Find the matrix associated with the linear map $f:R^2 \rightarrow R^2$ defined by $f(x,y)=(3x-y,y-x)$ with respect to the ordered basis ${(1,0),(1,1)}$
Let the matrix be $A$ and let $f(x)=AX$ where
$$X=\begin{bmatrix}
x \\
y
\end{bmatrix}$$
and
$$A=\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}$$
I tried solving for $a,b,c,d$ by using the basis vectors as $(x,y)$. That is, I took $(x,y)=(1,0)$ and $(x,y)=(1,1)$ to find $a,b,c,d$. But I am not getting the given answer.
Best Answer
Regarding to the standard basis $e_i$ the map has the matrix $$ A = \begin{pmatrix} 3 & -1 \\ -1 & 1 \end{pmatrix} $$ because $$ f(x) = x A = (x_1 e_1 + x_2 e_2) A = x_1 e_1 A + x_2 e_2 A = (x_1, x_2) \begin{pmatrix} e_1 A \\ e_2 A \end{pmatrix} \\ e_1 = (1,0):\quad f(1,0) = (3,-1) = e_1 A = (1,0) A = (a_{11} a_{12})\\ e_2 = (0,1):\quad f(0,1) = (-1,1) = e_2 A = (0,1) A = (a_{21} a_{22}) $$ What you are looking for is $$ B = M A M^{-1} $$ $M$ first transforms a vector regarding the basis $b_1 = (1,0)$ and $b_2 = (1,1)$ into standard coordinates (basis $e_i$). Then it applies the function $f$ via the matrix $A$ (representing $f$ if you use it with vectors regarding standard coordinates) and finally transforms the result into the same vector regarding the basis $b_i$ via $M^{-1}$.
The combined matrices have the same effect as $f$ for vectors regarding the basis $b_i$.
As the $b_i$ regarding themselves have the coordinates $(1,0)$ and $(0,1)$ these vectors must be mapped by $M$ to $(1,0)$ and $(1,1)$, thus $$ M = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \quad M^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} $$ and we calculate \begin{align} B &= M A M^{-1} \\ &= \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 3 & -1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \\ &= \begin{pmatrix} 3 & -1 \\ 2 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \\ &= \begin{pmatrix} 4 & -1 \\ 2 & 0 \end{pmatrix} \end{align}
Test:
$e_1 = (1,0)$ regarding basis $b_i$ still has coordinates $(1,0)$, as $(1,0) = 1 \cdot b_1 + 0 \cdot b_2$. Then $$ e_1 B = (1,0) \begin{pmatrix} 4 & -1 \\ 2 & 0 \end{pmatrix} = (4,-1) $$ And indeed $$ 4 b_1 - b_2 = 4 (1,0) - (1,1) = (4,0)-(1,1) = (3,-1) $$ Further $e_2 = (0,1)$ regarding standard coordinates has coordinates $(-1,1)$ regarding the $b_i$: $$ -1 \cdot b_1 + 1 \cdot b_2 = (-1,0) + (1,1) = (0,1) $$ This is mapped as: $$ (-1,1) \begin{pmatrix} 4 & -1 \\ 2 & 0 \end{pmatrix} = (-2, 1) $$ Calculating back to standard coordinates: $$ -2 b_1 + b_2 = -2 (1,0) + (1,1) = (-2,0) + (1,1) = (-1,1) $$ which is where $e_2$ should mapped to by $f$.
Note:
The above assumed from your question that you have to use row vectors. If you use column vectors you have to transpose the matrices and vectors: $$ f(x) = A x $$ as $A = A^T$ is symmetric. Further $$ B = (MAM^{-1})^T = (M^{-1})^T A^T M^T = (M^T)^{-1} A M^T = \begin{pmatrix} 4 & 2 \\ -1 & 0 \end{pmatrix} $$