LINEAR-ALGEBRA · Unit 11 · Video 3 · Interactive Practice

Rank-One Matrices: The Building Blocks

IKey Formulas

Formula	Name	Description
$A = \mathbf{u}\mathbf{v}^T$	Outer product	Column $\times$ row $\to$ matrix
$A_{ij} = u_i \cdot v_j$	Entry formula	Each entry is the product of corresponding components
$A = \sum_{k=1}^{r} \mathbf{u}_k \mathbf{v}_k^T$	Rank decomposition	A rank- $r$ matrix is a sum of $r$ rank-one matrices
$\text{rank}(A + B) \leq \text{rank}(A) + \text{rank}(B)$	Rank inequality	The rank of a sum cannot exceed the sum of ranks

IIVisualization 1 — The Outer Product Table

Every entry of $A = \mathbf{u}\mathbf{v}^T$ is a single product: $A_{ij} = u_i \times v_j$ .

u₁ 1 u₂ 2 v₁ 1 v₂ 4 v₃ 5

Every row of $A$ is a multiple of $\mathbf{v}^T$ and every column a multiple of $\mathbf{u}$ — that single-direction structure is exactly what rank one means.

IIIVisualization 2 — Columns and Rows Collapse to Lines

For $A = \mathbf{u}\mathbf{v}^T$ , every column is a multiple of $\mathbf{u}$ and every row a multiple of $\mathbf{v}$ .

u₁ 1 u₂ 2 v₁ 3 v₂ 1

Send $\mathbf{u}$ or $\mathbf{v}$ to the origin and $A$ becomes the zero matrix — rank $0$ , and both lines disappear.

IVVisualization 3 — When Does Rank Add Up?

Adding two rank-one matrices, $\text{rank}(M_1 + M_2)$ can hit the bound $2$ — or fall below it.

Since the sum can land at rank $0$ , $1$ , or $2$ , rank-one matrices are not closed under addition — they do not form a subspace.

VQuiz Questions

Question 1 · Outer Product Entry

Given $\mathbf{u} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$ and $\mathbf{v}^T = \begin{bmatrix} 2 & 4 \end{bmatrix}$ , what is entry $(2, 1)$ of the matrix $A = \mathbf{u}\mathbf{v}^T$ ?

✅ Correct! Entry (2,1) = u₂ · v₁ = (−1)(2) = −2.

❌ Not quite. Remember: entry (i,j) = uᵢ × vⱼ. For position (2,1), use u₂ = −1 and v₁ = 2.

Show solution

Solution:

Entry $(i, j)$ of $A = \mathbf{u}\mathbf{v}^T$ is $u_i \cdot v_j$ .

For entry $(2, 1)$ : $u_2 \cdot v_1 = (-1) \times 2 = -2$

The full matrix: $A = \begin{bmatrix} 3 \\ -1 \end{bmatrix} \begin{bmatrix} 2 & 4 \end{bmatrix} = \begin{bmatrix} 3 \times 2 & 3 \times 4 \\ (-1) \times 2 & (-1) \times 4 \end{bmatrix} = \begin{bmatrix} 6 & 12 \\ -2 & -4 \end{bmatrix}$

Common mistakes:

Choosing 6 means you used entry $(1,1) = u_1 \cdot v_1$ — wrong row
Choosing 12 means you used entry $(1,2) = u_1 \cdot v_2$ — wrong row and column
Choosing -4 means you used entry $(2,2) = u_2 \cdot v_2$ — wrong column

Question 2 · Subspace Property

True or False: The set of all $2 \times 2$ rank-one matrices forms a subspace of the vector space of all $2 \times 2$ matrices.

✅ Correct! Rank-one matrices are not closed under addition, so they cannot form a subspace.

❌ Not quite. Consider adding [[1,0],[0,0]] and [[0,0],[0,1]]. What rank does the sum have?

Show solution

Solution: False.

Rank-one matrices are not closed under addition, which is required for a subspace.

Counterexample:

$M_1 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \;\text{(rank 1)}, \qquad M_2 = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \;\text{(rank 1)}$

$M_1 + M_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I_2 \;\text{(rank 2, not rank 1!)}$

Since the sum of two rank-one matrices can have rank 2, the set fails the closure-under-addition requirement for a subspace.

(Note: the set also doesn't contain the zero matrix, which has rank 0, not rank 1.)

Question 3 · Rank after Cancellation

Matrix $A$ has rank 3. Let $B = -A$ . What is $\text{rank}(A + B)$ ?

✅ Correct! A + (−A) = 0, the zero matrix, which has rank 0. The rank inequality permits dramatic cancellation.

❌ Not quite. What is A + (−A)? And what is the rank of that resulting matrix?

Show solution

Solution:

$A + B = A + (-A) = \mathbf{0}$

The zero matrix has rank 0 — it has no nonzero columns at all.

This shows the rank inequality can be extremely strict:

$\text{rank}(A + B) = 0 \;\ll\; \text{rank}(A) + \text{rank}(B) = 3 + 3 = 6$

The upper bound allows rank up to 6, but the actual rank is 0. Complete cancellation is always possible when $B = -A$ , regardless of what $A$ looks like.

Question 4 · Rank-One Decomposition

A $10 \times 10$ matrix $A$ has rank 4. What is the minimum number of rank-one matrices needed to write $A$ as their sum?

✅ Correct! A rank-r matrix decomposes into exactly r rank-one pieces, regardless of the matrix dimensions.

❌ Not quite. The number of rank-one pieces equals the rank, not the matrix size or number of entries.

Show solution

Solution:

A matrix of rank $r$ decomposes into exactly $r$ rank-one matrices — no fewer.

Since $\text{rank}(A) = 4$ :

$A = \mathbf{u}_1\mathbf{v}_1^T + \mathbf{u}_2\mathbf{v}_2^T + \mathbf{u}_3\mathbf{v}_3^T + \mathbf{u}_4\mathbf{v}_4^T$

The matrix dimensions ( $10 \times 10 = 100$ entries) are irrelevant — only the rank determines the count. The specific vectors $\mathbf{u}_k, \mathbf{v}_k$ are not unique (many valid decompositions exist), but the minimum number of rank-one pieces always equals the rank.

1 is too few (would give a rank-1 matrix, not rank 4)
10 confuses the matrix dimension with the rank
100 confuses the number of entries with the rank

Solved: 0 / 4