## Motivating example

Consider the system $$Ax = b$$ defined over the real numbers where $$A=\begin{bmatrix} 1 & 2 & 0 & -1 & 0 & 3 \\ 0 & 0 & 1 & -1 & 0 & 4 \\ 0 & 0 & 0 & 0 & 1 & 5 \end{bmatrix}$$ and $$b = \begin{bmatrix} 6\\7\\8\end{bmatrix}$$. Note that $$x$$ is $$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6\end{bmatrix}$$ here.

The augmented matrix representing the system is $$\left[\begin{array}{cccccc|c} 1 & 2 & 0 & -1 & 0 & 3 & 6 \\ 0 & 0 & 1 & -1 & 0 & 4 & 7 \\ 0 & 0 & 0 & 0 & 1 & 5 & 8 \end{array}\right],$$ which is already in reduced row-echelon form with columns 1, 3, and 5 as pivot columns. Hence, the free variables are $$x_2$$, $$x_4$$ and $$x_6$$.

Setting $$x_2 = s$$, $$x_4 = t$$, and $$x_6 = u$$, we obtain the solution $$\begin{bmatrix} x_1\\x_2\\x_3\\x_4\\x_5\\x_6\end{bmatrix} = \begin{bmatrix} 6-2s+t-3u\\ s\\ 7+t- 4u\\ t\\ 8-5u\\ u\end{bmatrix}= \begin{bmatrix} 6\\0\\ 7\\ 0\\ 8\\ 0\end{bmatrix}+ s~\begin{bmatrix} -2\\1\\ 0\\ 0\\ 0\\ 0\end{bmatrix}+ t~\begin{bmatrix} 1\\0\\ 1\\ 1\\ 0\\ 0\end{bmatrix}+ u~\begin{bmatrix} -3\\0\\ -4\\ 0\\ -5\\ 1\end{bmatrix}$$ for any choice of real numbers $$s$$, $$t$$, and $$u$$.

Notice that each of $$\begin{bmatrix} -2\\1\\ 0\\ 0\\ 0\\ 0\end{bmatrix}$$, $$\begin{bmatrix} 1\\0\\ 1\\ 1\\ 0\\ 0\end{bmatrix}$$, and $$\begin{bmatrix} -3\\0\\ -4\\ 0\\ -5\\ 1\end{bmatrix}$$ is a solution to the homogeneous system $$Ax = 0$$. (A homogeneous system of linear equations has all the right-hand side values equal 0.)

In general, knowing the solutions of $$Ax=0$$ tells us a lot about the solutions to $$Ax = b$$. Suppose that both $$x^*$$ and $$x'$$ are solutions to $$Ax=b$$; i.e. $$Ax^*=b$$ and $$Ax'=b$$. Then $$A(x^* - x') = b - b = 0$$. (Why?) Hence, if $$\tilde{x} = x^* - x'$$, then $$\tilde{x}$$ is a solution to $$Ax = 0$$.

As a result, once we have one solution to $$Ax = b$$, say $$x^*$$, then every other solution differs from $$x^*$$ by a solution to $$Ax = 0$$. In other words, the set of solutions to $$Ax = b$$ is given by $$\{ x^* + d : Ad = 0\}$$.

In short, knowing one solution to $$Ax=b$$ and all the solutions to $$Ax = 0$$ gives us all the solutions to $$Ax = b$$.

Finally, note that $$Ax=0$$ always has the tuple of 0's as a solution. This solution is called the trivial solution. Any other solution to $$Ax = 0$$ is called a nontrivial solution.

## Exercise

1. Let $$A = \begin{bmatrix} 1 & 2 & -1 \\ 0 & -1 & 3\\ 1 & 1 & 2\end{bmatrix}$$. Does the homogenous system $$Ax = 0$$ have a nontrivial solution?

2. Let $$A \in \mathbb{F}^{m\times n}$$ where $$\mathbb{F}$$ denotes a field. Let $$N(A)$$ denote the set $$\{d \in \mathbb{R} : Ad = 0\}$$. Let $$x^*, x' \in \mathbb{F}^n$$.

1. Show that $$A(x^* - x') = Ax^* - Ax'$$.

2. Let $$z = x^* + x'$$. Show that if $$x^*, x' \in N(A)$$, then $$z \in N(A)$$.

3. Let $$\alpha \in \mathbb{F}$$. Let $$u = \alpha x^*$$. Show that if $$x^* \in N(A)$$, then $$u \in N(A)$$.