Confused by notation in this linear algebra exercise

Question

I really hate it when an author randomly brings up new or confusing notation without explanation.

This is an exercise in my linear algebra book on a section of matrix multiplication:

Let $A$ and $B$ be matrices for which the product matrix $AB$ is defined, and let $u_j$ and $v_j$ denote the $j$th columns of $AB$ and $B$, respectively. If $v_p = c_1 v_{j_1} + c_2 v_{j_2} + \cdots c_k v_{j_k}$ for some scalars, $c_1, c_2, \dots, c_k$, prove that $u_p =c_1 u_{j_1} + c_2 u_{j_2} + \cdots + c_k u_{j_k}$.

What's up with $v_{j_i}$ (and $u_{j_i}$)? I don't understand what this represents. My first thought was that it was the $i$th entry of the column vector $v_j$. But I'm very sure that these are supposed to be vectors, it's just that I don't know which vectors they represent. Can anyone help me by clarifying what it means?

Answer 1

Phrased (hopefully!) a bit more clearly, the question is asking this:

We have collections of vectors $\{v_1,\cdots,v_n\}$ forming the columns of $B$, and $\{u_1,\cdots,u_n\}$ forming the columns of $AB$. Suppose a $p$th vector can be given by $$ v_p = c_1 v_{j_1} + \cdots + c_k v_{j_k} $$ where $\{v_{j_1}, \cdots , v_{j_k}\} \subseteq \{v_1,\cdots,v_n\}$. Then prove that $$ u_p = c_1 u_{j_1} + \cdots + c_k u_{j_k} $$

What this ultimately means is that we are taking $k$-many columns of the $v_i$; we don't necessarily know which ones, though, so instead of indexing them with $1,2,\cdots,k$, we instead use indices $j_1,j_2,\cdots,j_k$, which helps us avoid dealing with order. We then suppose $v_p$ for some $p$ happens to be equal to $\sum c_i v_{j_i}$ for some choice of scalars $c_i$, and seek to prove a likewise relationship for the $u_i$ (using the same scalars, and the same indices -- so if it turns out that $j_5 = 2$, i.e. $v_{j_5}$ is the second column of $B$, then we will be including $u_{j_5} = u_2$, the second column of $AB$).

Answer 2

The subindexing means that we are picking certain indices and ignoring the others. For example, if we meant to consider the first, third and fourth columns of $B$, then we would take $j_1 = 1$, $j_2 = 3$, $j_3 = 4$. Then $\{v_{j_1},v_{j_2},v_{j_3}\} = \{v_1,v_3,v_4\}$.