The many meanings of $A\cdot\boldsymbol{x} = \boldsymbol{0}$

We reviewed what we had done in the previous class and how the matrix-vector product equation $$ A\cdot\boldsymbol{x} = \boldsymbol{0} \ \ \text{ [Note: the bold face "$\boldsymbol{0}$" is the zero vector, i.e. a vector of all zeros!]} $$ has at least three different, but equivalent interpretations that we had discussed. That was encapsulated with this diagram on the whiteboard:

So by studying the matrix-vector product equation, we qre also studying systems of linear equations, linear combinations of column vectors and linear transformations.

Elementary row operations

Here are three very important kinds of operations we can perform on a matrix.
Let $A$ be an $m\times n$ matrix with row vectors $\boldsymbol{r_1},\ldots, \boldsymbol{r_m}$. The elementary row operations are:
  1. replace $\boldsymbol{r_i}$ with $s\boldsymbol{r_i}$, where $s$ is a non-zero scalar
  2. swap row $i$ with row $j$
  3. replace $\boldsymbol{r_i}$ with $\boldsymbol{r_i} + s\boldsymbol{r_j}$, where $i\neq j$

Below is an example of each type of elementary row operation.

matrix $A$row operationnew matrix $A'$
$$ \begin{bmatrix} -1&0&6\\ 0&3&3\\ 2&3&-9 \end{bmatrix} $$

 Type i
 multiply row 3 by scalar 5 		 $$ \begin{bmatrix} -1&0&6\\ 0&3&3\\ 10&15&-45 \end{bmatrix} $$
$$ \begin{bmatrix} -1&0&6\\ 0&3&3\\ 2&3&-9 \end{bmatrix} $$

 Type ii
 swap row 1 and row 3 		 $$ \begin{bmatrix} 2&3&-9\\ 0&3&3\\ -1&0&6\\ \end{bmatrix} $$
$$ \begin{bmatrix} -1&0&6\\ 0&3&3\\ 2&3&-9 \end{bmatrix} $$

 Type iii
 add 5 times row 1 to row 3 		 $$ \begin{bmatrix} 2&3&-9\\ 0&3&3\\ 9&15&-39\\ \end{bmatrix} $$

The crucial thing about these elementary row operations is that although you are changing the matrix $A$, you are not changing the solutions of $A\cdot \boldsymbol{x}=\boldsymbol{0}$. This is made precise in the following theorem.

Let $A$ be an $m\times n$ matrix, and let $A'$ be the same matrix after performing one elementary row operation, then
   $\boldsymbol{x}$ is a solution to $A\cdot \boldsymbol{x}=\boldsymbol{0}$ if and only if $\boldsymbol{x}$ is a solution to $A'\cdot \boldsymbol{x}=\boldsymbol{0}$.

ACTIVITY
Break into groups of three or four, each group going to the board. Each group gets one of the elementary row operations.

  1. Consider the equation $A\cdot \boldsymbol{x}=\boldsymbol{0}$ with the 3x3 matrix from the above examples as $A$, and the solution vector [6 -1 1]. In other words: $$ \begin{bmatrix} -1&0&6\\ 0&3&3\\ 2&3&-9 \end{bmatrix} \cdot \begin{bmatrix} 6\\ -1\\ 1 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} $$

    Apply your row operation to this matrix to produce new matrix $A'$, and verify that [6 -1 1] is indeed a solution to $A'\cdot \boldsymbol{x}=\boldsymbol{0}$.

  2. Hopefully you gained some insight from this for why the theorem is true for your row operation - not just on this example matrix, but for any matrix. So now prove the theorem for your row operation!

So collectively, you and your classmates proved this theorem!

We prove that, for each of the three types of elementary row operations, the solutions to $A\cdot\boldsymbol{x} = \boldsymbol{0}$ are the same before and after the operation. Let $A$ be the original matrix, and let $A'$ be the same matrix after one elemenatry row operation. We will denote the row vectors of $A$ as $\boldsymbol{r_1},\boldsymbol{r_2},\cdots,\boldsymbol{r_m}$. So let's consider the three types of elementary row operations:
  1. We modify $A$ to produce $A'$ by replacing $\boldsymbol{r_i}$ with $s\cdot\boldsymbol{r_i}$, where $s\neq 0$. All the rows of $A' \cdot \boldsymbol{x}$ are identical to $A \cdot \boldsymbol{x}$ except row $i$. Row $i$ of $A \cdot \boldsymbol{x}$ is $\boldsymbol{r_i}\cdot\boldsymbol{x}$. Row $i$ is $A' \cdot \boldsymbol{x}$ is $(s\cdot\boldsymbol{r_i})\cdot\boldsymbol{x}$. By point (v) of $$ (s\cdot\boldsymbol{r_i})\cdot\boldsymbol{x} = s\cdot(\boldsymbol{r_i}\cdot\boldsymbol{x}) $$ and since $s \neq 0$, $\boldsymbol{r_i}\cdot\boldsymbol{x} = 0$ if and only if $s\cdot(\boldsymbol{r_i}\cdot\boldsymbol{x}) = 0$.
    Note: This does require that there are no "zero-divsors" in the underlying ring $R$, i.e. the product of non-zero elements is never zero. However, rings in which all non-zero elements have multiplicative inverses (which is a requirement for vector spaces) have no zero divisors.
  2. We modify $A$ to produce $A'$ by swapping row $i$ and row $j$. This changes the order of the rows, but not the rows themselves. So all rows of $A$ have the property that their dot products with $\boldsymbol{x}$ are zero if and only if all rows of $A'$ have the property that their dot products with $\boldsymbol{x}$ are zero.
  3. We modify $A$ to produce $A'$ by replacing $\boldsymbol{r_i}$ with $\boldsymbol{r_i} + s\cdot\boldsymbol{r_j}$. All the rows of $A' \cdot \boldsymbol{x}$ are identical to $A \cdot \boldsymbol{x}$ except row $i$. Row $i$ is $A \cdot \boldsymbol{x}$ is $\boldsymbol{r_i}\cdot\boldsymbol{x}$. Row $i$ of $A' \cdot \boldsymbol{x}$ is
    $ (\boldsymbol{r_i} + s\cdot\boldsymbol{r_j})\cdot\boldsymbol{x} = \boldsymbol{r_i}\cdot\boldsymbol{x} + s\cdot(\boldsymbol{r_j}\cdot\boldsymbol{x}) $, by points (iii) and (v) of .
    If all rows of $A\cdot\boldsymbol{x}$ are zero, then row $i$ of $A'\cdot\boldsymbol{x}$ is $\boldsymbol{r_i}\cdot\boldsymbol{x} + s\cdot(\boldsymbol{r_j}\cdot\boldsymbol{x}) = 0 + s\cdot 0 = 0$.
    If all rows of $A'\cdot\boldsymbol{x}$ are zero, then row $i$ of $A'\cdot\boldsymbol{x}$ is zero, so $\boldsymbol{r_i}\cdot\boldsymbol{x} + s\cdot(\boldsymbol{r_j}\cdot\boldsymbol{x}) = 0 = \boldsymbol{r_i}\cdot\boldsymbol{x} + s\cdot 0$. Therefore, $\boldsymbol{r_i}\cdot\boldsymbol{x} = 0$.
    So, $A\cdot\boldsymbol{x} = \boldsymbol{0}$ if and only if $A'\cdot\boldsymbol{x} = \boldsymbol{0}$.
Christopher W Brown