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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.

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    $\begingroup$ Is there anything unsatisfactory about the answer "the numbers $A_{ij}$ are the elements of the matrix $A$?" If so, what are you confused about? $\endgroup$ Jan 29 '20 at 18:51
  • $\begingroup$ It’s easier to have intuition about linear algebra if you consider the case where $V$ and $W$ are the same space. Start by thinking about transformations 3D Euclidean space which rotate it or stretch/shrink it along an axis. $\endgroup$
    – G. Smith
    Jan 29 '20 at 18:55
  • $\begingroup$ After you are comfortable thinking about linear transformations of one vector space, think about linear transformations between two spaces. $\endgroup$
    – G. Smith
    Jan 29 '20 at 19:01
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    $\begingroup$ That’s right. In general it will change each basis vector, making it point in a different direction and/or stretching/shrinking it. Write some arbitrary $3\times 3$ matrix (for simplicity, just use random integer matrix entries) and let it act on the column vectors $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. Now let it act on a random vector like $(2,3,4)$ and see how this is 2 times the first result plus 3 times the second plus 4 times the third: linearity in action! $\endgroup$
    – G. Smith
    Jan 29 '20 at 22:23
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    $\begingroup$ You answered your own question already. For each $j$ there must exist $n$ complex numbers that allow you to decompose $A | v_j\rangle$ into that basis. The resulting set of numbers $A_{i,j}$ Is what we call the matrix representation of our operator in those bases. It's a definition. $\endgroup$
    – lcv
    Jan 30 '20 at 11:47
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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 $|w_m\rangle$.

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  • $\begingroup$ In this answer you have taken $W=V$. In the OP’s question there are two different bases, possibly with two different numbers of basis vectors. $\endgroup$
    – G. Smith
    Jan 29 '20 at 18:57
  • $\begingroup$ @ G. Smith. Ouch. Yes! I'll amend my answer. $\endgroup$
    – mike stone
    Jan 29 '20 at 19:03
  • $\begingroup$ This says what the OP says in a different notation. Moreover you don't need a scalar product to define the matrix representation of an operator (which is what you are doing in the last line). Not mentioning the fact that the bases don't need to be orthonormal. $\endgroup$
    – lcv
    Jan 30 '20 at 11:59
  • $\begingroup$ @Icv It is not the same. His matrix is the transpose of the correct one. I use the QM example to show that we are all familiar with the idea that the matrix must act to the left when acting on a basis if you want to act to the right when acting on components. $\endgroup$
    – mike stone
    Jan 30 '20 at 13:32
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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 $A_{ij}$.

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  • $\begingroup$ The same comment as for the other answer applies. You don't need a scalar product to define the matrix representation of an operator. Moreover your answer only works if the basis is orthonormal which obviously doesn't need to be. The OP already said things more precisely $\endgroup$
    – lcv
    Jan 30 '20 at 11:52

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