Need help understanding matrix representation of a linear operator

Question

I'm struggling with linear algebra. Specifically, understanding the following:

$\newcommand{\ket}[1]{|#1\rangle}$ Suppose $A:V \rightarrow W$ is a linear operator between vector spaces $V$ and $W$. Suppose $\ket{v_1},\ldots,\ket{v_m}$ is a basis for $V$ and $\ket{w_1}, \ldots, \ket{w_n}$ is a basis for $W$. Then for each $j$ in the range $1,\ldots,m$, there exist complex numbers $A_{1j}$ through $A_{nj}$ such that

$$ A\ket{v_j} = \sum_i A_{ij} \ket{w_i}. $$

I understand that $A \ket{v_j}$ is a vector in $W$. I also understand that we can write any vector in $W$ as a linear combination of the basis vectors $\ket{w_1}, \ldots, \ket{w_n}$. I don't understand how that corresponds to the matrix form of $A$ and am overall lacking intuition for what's going on here.

Can someone help give me intuition for what the above means? Also, suggestions on books/videos/lectures/etc are also appreciated.

Answer 1

If you let the linear operator act on a basis of $V$, you should write the matrix behind the basis vectors of $W$: $$ A{\bf e}_i = {\bf f}_j {A^j}_i. $$ Suppose that ${\bf y}= A{\bf x}$ where ${\bf x}= x^i {\bf e}_i$, and ${\bf y}= y^j {\bf f}_j$. We have $$ A{\bf x}= x^i (A{\bf e}_i)= x^i({\bf f}_j {A^j}_i)= {\bf f}_j ({A^j}_i x^i) $$ comparing with ${\bf y}= y^j {\bf f}_j$ gives $$ y^j= \sum_{i=1}^{{\rm dim\,}V} {A^j}_i x^i $$ which is the usual action of a matrix on a column vector where we sum over adjacent indices.
One can think of this as it appears in Quantum Mechanics $$ A|v_n\rangle = \sum_{m=1}^{{\rm dim} W} |w_m\rangle \langle w_m|A|v_n\rangle, $$ where the matrix elements naturally appear behind the basis vector $|w_m\rangle$.

Answer 2

In quantum mechanics there’s a trick we use when we are stuck. It’s called completeness, which is basically a change of reference in the form of basis change. There is a powerful property from linear algebra that says: $$\sum_{i}|w_i\rangle\langle w_i|=\mathbb{1}$$ where the $|w_i\rangle$s form a basis. Using this we have $$A|v_j\rangle=\sum_{i} |w_i\rangle\langle w_i|A|v_j\rangle$$ and the number $\langle w_i|A|v_j\rangle$ is just the matrix element $A_{ij}$.