Subspaces — four fundamental subspaces of a matrix: Worked examples

The scenario matrix

Every matrix falls into one of these case classes. The columns are the questions that change the answer; the last column names the example that covers it.

#	Case class	Shape $m \times n$	Rank $r$	What's special	Covered by
1	Wide, rank-deficient	$2 \times 3$	$r < m, r < n$	free variables and unreachable outputs	Ex 1
2	Tall, full column rank	$3 \times 2$	$r = n < m$	$N (A) = {0}$ , big left null space	Ex 2
3	Square invertible	$n \times n$	$r = n = m$	all subspaces trivial or everything	Ex 3
4	The zero matrix	$2 \times 2$	$r = 0$	the degenerate limit	Ex 4
5	Repeated / dependent rows	$3 \times 3$	$r < m, r < n$	both null spaces nontrivial	Ex 5
6	Word problem (networks)	$3 \times 3$	$r = 2$	left null space = a conservation law	Ex 6
7	Exam twist: solvability of $A x = b$	$3 \times 3$	$r = 2$	uses $b \in C (A)$ + $N (A^{T})$	Ex 7
8	Single column, single row	$3 \times 1$ , $1 \times 3$	$r = 1$	rank-one limiting shapes	Ex 8

The two "control knobs" are always: how many pivots ( $r$ ) and which dimension is bigger ( $m$ vs $n$ ). Every cell above is a different setting of those two knobs.

Figure — Subspaces — four fundamental subspaces of a matrix

Example 1 — Wide, rank-deficient (cell 1)

Worked example

$A = [122538]$ , size $2 \times 3$ Forecast: guess $r$ , then $dim C (A), dim C (A^{T}), dim N (A), dim N (A^{T})$ before reading on. (Hint: two rows, are they parallel?)

Step 1 — Row reduce. $R_{2} \to R_{2} - 2 R_{1}$ gives $[102132]$ , then $R_{1} \to R_{1} - 2 R_{2}$ gives the RREF $[1001 - 1 2]$ . Why this step? RREF exposes the pivots, and the pivot count is the rank — the one number that fixes all four dimensions. Here pivots sit in columns 1 and 2, so $r = 2$ .

Step 2 — Column space $C (A) \subseteq R^{2}$ . Pivot columns are 1 and 2, so take the original columns $(1, 2)$ and $(2, 5)$ : $C (A) = span {(1, 2), (2, 5)} = R^{2}$ . Why this step? Pivot columns are independent and span the column space. Two independent vectors in $R^{2}$ fill it: every output is reachable.

Step 3 — Row space $C (A^{T}) \subseteq R^{3}$ . The two nonzero RREF rows: $span {(1, 0, - 1), (0, 1, 2)}$ , dimension $r = 2$ .

Step 4 — Null space $N (A) \subseteq R^{3}$ . From RREF: $x_{1} = x_{3}$ , $x_{2} = - 2 x_{3}$ . Free variable $x_{3}$ . Set $x_{3} = 1$ : special solution $(1, - 2, 1)$ . So $N (A) = span {(1, - 2, 1)}$ , dimension $n - r = 3 - 2 = 1$ . Why this step? Each free variable gives one independent direction the machine crushes to zero.

Step 5 — Left null space $N (A^{T}) \subseteq R^{2}$ . Dimension $m - r = 2 - 2 = 0$ , so $N (A^{T}) = {0}$ .

Verify: dims sum correctly: $r + (n - r) = 2 + 1 = 3 = n$ ✓ and $r + (m - r) = 2 + 0 = 2 = m$ ✓. Orthogonality: $(1, - 2, 1) \cdot (1, 0, - 1) = 1 - 1 = 0$ ✓ and $(1, - 2, 1) \cdot (0, 1, 2) = - 2 + 2 = 0$ ✓ — the null vector is perpendicular to every row.

