LINEAR-ALGEBRA · Unit 13 · Video 2 · Interactive Practice

How Ax = 0 Splits Rⁿ into Two Perpendicular Pieces

IKey Formulas

Formula	Name	Description
$Ax = 0 \;\Longleftrightarrow\; \text{row}_i \cdot x = 0\ \forall\, i$	Null space condition	$x$ is perpendicular to every row of $A$
$\text{Null}(A) = \text{Row}(A)^{\perp}$	Fundamental Theorem (Part 2)	Null space is the orthogonal complement of the row space
$r + (n - r) = n$	Rank–Nullity Theorem	$\dim(\text{Row space}) + \dim(\text{Null space}) = n$
$x = x_{\text{row}} + x_{\text{null}},\; x_{\text{row}} \perp x_{\text{null}}$	Orthogonal decomposition	Every $x \in \mathbb{R}^n$ splits into $\perp$ components

IIVisualization 1 — Finding a Null-Space Vector

For $A=\begin{bmatrix}1&2&5\\2&4&10\end{bmatrix}$ , $Ax=0$ holds exactly when both rows are perpendicular to $x$ — that is, when both dot products vanish.

x₁ 1 x₂ 1 x₃ 1

💡 Challenge: drive both bars to $0$ — e.g. $x=(-2,1,0)$ . Because the null space is a whole plane, infinitely many $x$ work.

IIIVisualization 2 — Null Space $\perp$ Row Space

A vector lies in the null space exactly when it is perpendicular to the row space — the two are orthogonal complements.

💡 In the video's $2\times3$ example the null space is a 2-D plane (dim $1+2=3$ ); this $\mathbb{R}^2$ companion $A=\begin{bmatrix}1&2\\2&4\end{bmatrix}$ collapses it to the perpendicular line (dim $1+1=2$ ) — the same orthogonal-complement principle.

IVVisualization 3 — Orthogonal vs. Orthogonal Complement

Two orthogonal subspaces are complements only when their dimensions add up to $n$ , leaving no perpendicular direction uncaptured.

dim(W) 1

VQuiz Questions

Question 1

For $A = \begin{bmatrix} 1 & 2 & 5 \\ 2 & 4 & 10 \end{bmatrix}$ , which vector is in $\text{Null}(A)$ ?

(Hint: check which vector satisfies $x_1 + 2x_2 + 5x_3 = 0$ .)

✅ Correct! We have $1 + 2(-3) + 5(1) = 1 - 6 + 5 = 0$ .

❌ Not quite. Compute $x_1 + 2x_2 + 5x_3$ for your chosen vector — it should equal zero.

Show solution

Solution:

Since row $_2$ = 2 $\cdot$ row $_1$ , we only need $x_1 + 2x_2 + 5x_3 = 0$ :

Vector	$x_1 + 2x_2 + 5x_3$	In Null(A)?
$(1, -3, 1)$	$1 - 6 + 5 = 0$	✅ Yes
$(5, -1, -1)$	$5 - 2 - 5 = -2$	❌ No
$(1, 2, 5)$	$1 + 4 + 25 = 30$	❌ No
$(2, 1, 0)$	$2 + 2 + 0 = 4$	❌ No

The answer is $x = (1, -3, 1)$ .

Question 2

True or False: If every vector in subspace $U$ is perpendicular to every vector in subspace $W$ , then $W$ must be the orthogonal complement $U^{\perp}$ .

✅ Correct! Orthogonal $\neq$ orthogonal complement. The complement must leave no perpendicular direction uncaptured, requiring dimensions to sum to $n$ .

❌ Not quite. Think about the counterexample: two perpendicular lines in $\mathbb{R}^3$ are orthogonal but miss the third dimension.

Show solution

Solution: False.

Orthogonality alone is not enough. The orthogonal complement $U^{\perp}$ must capture all perpendicular directions.

Counterexample (from the video): In $\mathbb{R}^3$ , let $U$ = the $x$ -axis and $W$ = the $y$ -axis.

Every vector in $U$ is $\perp$ to every vector in $W$ ✓
But $\dim(U) + \dim(W) = 1 + 1 = 2 \neq 3$
The $z$ -axis direction $(0,0,1)$ is perpendicular to $U$ but is not in $W$

The actual complement $U^{\perp}$ is the entire $yz$ -plane (dimension 2). Being a complement requires $\dim(U) + \dim(W) = n$ .

Question 3

A $4 \times 6$ matrix $A$ has rank $2$ . What is the dimension of $\text{Null}(A)$ ?

✅ Correct! By rank-nullity: nullity $= n - r = 6 - 2 = 4$ .

❌ Not quite. Remember: nullity $= n - r$ , where $n$ is the number of columns (not rows).

Show solution

Solution:

Apply the Rank–Nullity Theorem: $\dim(\text{Row space}) + \dim(\text{Null space}) = n \quad (\text{number of columns})$ $2 + \dim(\text{Null space}) = 6$ $\dim(\text{Null space}) = 6 - 2 = \boxed{4}$

Key point: Use $n = 6$ (number of columns), not $m = 4$ (number of rows). The null space lives in $\mathbb{R}^n = \mathbb{R}^6$ .

Question 4

The Fundamental Theorem of Linear Algebra (Part 2) identifies two orthogonal complement pairs. In $\mathbb{R}^m$ , which subspace is the orthogonal complement of the column space?

✅ Correct! $\text{Null}(A^T)$ contains every vector in $\mathbb{R}^m$ perpendicular to the column space, with dimensions $r + (m-r) = m$ .

❌ Not quite. Think about which subspace lives in $\mathbb{R}^m$ and consists of vectors perpendicular to every column of $A$ . What does $A^T y = 0$ tell you?

Show solution

Solution:

The four fundamental subspaces form two orthogonal complement pairs:

Pair	Space 1	Space 2	Lives in
1	Row space (dim $r$ )	Null space (dim $n-r$ )	$\mathbb{R}^n$
2	Column space (dim $r$ )	Left null space (dim $m-r$ )	$\mathbb{R}^m$

The left null space $\text{Null}(A^T)$ is the orthogonal complement of the column space.

Why? The rows of $A^T$ are the columns of $A$ . So $A^T y = 0$ means $y$ is perpendicular to every column of $A$ — i.e., perpendicular to the entire column space. Dimension check: $r + (m - r) = m$ ✓.

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How Ax = 0 Splits Rⁿ into Two Perpendicular Pieces

IKey Formulas

IIVisualization 1 — Finding a Null-Space Vector

IIIVisualization 2 — Null Space ⊥\perp⊥ Row Space

IVVisualization 3 — Orthogonal vs. Orthogonal Complement

VQuiz Questions

IIIVisualization 2 — Null Space $\perp$ Row Space