Beginning with the set that is a basis of $\mathbb{R}^2$: $b1 = \begin{bmatrix}1 \\-2\end{bmatrix}$ and $b2 = \begin{bmatrix}-3 \\-5\end{bmatrix}$

Let $T:\mathbb{R}^2→\mathbb{R}^2$ is a linear transformation such that $T(b1)=4b1+5b2$ and $T(b2)=6b1+3b2$ Then the matrix of $T$ relative to this basis is $$\begin{bmatrix}4 & 6 \\5 & 3 \end{bmatrix}$$

I understand this part, but I am unsure of how I would represent in terms of the standard basis. I attempted to solve and got $$\begin{bmatrix}-19 & -15\\-33 & -27\end{bmatrix}$$ I did this by taking $4b1+5b2$ as the first column, however, I don't think this is correct.


Assuming that $b1 = \begin{bmatrix}1 \\-2\end{bmatrix}$ and $b2 = \begin{bmatrix}-3 \\-5\end{bmatrix}$ are in terms of the standard basis, the matrix that change your basis to the standard one is \begin{equation} P = \left(\begin{matrix} 1 & -3\\ -2 & -5 \end{matrix}\right). \end{equation} This implies that the matrix of $T$ expressed in the standard basis must be following product: \begin{equation} P\left(\begin{matrix}4 & 6 \\5 & 3 \end{matrix}\right) P^{-1} = \left(\begin{matrix}1 & -3 \\-2 & -5 \end{matrix}\right) \left(\begin{matrix}4 & 6 \\5 & 3 \end{matrix}\right) \left(\begin{matrix}5/11 & -3/11 \\-2/11 & -1/11 \end{matrix}\right) \end{equation}

This is because the input vector, that must be multiplicated by the right hand side, must be changed from the standard basis to the original one (using $P^{-1}$) and the output vector must be changed from the original basis to the standard one (using $P$).


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