Linear constrained form to free parameter form

Question

I came across this fact that for a linear system of equation $Ax=b$ one can write a free parameter form as $x=Fz+x_0$, where the columns of $F$ span the null space of A, and $x_0$ is some specific solution.

I am wondering if this is always true? I am also wondering why I did not encounter this in basic linear algebra course? Any reference is appreciated.

Answer 1

Whenever you have a linear equation system $Ax = b$, the set of all solutions $\mathbb{L}$ can be written as

$$\mathbb{L} = \mathbb{L}_{\text{h}} + \{ y_{\text{p}}\} = \{x_0 + y_{\text{p}} : x_0 \in \mathbb{L}_h \}, \tag{$\ast$}$$

with $\mathbb{L}_{\text{h}} = \{ x_0 : Ax_0 = 0 \}$ and $y_p$ any solution of $Ax = b$. This means that the general solution of a linear equation system is the sum of the general solution of the associated homogeneous system and a particular solution of the inhomogeneous system.

Now assume for the moment that $(\ast)$ holds. Then it follows that every solution $x$ of $Ax = b$ has a representation $x = \sum_{i} \lambda_i x_i + y_{\text{p}}$, where the vectors $x_1, \dots, x_n$ span the kernel of $A$. But this equation can be rewritten in matrix form as

$$x = \left(\begin{array}{ccc}x_1^1 & \dots & x_1^k \\ \vdots &\ddots & \vdots \\ x_n^1 & \dots & x_n^k\end{array}\right) \begin{pmatrix} \lambda_1 \\ \vdots \\ \lambda_n \end{pmatrix} + y_p.$$

This equation has the desired form $x = Fz + x_0$.

---

For the sake of completeness, we now show $(\ast)$. Two inclusions have to be shown. Let $x, y, z \in \mathbb{K}^n$.

• "$\supseteq$": Let $y \in \mathbb{L}$ and $x \in \mathbb{L}_{\text{h}}$. Then $y + x$ solves the system since

$$A(y + x) = Ay + Ax = b + 0 = b.$$

• "$\subseteq$": Let $z \in \mathbb{L}$ and $y \in \mathbb{L}$. Then $$A(z - y) = Az - Ay = b - b = 0$$ implies that $z - y \in \mathbb{L}_{\text{h}}$. But then every $z \in \mathbb{L}$ has a representation $$z = y + (z - y) \in \{ y \} + \mathbb{L}_{\text{h}}.$$

Answer 2

Let us denote by $\{v_1,\dots,v_k\}$ a basis for the null space. We know that given a particular solution to $Ax=b$, denote it as $x^p$, every other solution is of the form $x=\alpha_1v^1+\dots+\alpha_kv^k+x^p$.

This can be written in a more compact form as follows: $$ \left(\begin{array}{c}x_1 \\ \vdots \\ x_n\end{array}\right)=\left(\begin{array}{c}\sum\limits_{i=1}^k\alpha_iv_1^i \\ \vdots \\ \sum\limits_{i=1}^k\alpha_iv_n^k\end{array}\right)+\left(\begin{array}{c}x_1^p \\ \vdots \\ x_n^p\end{array}\right) $$ The middle term is exactly the Matrix-Vector product: $$ \left(\begin{array}{ccc}v_1^1 & \dots & v_1^k \\ \vdots &\ddots & \vdots \\ v_n^1 & \dots & v_n^k\end{array}\right)\left(\begin{array}{c}\alpha_1 \\ \vdots \\ \alpha _k\end{array}\right):=V\alpha $$ Which in your case you denoted as $Fz$, which is the same thing.