Example 2 — Tall, full column rank (cell 2)

Worked example

$A = 111012$ , size $3 \times 2$ Forecast: columns clearly independent (not multiples). So $r = 2$ . Where does the "missing" dimension go — into $N (A)$ or $N (A^{T})$ ?

Step 1 — Rank. Columns $(1, 1, 1)$ and $(0, 1, 2)$ are independent, so $r = 2 = n$ . Why this step? Full column rank means no input is wasted.

Step 2 — Null space. $dim N (A) = n - r = 2 - 2 = 0$ , so $N (A) = {0}$ : nothing gets crushed.

Step 3 — Column space $C (A) \subseteq R^{3}$ . $span {(1, 1, 1), (0, 1, 2)}$ , a plane through the origin in $R^{3}$ , dimension $2$ .

Step 4 — Left null space $N (A^{T}) \subseteq R^{3}$ . Dimension $m - r = 3 - 2 = 1$ . Solve $A^{T} y = 0$ : $y_{1} + y_{2} + y_{3} = 0$ and $y_{2} + 2 y_{3} = 0$ . From the second $y_{2} = - 2 y_{3}$ ; from the first $y_{1} = - y_{2} - y_{3} = 2 y_{3} - y_{3} = y_{3}$ . Set $y_{3} = 1$ : $(1, - 2, 1)$ . So $N (A^{T}) = span {(1, - 2, 1)}$ .

Verify: this $(1, - 2, 1)$ must be perpendicular to the whole column space (the plane). $(1, - 2, 1) \cdot (1, 1, 1) = 0$ ✓ and $(1, - 2, 1) \cdot (0, 1, 2) = - 2 + 2 = 0$ ✓. So $N (A^{T})$ is exactly the normal line to the plane $C (A)$ — this is what " $N (A^{T}) ⊥ C (A)$ in $R^{m}$ " looks like, and it is the vector least squares uses.

Example 3 — Square invertible (cell 3)

Example 4 — The zero matrix (degenerate limit, cell 4)

Example 5 — Dependent rows, both null spaces nontrivial (cell 5)

Worked example

$A = 121120121$ , size $3 \times 3$ Forecast: row 2 is $2 \times$ row 1. So the rows can't all be independent — guess $r$ , then the four dims.

Step 1 — Row reduce. $R_{2} \to R_{2} - 2 R_{1}$ kills row 2 to zeros. $R_{3} \to R_{3} - R_{1} = (0, - 1, 0)$ . Reorder/clean to RREF $100010100$ . Two pivots $\Rightarrow r = 2$ .

Step 2 — Row space $C (A^{T}) \subseteq R^{3}$ . $span {(1, 0, 1), (0, 1, 0)}$ , dimension $2$ .

Step 3 — Null space $N (A) \subseteq R^{3}$ . From RREF: $x_{1} = - x_{3}$ , $x_{2} = 0$ , $x_{3}$ free. Set $x_{3} = 1$ : $(- 1, 0, 1)$ . So $N (A) = span {(- 1, 0, 1)}$ , dimension $n - r = 1$ .

Step 4 — Column space $C (A) \subseteq R^{3}$ . Pivot columns 1,2 of the original $A$ : $(1, 2, 1)$ and $(1, 2, 0)$ . Dimension $2$ .

Step 5 — Left null space $N (A^{T}) \subseteq R^{3}$ . Dimension $m - r = 1$ . Solve $y_{1} (1, 1, 1) + y_{2} (2, 2, 2) + y_{3} (1, 0, 1) = 0$ column-wise (i.e. $A^{T} y = 0$ ): the dependency row-2 $= 2 \times$ row-1 forces $y_{1} + 2 y_{2} = 0$ with $y_{3} = 0$ . Set $y_{2} = 1$ : $y = (- 2, 1, 0)$ . So $N (A^{T}) = span {(- 2, 1, 0)}$ .

