcal{B}=(v_1,\ldots, v_n)$ and $W$ has a basis $\mathcal{B}'=(w_1,\ldots, w_m)$. The matrix representation of $T$ with respect to $\mathcal{B}$ and $\mathcal{B}'$ is an $m\times n$ matrix. To find the entries of the $j^{th}$ column, for $1\le j \le n$, we take the (unique) expression $$T(v_j)=\sum_{i=1}^m a_{ij}w_i.$$ Then, the $j^{th}$ column of the matrix representation $\mathcal{M}(T,\mathcal{B}, \mathcal{B'})$ is $$ \begin{bmatrix} a_{1j}\\ \vdots\\ a_{mj} \end{bmatrix}$$ Let's apply this procedure to our current situation. Let's label $$v_1=(1,1,1), v_2=(1,0,0), v_3=(0,0,1)$$ and $$w_1=(1,2), w_2=(3,3).$$ We have a linear transformation given by $$F(x,y,z)=(x-y+z,x+y-z) $$ Then $$F(v_1)=F(1,1,1)=(1,1)=\frac{1}{3}w_2$$ $$ F(v_2)=F(1,0,0)=(1,1)=\frac{1}{3}w_2$$ $$ F(v_3)=F(0,0,1)=(1,-1)=w_2-2w_1.$$ So, our matrix is $$ \mathcal{M}(F,\mathcal{B},\mathcal{B}')=\begin{bmatrix} 0&0&-2\\ \frac{1}{3}&\frac{1}{3}&1 \end{bmatrix}.$$