Let $A = \left[\begin{matrix} 2 & 3 & 4 \\ 8 & 5 & 1\end{matrix}\right]$ and consider $A$ as a linear transformation from $\mathbb{R}^3$ into $\mathbb{R}^2$ using the standard bases. Find the matrix representation of $A$ with resepect to the basis $\left\{\space\left[\begin{matrix}1 \\ 1 \\ 0\end{matrix}\right], \left[\begin{matrix}0 \\ 1 \\ 1\end{matrix}\right], \left[\begin{matrix}1 \\ 0 \\ 1\end{matrix}\right]\space\right\}$ for $\mathbb{R}^3$ and $\left\{ \left[\begin{matrix}3 \\ 1\end{matrix}\right], \left[\begin{matrix}2 \\ 1\end{matrix}\right] \right\}$ for $\mathbb{R}^2$.
Here is my attempt taking from these two posts post_1 and post_2: Let $C = \left[\begin{matrix}\mathbf{c}_1 & \mathbf{c}_2 & \mathbf{c_3}\end{matrix}\right]$, that is, the matrix representation of the vectors of the given basis for $\mathbb{R}^3$. Similarly represent the basis vectors for $\mathbb{R}^2$ as $B = \left[\begin{matrix}\mathbf{b}_1 & \mathbf{b}_2 \end{matrix}\right]$. Then, apply the transformation $T(\mathbf{c}_1) = A\mathbf{c}_1 = \mathbf{v}_1$. We have a new vector now in $\mathbb{R}^2$.
Side note: formula for coordinate vector of $\mathbf{v}$ with respect to the basis B.
$$(1)\quad\quad \mathbf{v} = B[\mathbf{v}]_B \implies B^{-1}\mathbf{v} = [\mathbf{v}]_B$$
Now using formula (1) to find $\mathbf{v}_1$ in terms of our basis for $\mathbb{R}^2$ we have $B^{-1}\mathbf{v}_1$. Thus, this is our first vector in our new matrix representation of A. Repeat this process for $\mathbf{c}_2$ and $\mathbf{c}_3$. Generalizing this process with the help of post_2 we get
$$(2)\quad\quad B^{-1}AC$$
This is our new matrix and our representation for matrix A. Is this correct? Is there something I am missing?
Furthermore, my intuition for this is that we are given matrix $A$ that maps vectors $\mathbb{R}^3 \to \mathbb{R}^2$. Then we need to map the given basis vectors (IDK … thb) for $\mathbb{R}^3$ into $\mathbb{R}^2$. But now we need to write these new vectors (hence new matrix) in terms of our given basis for $\mathbb{R}^2$. So we take each vector $\mathbf{v}_i$, $1 \leq i \leq 3$, and do $\mathbf{v}_i = B[\mathbf{v}_i]_B \implies B^{-1}\mathbf{v}_i = [\mathbf{v}_i]_B$. That is now we have each vector represented now in terms of the basis in $\mathbb{R}^2$ which gives us the matrix representation.
Let me know if what I have done or my intuition needs correcting! I would love all the help!
Best Answer
Yes, I think you understand this well.
To sum up: each transition matrix $B = [ \mathbf{b}_1 \: \mathbf{b}_2]$ and $C = [ \mathbf{c}_1 \: \mathbf{c}_2 \: \mathbf{c}_3 ]$ as a linear transformation converts from coordinates in the custom basis to coordinates in the standard basis. So given $A$ mapping from standard basis of $\mathbb{R}^3$ to standard basis of $\mathbb{R}^2$, the transformation from coordinates in the custom basis of $\mathbb{R}^3$ to coordinates in the custom basis of $\mathbb{R}^3$ is $B^{-1} A C$.
$$ \mathbb{R}_C^3 \stackrel{C}{\longrightarrow} \mathbb{R}^3 \stackrel{A}{\longrightarrow} \mathbb{R}^2 \stackrel{B^{-1}}{\longrightarrow} \mathbb{R}_B^2$$