## Subsection6.3.4Matrix-vector multiplication

Assume $A \in \Rmxn$ and partition

\begin{equation*} A = \left( \begin{array}{c} a_0^T \\ a_1^T \\ \vdots \\ a_{m-1}^T \end{array} \right) \quad \mbox{and} \quad \left( \begin{array}{c} \psi_0 \\ \psi_1 \\ \vdots \\ \psi_{m-1} \end{array} \right). \end{equation*}

Then

\begin{equation*} \left( \begin{array}{c} \psi_0 \\ \psi_1 \\ \vdots \\ \psi_{m-1} \end{array} \right) := \left( \begin{array}{c} a_0^T x \\ a_1^T x\\ \vdots \\ a_{m-1}^T x\\ \end{array} \right). \end{equation*}

From R-1B 6.3.3.2 regarding the dot product we know that

\begin{equation*} \begin{array}{l} \check y = \left( \begin{array}{c} \check \psi_0 \\ \check \psi_1 \\ \vdots \\ \check \psi_{m-1} \end{array} \right) = \left( \begin{array}{c} ( a_0 + \delta\!a_0)^T x \\ ( a_1 + \delta\!a_1)^T x\\ \vdots \\ ( a_{m-1} + \delta\!a_{m-1})^T x\\ \end{array} \right) \\ ~~~= \left( \left( \begin{array}{c} a_0^T \\ a_1^T\\ \vdots \\ a_{m-1}^T \end{array} \right) + \left( \begin{array}{c} \delta\!a_0^T \\ \delta\!a_1^T \\ \vdots \\ \delta\!a_{m-1}^T \end{array} \right) \right) x = ( A + \Delta\!\!A ) x , \end{array} \end{equation*}

where $\vert \delta\!a_i \vert \leq \gamma_n \vert a_i \vert\text{,}$ $i = 0, \ldots , m-1 \text{,}$ and hence $\vert \Delta\!\!A \vert \leq \gamma_n \vert A \vert \text{.}$ .

Also, from R-1B 6.3.3.2 regarding the dot product we know that

\begin{equation*} \check y = \left( \begin{array}{c} \check \psi_0 \\ \check \psi_1 \\ \vdots \\ \check \psi_{m-1} \end{array} \right) = \left( \begin{array}{c} a_0^T x + \delta\!\psi_0\\ a_1^T x + \delta\!\psi_1\\ \vdots \\ a_{m-1}^T x + \delta\!\psi_{m-1} \end{array} \right) = \left( \begin{array}{c} a_0^T \\ a_1^T\\ \vdots \\ a_{m-1}^T \end{array} \right) x + \left( \begin{array}{c} \delta\!\psi_0 \\ \delta\!\psi_1 \\ \vdots \\ \delta\!\psi_{m-1} \end{array} \right) = A x + \deltay . \end{equation*}

where $\vert \delta\!\psi_i \vert \leq \gamma_n \vert a_i \vert^T \vert x \vert$ and hence $\vert \deltay \vert \leq \gamma_n \vert A \vert \vert x \vert \text{.}$

The above observations can be summarized in the following theorem:

###### Ponder This6.3.4.1.

In the above theorem, could one instead prove the result

\begin{equation*} \check y = A ( x + \deltax ), \end{equation*}

where $\deltax$ is "small"?

Solution

The answer is "sort of". The reason is that for each individual element of $y$

\begin{equation*} \check \psi_i = a_i^T ( x + \deltax ) \end{equation*}

which would appear to support that

\begin{equation*} \left( \begin{array}{c} \check \psi_0 \\ \check \psi_1 \\ \vdots \\ \check \psi_{m-1} \end{array} \right) = \left( \begin{array}{c} a_0^T ( x + \deltax ) \\ a_1^T ( x + \deltax ) \\ \vdots \\ a_{m-1}^T ( x + \deltax ) \end{array} \right). \end{equation*}

However, the $\deltax$ for each entry $\check \psi_{i}$ is different, meaning that we cannot factor out $x + \deltax$ to find that $\check y = A ( x + \deltax ) \text{.}$

However, one could argue that we know that $\check y = A x + \delta\!y$ where $\vert \deltay \vert \leq \gamma_{n} \vert A \vert \vert x \vert \text{.}$ Hence if $A \delta\!x = \delta\!y$ then $A ( x + \deltax ) = \check y \text{.}$ This would mean that $\delta\!y$ is in the column space of $A \text{.}$ (For example, if $A$ is nonsingular). However, that is not quite what we are going for here.