Prove that if the first and second column of B are equal, then the first and second column of AB are equal I need to prove that if we have a matrix B with 2 equal columns then the product AB will have also equal columns and those columns will be the same (or in the same position).
I have written the product using generic matrices and I can see that the statement is true but I can't find a way to prove it. Thank you very much.
 A: You can use the definition of matrix multiplication.
Recall that, given an matrix $m \times n$ matrix $M$, we denote the entry in the $i$th row and $j$th column by $(M)_{ij}$ (sometimes, the parentheses are omitted). Beware: the index system "goes down" by $i$ first, then "across right" $j$, whereas indexing in the Cartesian plane goes left-right first, then up-down second. The index system has $1 \le i \le m$ and $1 \le j \le n$.
Using this notation, the $j_0$th column is the vector $((M)_{1j_0}, (M)_{2j_0}, \ldots, (M)_{mj_0})$. To say that the $j_0$th column is the same as the $j_1$th column is to say:
$$(M)_{ij_0} = (M)_{ij_1}, \quad \text{ for all } 0 \le i \le m.$$
Matrix multiplication is defined as follows: if $A$ is $m \times n$ and $B$ is $n \times p$, then $AB$ is the $m \times p$ matrix defined by:
$$(AB)_{ij} = \sum_{k=1}^n (A)_{ik} (B)_{kj}.$$
This should match up with your intuition: this is a dot product between the $i$th row of $A$ and the $j$th column of $B$, which is precisely what you compute when you do the full matrix multiplication process.
So, we can assume that $B$ has two columns the same: let's say $j_0$ and $j_1$ (here, $j_0$ and $j_1$ lie between $1$ and $p$). That is,
$$(B)_{kj_0} = (B)_{kj_1}, \quad \text{ for all } 1 \le k \le n.$$
So,
$$(AB)_{ij_0} = \sum_{k=1}^n (A)_{ik} (B)_{kj_0} = \sum_{k=1}^n (A)_{ik} (B)_{kj_1} = (AB)_{ij_1},$$
for all $1 \le i \le m$. This proves that he $j_0$th and $j_1$th columns of $AB$ are also the same.
A: Edit: this sounds like a homework question, so I will just say:
HINT: Recall that the $i$-th column of $A$ is $Ae_i$ where $e_i$ is the $i$-th standard basis vector (which looks like $(0,0,\ldots, 0, 0, 1, 0, 0, \ldots, 0)$ with a $1$ in the $i$-th place and all other entries $0$.)
A: There are a lot of different (but equivalent) ways to define matrix multiplication, so the precise argument you use will need to conform to the conventions and definitions in use in your class and/or textbook. However, one way to define matrix multiplication is as follows:

Suppose the $k$th column of $B$ is $\vec v_k$.  Then $AB$ is the
matrix whose $k$th column is $A\vec v_k$.

With this definition, the proof is trivial.  So either confirm that this is how your book defines matrix multiplication, or prove the statement above as a general theorem and then reach the property you want as a trivial corollary.
