How do columns and outer products build a matrix?

Source scope: §3.17.1–3.17.3, PDF pages 26. Numerical examples, experiments, added derivations, and practice are teaching supplements unless identified as source results.

A column is the image of a coordinate direction

For $A=\begin{bmatrix}1&2\\3&4\end{bmatrix}$ and $x=(2,-1)^T$,

Proposition 10. If $A\in\mathbb R^{m\times n}$ has columns $a_1,\ldots,a_n$ and $x\in\mathbb R^n$, then $Ax=\sum_{i=1}^n x_i a_i$.

The source asks the reader to verify this entry by entry. Added solution: the $j$th entry of both sides is $\sum_i A_{ji}x_i$ by the definition of matrix multiplication. Thus all entries agree.

An outer product writes one pattern

For $a\in\mathbb R^m$ and $b\in\mathbb R^n$, the outer product $ab^T$ is $m\times n$ with $[ab^T]_{ij}=a_i b_j$. Each column is a multiple of $a$, so its rank is at most one.

Proposition 11. For $a_1,\ldots,a_k\in\mathbb R^m$ and $b_1,\ldots,b_k\in\mathbb R^n$, let $A=[a_1\ \cdots\ a_k]$ and $B=[b_1\ \cdots\ b_k]$. Then

Source proof. At each position $(i,j)$,

The penultimate expression is the dot product of row $i$ of $A$ and row $j$ of $B$, equivalently column $j$ of $B^T$. This proves equality at every entry. Notice the transpose: stacking $b_\ell$ as columns does not make the product $AB$.

Expand a quadratic form

For square $A\in\mathbb R^{n\times n}$,

First form $(Ax)_i=\sum_jA_{ij}x_j$, then take $\sum_i x_i(Ax)_i$. The source introduces quadratic forms for symmetric matrices but explicitly notes that this identity needs only a square matrix. For the running nonsymmetric matrix,

At $(2,-1)$ this is $4-10+4=-2$, matching $x^T(Ax)=(2,-1)(0,2)^T$. Symmetry is unnecessary for the expansion; it matters for later spectral and definiteness claims.

Transfer

If $a=(1,2)^T$ and $b=(3,-1,0)^T$, the outer product is $\begin{bmatrix}3&-1&0\\6&-2&0\end{bmatrix}$. Its three columns are multiples of one vector, so it has rank one despite having several nonzero entries.

Explore the relationship