I was trying to solve this linear algebra exercise. it's about finding the matrix of this linear transformation:
matrix with respect to canonical basis of R2 (it's equal to the linear transformation "x-y, -x+3y") \begin{bmatrix} 1&-1\\-1&3\end{bmatrix}
but the matrix I have to find has a different basis of vectors. this is what I'm talking about: B = ( (1, 2), (2, -2) ). (1,2) is f1. (2, -2) is f2.
the exercise asks to find the matrix using the matrix similarity concept (of similar matrices), which means finding an E matrix that is as follows: first vector is like the span of the other basis, and so on. I did this T( (1, 2) ) = (vector after the linear transformation) (-1, 5) = (with respect to B) I found two scalars as follows alpha * f1 + beta * f2. using the linear system of equations, I found it: (-1, 5). same for the second vector, and it's equal to: (4, -8). now, I find the inverse of E, like this:
(1) by computing the determinant of E, (2) by doing 1 / det(E) (1 divided by det(E)) (3) then doing 1/det(E) * matrix of E but each entry of it is replaced by (-1)^(i+j) * original matrix without the row and the column with the current entry.
now, I find the matrix (with respect to B) by computing E^-1 * A (with respect to the canonical basis) * E.
but I'm afraid the result isn't correct, because my result is as follow:
matrix (with respect to B = ( (1, 2), (2, -2) ): \begin{bmatrix} 320 & 0.6667 \\ 7.0833 & 2.9167\end{bmatrix}
I used the theory from my linear algebra lecture notes.
Let $T: \mathbb R^2 \rightarrow \mathbb R^2:(x,y) \mapsto (x-y,-x+3y)$ be the linear transformation, $\{e_1,e_2\}$ the canonical basis and $f_1,f_2$ the basis $\mathcal B$. In the canonical basis the matrix for $T$ is $$ A = \begin{bmatrix}1 & -1 \\ -1 &3 \end{bmatrix}.$$ To find the change of basis matrix we write the $f_i$ as linear combinations of the canonical basis vectors
$$ f_1 = \color{orange}1e_1+\color{orange}2 e_2, \quad f_2= \color{blue}2e_1+(\color{blue}{-2})e_2,$$
So the change of basis matrix $E$ is $$ E = \begin{bmatrix} \color{orange}1 & \color{blue}2 \\ \color{orange}2 & \color{blue}{-2}\end{bmatrix}$$ It's inverse is $$ E^{-1} = \frac{-1}{6} \begin{bmatrix} -2 &-2 \\ -2 & 1\end{bmatrix}.$$
Therefore the matrix of $T$ in the basis $\mathcal B$ is
$$ [T]_\mathcal B = E^{-1}AE = \begin{bmatrix} 4/3 & -4/3 \\ -7/6 & 8/3\end{bmatrix}.$$
Let's check that we didn't make a mistake. By definition of the matrix of a linear transformation in a given basis, the last equation tells us that we should have $$ T(f_1) = 4/3 f_1-7/6f_2.$$ We can see that this equality does hold \begin{align*} 4/3f_1-7/6f_2 &= 4/3(e_1+2e_2)-7/6(2e_1-2e_2) \\ &= (4/3 -7/3)e_1 + (8/3 + 7/3)e_2 \\ &= -e_1 + 5e_2. \end{align*} You can also double check that $T(f_2) = -4/3f_1+8/3f_2$.