Verify: $N (A) ⊥$ rows: $(- 1, 0, 1) \cdot (1, 0, 1) = 0$ ✓, $(- 1, 0, 1) \cdot (0, 1, 0) = 0$ ✓. $N (A^{T}) ⊥$ columns: $(- 2, 1, 0) \cdot (1, 2, 1) = - 2 + 2 = 0$ ✓, $(- 2, 1, 0) \cdot (1, 2, 0) = - 2 + 2 = 0$ ✓. Sums: $2 + 1 = 3 = n$ ✓, $2 + 1 = 3 = m$ ✓.

Example 6 — Word problem: a network conservation law (cell 6)

Worked example Currents in a triangle. Three nodes, three wires. The

incidence matrix for edges $1 \to 2, 2 \to 3, 1 \to 3$ is $A = - 1 0 - 1 1 - 1 0 011 .$ Each row says " $(value at head) - (value at tail)$ " for that edge. We seek: which node-potential vectors give zero voltage across every edge, and what the left null space tells us physically. Forecast: all three nodes at the same potential should give zero everywhere — so $N (A)$ is at least a line. Guess $r$ .

Step 1 — Rank. Add: row3 $=$ row1 $+$ row2 (check: $(- 1, 1, 0) + (0, - 1, 1) = (- 1, 0, 1)$ ✓). So rows are dependent, $r = 2$ . Why this step? A dependent row means one edge's voltage is forced by the other two — a loop law.

Step 2 — Null space $N (A) \subseteq R^{3}$ (potentials). $A x = 0$ means every edge sees equal endpoint potentials, i.e. $x_{1} = x_{2} = x_{3}$ . So $N (A) = span {(1, 1, 1)}$ , dimension $n - r = 1$ . Physical meaning: shifting all potentials by a constant changes no voltage.

Step 3 — Left null space $N (A^{T}) \subseteq R^{3}$ (edge currents). Dimension $m - r = 1$ . Solve $A^{T} y = 0$ : this says the currents $y$ satisfy Kirchhoff's current law at each node. The dependency row3=row1+row2 gives $y = (1, 1, - 1)$ (current around the loop).

Verify: $A x = 0$ for $x = (1, 1, 1)$ : each row sums its $\pm 1$ pattern to $0$ ✓. $A^{T} y = 0$ for $y = (1, 1, - 1)$ : node 1 column $(- 1, 0, - 1) \cdot (1, 1, - 1) = - 1 + 1 = 0$ ✓; node 2 $(1, - 1, 0) \cdot (1, 1, - 1) = 1 - 1 = 0$ ✓; node 3 $(0, 1, 1) \cdot (1, 1, - 1) = 1 - 1 = 0$ ✓. So the left null space is the conservation law.

Example 7 — Exam twist: is $A x = b$ solvable? (cell 7)

Worked example With

$A = 101202011$ , decide whether $b = (3, 1, 4)$ and $b^{'} = (3, 1, 5)$ are reachable. Forecast: row 3 $=$ row 1 $+$ row 2 of $A$ . What constraint does that force on any reachable $b$ ?

Step 1 — Rank and left null space. row3 $=$ row1 $+$ row2 $\Rightarrow r = 2$ , and $dim N (A^{T}) = m - r = 1$ . Solving $A^{T} y = 0$ gives the dependency vector $y = (1, 1, - 1)$ . Why this step? $Ax=b$ is solvable $⟺ b \in C (A) ⟺ b ⊥ N (A^{T})$ . So the single condition is $y \cdot b = 0$ .

Step 2 — Test $b = (3, 1, 4)$ . $y \cdot b = (1, 1, - 1) \cdot (3, 1, 4) = 3 + 1 - 4 = 0$ ✓ — solvable.

Step 3 — Test $b^{'} = (3, 1, 5)$ . $y \cdot b^{'} = 3 + 1 - 5 = - 1 \neq = 0$ — not solvable, $b^{'} \in / C (A)$ .

