18.1 Template for Complete Solutions to GLSP

Types: N/A
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Generalizations: 17.1 The General Linear-Systems Problem

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The Complete Solution to the GLSP

The complete solution to a linear-systems problem takes the form

\vec{x} = \vec{p} + \vec{z}

The vector $\vec{p}$ is a particular solution with

A \vec{p} = \vec{b}

For now, we will call the vector $\vec{z}$ a trivial solution to the homogenous linear system that sends $A$ to zero with

A \vec{z} = 0

When combined linearly together, these vectors produce our complete solution

A (\vec{p} + \vec{z}) = \vec{0} .

Let's take a look at what particular and trivial solutions are. Recall Integrals (Antiderivatives) from Calculus. Imagine we wanted to solve this problem.

Example

$\frac{d}{d x} [F (x)] = x^{2}$

To solve for $F (x)$ , we would take the antiderivative of $x^{2}$ .

\begin{aligned} F (x) & = \int x^{2} d x \\ = \underset{particular solution}{\underset{⏟}{\frac{x^{3}}{3}}} + \underset{trivial solution}{\underset{⏟}{C}} \end{aligned}

If we "hit" our particular solution with $\frac{d}{d x}$ , we get back to $x^{2}$ . If we "hit" our trivial solution with $\frac{d}{d x}$ , we get $0$ .

Similarly in Linear Algebra, if we "hit" our particular solution with matrix $A$ , we get $\vec{b}$ and if we "hit" our trivial solution with matrix $A$ , we get $0$ .

Example

$[\begin{matrix} 1 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = 8$

To find the values of $\vec{x}$ , we can take the 6.1 Linear Combination of Vectors.

1 x_{1} + 0 x_{2} = 8

Because $0$ annihilates $x_{2}$ , this variable can be anything. However, $x_{1}$ has to be a particular solution.

\begin{aligned} x_{1} & = 8 \\ ⟹ \vec{x} & = [\begin{array}{c} 8 \\ x_{2} \end{array}] = \underset{\vec{p}}{\underset{⏟}{[\begin{array}{c} 8 \\ 0 \end{array}]}} + \underset{\vec{z}}{\underset{⏟}{[\begin{array}{c} 0 \\ x_{2} \end{array}]}} \end{aligned}