I am working on learning how to solve matrix problems similar to this one and would appreciate an example if someone were willing to walk me through it.
Let T: $R^2$ -> $R^3$ Have the standard matrix $$ \begin{bmatrix} 1\space\space\space\space\ \ 2 \\ 2\space\ -1 \\ 1 \space\space\space\space\space\ \ 0 \end{bmatrix} $$ B is the basis $B$ = $ \{\ \begin{bmatrix} 1 \\ 1 \end{bmatrix} \ \ ,\ \begin{bmatrix} 1 \\ -1 \end{bmatrix} \ \}\ of \space R^2 $$ \quad \text {C is the Basis} \ C = $$ \{ \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} , \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} , \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}\} \space of \space R^3$ Find the matrix for $T$ relative to the Bases $B$ and $C$
Any suggestions will be appreciated
You know that $T(1,1)=T(1,0)+T(0,1)=(1,2,1)+(2,-1,0)=(3,1,1)$. Now you need to find the vector $(3,1,1)$ w.r.t. the basis $C$. Thus you need to solve the following system of equations:
$$\begin{pmatrix} 1&0&1&\mid&3\\ 0&1&1&\mid&1\\ 1&1&0&\mid&1 \end{pmatrix}=\begin{pmatrix} 1&0&1&\mid&3\\ 0&1&1&\mid&1\\ 0&1&-1&\mid&-2 \end{pmatrix}=\begin{pmatrix} 1&0&0&\mid&\frac{3}{2}\\ 0&1&0&\mid&-\frac{1}{2}\\ 0&0&1&\mid&\frac{3}{2} \end{pmatrix}$$
Hence $(3,1,1)=\frac{3}{2}(1,0,1)-\frac{1}{2}(0,1,1)+\frac{3}{2}(1,1,0)$. It follows that the matrix of $T$ w.r.t. the bases $B$ and $C$ is given by $$\begin{pmatrix} \frac{3}{2}&*\\ -\frac{1}{2}&*\\ \frac{3}{2}&* \end{pmatrix}$$.
Find the second column yourself. The key is to remember that the matrix of a transformation $P$ w.r.t. the bases $A$ and $B$ is given by the following procedure:
Find $P(a_i)$ ($P$ applied to the $i$-th basisvector of $A$). Write $P(a_i)=\sum_{j}\lambda_{ji}b_j$. That is, write $P(a_i)$ w.r.t. the basis $B$. The coefficients $\lambda_{ji}$ make up the $i$-th column of the matrix that we are looking for.