Step 4 — General solution for the solvable $b$ . RREF of $[A ∣ b]$ gives $x_{1} + 2 x_{2} = 3, x_{3} = 1$ . Particular: $(3, 0, 1)$ . Homogeneous part $N (A) = span {(- 2, 1, 0)}$ . So all solutions: $x = (3, 0, 1) + t (- 2, 1, 0)$ .

Verify: $A (3, 0, 1) = (3, 1, 3 + 1) = (3, 1, 4) = b$ ✓. And $A (- 2, 1, 0) = (- 2 + 2, 0, - 2 + 2) = (0, 0, 0)$ ✓ so adding any $t (- 2, 1, 0)$ stays a solution. The exam trap — "which right-hand sides work?" — is answered entirely by the left null space.

Example 8 — Single column and single row (rank-one limits, cell 8)

Worked example Column vector

$A = 2 - 1 4$ ( $3 \times 1$ ) and row vector $B = [2 - 1 4]$ ( $1 \times 3$ ). Forecast: both have a single nonzero entry-pattern, so $r = 1$ for each. Which spaces collapse to points, which fill lines/planes?

Step 1 — For $A$ ( $m = 3, n = 1, r = 1$ ). $C (A) = span {(2, - 1, 4)} \subseteq R^{3}$ (a line). $N (A)$ : $dim = n - r = 1 - 1 = 0$ , so $N (A) = {0}$ in $R^{1}$ . $N (A^{T})$ : $dim = m - r = 3 - 1 = 2$ , a plane in $R^{3}$ — all vectors perpendicular to $(2, - 1, 4)$ .

Step 2 — For $B = A^{T}$ ( $m = 1, n = 3, r = 1$ ). By symmetry the roles swap: $C (B^{T}) =$ the same line $span {(2, - 1, 4)} \subseteq R^{3}$ ; $N (B)$ : $dim = n - r = 3 - 1 = 2$ , the plane perpendicular to $(2, - 1, 4)$ ; $C (B) = R^{1}$ ; $N (B^{T}) = {0}$ .

Step 3 — Basis of the perpendicular plane. Need two independent vectors with $2 u_{1} - u_{2} + 4 u_{3} = 0$ . Set $(u_{2}, u_{3}) = (2, 0) \Rightarrow u_{1} = 1$ : $(1, 2, 0)$ . Set $(u_{2}, u_{3}) = (0, - 1) \Rightarrow 2 u_{1} = 4 \Rightarrow (2, 0, - 1)$ ... check: $2 \cdot 2 - 0 + 4 (- 1) = 0$ ✓. Plane $= span {(1, 2, 0), (2, 0, - 1)}$ .

Verify: $(2, - 1, 4) \cdot (1, 2, 0) = 2 - 2 + 0 = 0$ ✓, $(2, - 1, 4) \cdot (2, 0, - 1) = 4 + 0 - 4 = 0$ ✓. Dims: for $A$ , $1 + 0 = 1 = n$ ✓ and $1 + 2 = 3 = m$ ✓. Transposing swaps input/output worlds — the orthogonal complement of a line in $R^{3}$ is a plane, exactly as the dimension count $n - r$ predicts.

Recall Self-test: name the four dims from shape and rank

$A$ is $5 \times 3$ with rank $2$ . Give all four dimensions. $dim C (A)$ ::: $r = 2$ $dim C (A^{T})$ ::: $r = 2$ $dim N (A)$ ::: $n - r = 3 - 2 = 1$ $dim N (A^{T})$ ::: $m - r = 5 - 2 = 3$

Connections

Rank of a Matrix — the knob $r$ that sets every cell in the scenario matrix.
Rank–Nullity Theorem — the sum-checks in every Verify block.
RREF and Pivots — the engine used in Examples 1, 5, 7.
Solving Ax=b — Example 7's solvability test via $b \in C (A)$ .
Orthogonal Complements — Examples 2 and 8's normal line/plane.
Least Squares & Projections — the left null space direction found in Example 2.
Subspaces — four fundamental subspaces of a matrix — the parent this drills.