LINEAR-ALGEBRA · Unit 3 · Video 1 · Interactive Practice

5 Ways to Think About Matrix Multiplication

IKey Formulas

Method	Formula	Key Insight
Entry way	$C_{ij} = \sum_{k=1}^{n} a_{ik}\, b_{kj}$	Dot product of row $i$ of $A$ with column $j$ of $B$
Column way	$\text{col}_j(C) = A \cdot \text{col}_j(B)$	Each column of $C$ is a linear combination of $A$ 's columns
Row way	$\text{row}_i(C) = \text{row}_i(A) \cdot B$	Each row of $C$ is a linear combination of $B$ 's rows
Col × Row	$AB = \sum_{k=1}^{n} \text{col}_k(A)\;\text{row}_k(B)$	Sum of $n$ rank-one outer products

IIVisualization 1 — The Entry Way

Each entry $C_{ij}$ is a single number: the dot product of row $i$ of $A$ with column $j$ of $B$ .

IIIVisualization 2 — The Column Way

Each column of $C$ is a linear combination of $A$ 's columns, weighted by a column of $B$ .

Column of B

IVVisualization 3 — The Row Way

Symmetrically, each row of $C$ is a linear combination of $B$ 's rows, weighted by a row of $A$ .

Row of A

VVisualization 4 — Column × Row

$AB$ equals a sum of $n$ rank-one outer products — one for each index of the shared dimension.

Terms included 1

💡 Each term is rank one — every column is a multiple of one vector, every row a multiple of another. This decomposition underlies matrix rank and low-rank approximation.

VIQuiz Questions

Question 1

Consider a $2 \times 3$ matrix $D$ with row 1 $= [3,\; -1,\; 2]$ and a $3 \times 2$ matrix $E$ with column 2 $= [4,\; 0,\; -1]^T$ .

Using the entry way, compute $(DE)_{1,2}$ .

✅ Correct! The dot product gives $12 + 0 - 2 = 10$ .

❌ Not quite. Multiply corresponding entries and watch the signs carefully.

Show solution

Solution:

$(DE)_{1,2}$ is the dot product of row 1 of $D$ with column 2 of $E$ :

$(DE)_{1,2} = 3 \cdot 4 + (-1) \cdot 0 + 2 \cdot (-1) = 12 + 0 - 2 = 10$

The answer is 10.

Question 2

In the column way of thinking about $C = AB$ , every column of $C$ is a linear combination of which set of vectors?

✅ Correct! Each column of $C$ lives in the column space of $A$ .

❌ Not quite. Think about what $A \cdot \text{col}_j(B)$ produces — which vectors of $A$ get scaled and added?

Show solution

Solution:

Column $j$ of $C$ equals $A \cdot \text{col}_j(B)$ , which expands to:

$\text{col}_j(C) = b_{1j}\,\text{col}_1(A) + b_{2j}\,\text{col}_2(A) + \cdots + b_{nj}\,\text{col}_n(A)$

So every column of $C$ is a linear combination of the columns of $A$ , with weights coming from column $j$ of $B$ .

This is why $\text{Col space}(C) \subseteq \text{Col space}(A)$ .

Question 3

The outer product of a nonzero column vector $\mathbf{u}$ ( $m \times 1$ ) and a nonzero row vector $\mathbf{v}^T$ ( $1 \times p$ ) produces an $m \times p$ matrix. What is the rank of this matrix?

✅ Correct! An outer product of two nonzero vectors always has rank 1.

❌ Not quite. How many linearly independent columns does the outer product matrix have?

Show solution

Solution:

The outer product $\mathbf{u}\mathbf{v}^T$ produces a matrix where:

Every column is a scalar multiple of $\mathbf{u}$
Every row is a scalar multiple of $\mathbf{v}^T$

Since all columns point in the same direction, the column space is one-dimensional. Therefore the rank is 1.

These rank-1 matrices are the simplest non-zero matrices — and every matrix product $AB$ decomposes into a sum of them.

Question 4

True or False: For $C = AB$ where $A$ is $m \times n$ and $B$ is $n \times p$ , the product $C$ can always be written as the sum of exactly $n$ rank-one matrices (one for each index in the shared dimension).

✅ Correct! The column × row decomposition always produces exactly $n$ rank-one terms.

❌ Not quite. Recall the col × row decomposition — how many column-row pairs are there?

Show solution

Solution:

True. The column × row decomposition gives:

$AB = \sum_{k=1}^{n} \text{col}_k(A)\;\text{row}_k(B)$

Each term $\text{col}_k(A)\;\text{row}_k(B)$ is an outer product — a rank-one (or zero) matrix. There are exactly $n$ such terms, where $n$ is the shared inner dimension.

Note: individual terms may be the zero matrix (if a column of $A$ or row of $B$ is zero), but the decomposition still has $n$ terms. The actual rank of $C$ may be less than $n$ .

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