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