Proving two simple matrix product properties

[TheSodesa]

Homework Statement

Let $A$ be an n × p matrix and $B$ be an p × m matrix with the following column vector representation,

$$B = \begin{bmatrix} b_1 , & b_2, & ... & ,b_m \end{bmatrix}$$

Prove that
$$AB = \begin{bmatrix} Ab_1 , & Ab_2, & ... & , Ab_m \end{bmatrix}$$

If $A$ is represented with help of its row vectors, prove that

$$AB = \begin{bmatrix} a^{T}_{1}\\ \vdots\\ a_{n}^{T} \end{bmatrix} B = \begin{bmatrix} a^{T}_{1} B\\ \vdots\\ a^{T}_{n} B \end{bmatrix}$$

Homework Equations

The matrix product:
If $A$ is an $m\times p$ matrix and $B$ is a $p\times n$ matrix, then
\begin{equation}
AB = C = (c_{ij})_{m \times n} = (\sum_{k=1}^{p} a_{ik}b_{kj})_{m \times n}
\end{equation}

The Attempt at a Solution

For starters what does proving in this context mean? Should I simply write out the matrix

$$C = \begin{bmatrix} \sum_{k=1}^{p} a_{1k} b_{k1} & \cdots & \sum_{k=1}^{p} a_{1k} b_{km}\\ \vdots & \ddots & \vdots\\ \sum_{k=1}^{p} a_{mk} b_{k1} & \cdots &\sum_{k=1}^{p} a_{mk} b_{km} \end{bmatrix}$$
and conclude that each column is essentially $Ab_{l}$ where $1 < l < m$, since each element of the matrix is the dot (inner) product of a row in $A$ and a column in $B$?

 

[TheSodesa]

Oops. This

$$C = \begin{bmatrix} \sum_{k=1}^{p} a_{1k} b_{k1} & \cdots & \sum_{k=1}^{p} a_{1k} b_{km}\\ \vdots & \ddots & \vdots\\ \sum_{k=1}^{p} a_{mk} b_{k1} & \cdots &\sum_{k=1}^{p} a_{mk} b_{km} \end{bmatrix}$$

should be this
$$C = \begin{bmatrix} \sum_{k=1}^{p} a_{1k} b_{k1} & \cdots & \sum_{k=1}^{p} a_{1k} b_{km}\\ \vdots & \ddots & \vdots\\ \sum_{k=1}^{p} a_{nk} b_{k1} & \cdots &\sum_{k=1}^{p} a_{mk} b_{km} \end{bmatrix}$$
since it's an $n \times m$ matrix, not an $m \times m$ matrix.

 

[EnumaElish]

1. Calculate C = AB.
2. Calculate D = [A b₁ ... A b_m]
3. Show D_ij = C_ij for any i,j