Matrix-Vector Multiplication

Suppose we have an $m \times n$ matrix $A$ and $n$-vector $\mathbf b$

  • How to calculate $\mathbf x = A \mathbf b$?
  • note that $\mathbf x \in \mathbb R^m$

There are two equivalent ways to do it:

  • Row at a time
  • Column at a time


Row at a Time

See $A$ as $m$ vectors along rows:

$A = \begin{bmatrix} — \mathbf a_1 \,— \\ — \mathbf a_2 \,— \\

...   \\ 

— \mathbf a_m \,— \end{bmatrix}$

And then multiply (using Dot Product) each row $(\mathbf a_i)^T$ with the vector $\bf x$:

  • $x_i = (\mathbf a_i)^T \mathbf b$
  • $\mathbf x = \begin{bmatrix}

— (\mathbf a_1)^T \mathbf b \,— \\ — (\mathbf a_2)^T \mathbf b \,— \\

...   \\ 

— (\mathbf a_m)^T \mathbf b \,— \end{bmatrix}$

  • Where dot product is $\mathbf a^T \mathbf b = \sum\limits_{i=1}^m a_i b_i$


Column at a Time

Another way to see $A$ is as $n$ vectors along columns:

$A = \begin{bmatrix} \mathop{a_1}\limits_|^| \ \mathop{a_2}\limits_|^| \ \cdots \ \mathop{a_n}\limits_|^| \end{bmatrix}$

When we multiply $A$ on a vector $\mathbf b$, it produces a Linear Combination of these column vectors:

$A \mathbf b = \begin{bmatrix} \mathop{a_1}\limits_|^| \ \mathop{a_2}\limits_|^| \ \cdots \ \mathop{a_n}\limits_|^| \end{bmatrix} \mathbf b =

 b_1 \begin{bmatrix} \mathop{a_1}\limits_|^| \end{bmatrix} 

+ b_2 \begin{bmatrix} \mathop{a_2}\limits_|^| \end{bmatrix} + \cdots + \ b_n \begin{bmatrix} \mathop{a_n}\limits_|^| \end{bmatrix}$


Example

$\begin{bmatrix} 2 & 5\\ 1 & 3 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 2 \end{bmatrix} $


Row at a time:

  • $[2 \ 5] \begin{bmatrix}

1 \\ 2 \end{bmatrix} = 2 \cdot 1 + 5 \cdot 2 = 12$

  • $[1 \ 3] \begin{bmatrix}

1 \\ 2 \end{bmatrix} = 1 \cdot 1 + 3 \cdot 2 = 7$

  • so $\begin{bmatrix}

2 & 5\\ 1 & 3 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 12 \\ 7 \end{bmatrix}$


Column at a time

  • $1 \begin{bmatrix}

2 \\ 1 \end{bmatrix} + 2 \begin{bmatrix} 5 \\ 3 \end{bmatrix} = \begin{bmatrix} 12 \\ 7 \end{bmatrix}$


Left Vector Multiplication

A vector may be on the left of the matrix as well

  • in such case $\mathbf b$ is a row vector, and thus the result $\mathbf x$ is as well a row vector
  • let $\mathbf b \in \mathbb R^{m}$ and $A \in \mathbb{R}^{m \times n}$
  • $\mathbf b^T A = \mathbf x^T$
  • Can transpose both parts and get $A^T \mathbf b = \mathbf x$
  • and we're back to the normal column-vector case


Sources

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