Lecture 13 Linear Combinations & Matrix Equations

Tags: #linear-algebra #vectors #matrices #span

1. The "Three Languages" of Linear Algebra

These three forms represent the exact same mathematical problem. If you can solve one, you can solve the others:

1. Linear System (Augmented Matrix)
2. Vector Equation: $x_1{\mathbf{a}}_1+x_2{\mathbf{a}}_2+\cdots +x_n{\mathbf{a}}_n=\mathbf{b}$
3. Matrix Equation: $A\mathbf{x}=\mathbf{b}$

2. Linear Combinations

A vector $\mathbf{b}$ is a linear combination of vectors ${\mathbf{a}}_1,\dots ,{\mathbf{a}}_n$ if you can scale them by some weights ($x_1,\dots ,x_n$) and add them together to get $\mathbf{b}$.

3. The Span

Definition: The Span{v1, v2, ...} is the set of ALL possible linear combinations you can make with those vectors.

Geometric Interpretation in ${\mathbb{R}}^3$:

• Span of 1 (non-zero) vector = A Line through the origin.
• Span of 2 (non-multiple) vectors = A Plane through the origin.
• Span of 3 (independent) vectors = The entire 3D space.

4. The Matrix-Vector Product ($A\mathbf{x}$)

There are two ways to look at $A\mathbf{x}$:

1. The Row-Column Rule (The "Calculator" View):
   Dot products of the matrix rows with the vector column.
2. The Column Combination Rule (The "Matrix Intuition" View):
   Multiplying a matrix by a vector $\mathbf{x}$ is just creating a linear combination of the columns of $A$, using the entries of $\mathbf{x}$ as the weights.$$A\mathbf{x}=\left[\begin{array}{cc}{\mathbf{a}}_1& {\mathbf{a}}_2\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=x_1{\mathbf{a}}_1+x_2{\mathbf{a}}_2$$

Standard Unit Vectors (${\mathbf{e}}_j$)

Multiplying a matrix $A$ by the $j$-th standard unit vector ${\mathbf{e}}_j$ (like $\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$) acts as a "selector switch"—it perfectly extracts the $j$-th column of $A$.

5. Important Theorems & Tricks

Existence of Solutions

For an $m\times n$ matrix $A$, the equation $A\mathbf{x}=\mathbf{b}$ has a solution for EVERY possible $\mathbf{b}$ in ${\mathbb{R}}^m$ if and only if:

1. The columns of $A$ span ${\mathbb{R}}^m$.
2. Matrix $A$ has a pivot position in every row. (No all-zero rows in the coefficient matrix!)

Properties of Linearity

Matrix equations are linear, meaning you can distribute and pull out scalars:

1. $A(\mathbf{u}+\mathbf{v})=A\mathbf{u}+A\mathbf{v}$
2. $A(c\mathbf{u})=c(A\mathbf{u})$

💡 Trick: Solving by Inspection (No Row Operations)

If you are given a relationship between vectors, you can easily find the solution vector $\mathbf{x}$ for a matrix equation without doing any math.

Example:
Given vectors $\mathbf{u},\mathbf{v},\mathbf{w}$ and the fact: $-3\mathbf{u}+1\mathbf{v}-\mathbf{w}=\mathbf{0}$.
Find $\mathbf{x}$ for the equation: $[\mathbf{u}\mathbf{v}]\mathbf{x}=\mathbf{w}$

Solution:

1. Rearrange the given fact to isolate the target vector ($\mathbf{w}$):
   $-3\mathbf{u}+1\mathbf{v}=\mathbf{w}$
2. Since the matrix equation is really just $x_1\mathbf{u}+x_2\mathbf{v}=\mathbf{w}$, we can just read the weights directly!
3. Therefore, $\mathbf{x}=\left[\begin{array}{c}-3\\ 1\end{array}\right]$.