# Proving two simple matrix product properties

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1. Sep 4, 2016

### TheSodesa

1. The problem statement, all variables and given/known data
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

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

2. Relevant equations

The matrix product:
If $A$ is an $m\times p$ matrix and $B$ is a $p\times n$ matrix, then

AB = C = (c_{ij})_{m \times n} = (\sum_{k=1}^{p} a_{ik}b_{kj})_{m \times n}

3. 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 $B$?

Last edited: Sep 4, 2016
2. Sep 4, 2016

### 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.

3. Sep 8, 2016

### EnumaElish

1. Calculate C = AB.
2. Calculate D = [A b1 ... A bm]
3. Show Dij = Cij for any i,j