The column picture turned a system into a question about combining vectors. This topic sets up the language for that, in any number of dimensions.
A vector in $\mathbb{R}^n$ is an ordered list of $n$ real numbers, written as a column: $\mathbf{v} = \begin{bmatrix} v_1 \\ \vdots \\ v_n \end{bmatrix}$. Two vectors are equal when every entry matches. We add vectors entry by entry and multiply a vector by a number (a scalar) by multiplying every entry. For $n = 2$ or $3$ you can draw them as arrows from the origin; for larger $n$ the algebra works the same even though the picture runs out.
In the plane, adding $\mathbf{u} + \mathbf{v}$ means walking along $\mathbf{u}$ and then along $\mathbf{v}$ (tip to tail); the sum is the diagonal of the parallelogram the two arrows make. Scaling by $c$ stretches an arrow by $|c|$ and flips it when $c < 0$. The usual rules hold because they hold in each entry: $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$, $c(\mathbf{u} + \mathbf{v}) = c\mathbf{u} + c\mathbf{v}$, $(c + d)\mathbf{u} = c\mathbf{u} + d\mathbf{u}$, and $\mathbf{u} + \mathbf{0} = \mathbf{u}$.
Given vectors $\mathbf{v}_1, \dots, \mathbf{v}_p$ and scalars $c_1, \dots, c_p$, the vector $$c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_p\mathbf{v}_p$$ is a linear combination of the $\mathbf{v}$'s with weights $c_1, \dots, c_p$. Weights can be any real numbers, including $0$ and negatives.
Worked example: Computing a combination
- Scale: $3\mathbf{u} = \begin{bmatrix} 6 \\ -3 \\ 0 \end{bmatrix}$ and $-2\mathbf{v} = \begin{bmatrix} -2 \\ -6 \\ 4 \end{bmatrix}$.
- Add entry by entry: $\begin{bmatrix} 4 \\ -9 \\ 4 \end{bmatrix} = \begin{bmatrix} 4 \\ -9 \\ 4 \end{bmatrix}$.
Answer: $\begin{bmatrix} 4 \\ -9 \\ 4 \end{bmatrix}$
Why it works
Worked example: Finding the weights
- Row reduce $\left[\begin{array}{cc|c} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 2 & -1 & 7 \end{array}\right]$.
- $R_3 \to R_3 - 2R_1$ gives row 3 $= (0, -5 \mid 5)$; then $R_3 \to R_3 + 5R_2$ gives $(0, 0 \mid 0)$; then $R_1 \to R_1 - 2R_2$: $\left[\begin{array}{cc|c} 1 & 0 & 3 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array}\right]$.
- Consistent, with $x_1 = 3$ and $x_2 = -1$.
- Check: $3\begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} - \begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 7 \end{bmatrix}$. ✓
Answer: Yes: $\mathbf{b} = 3\mathbf{a}_1 - \mathbf{a}_2$.
Key idea