Subspaces — four fundamental subspaces of a matrix

1. What is a subspace? (foundation first)

WHY these rules? A subspace must be a "flat thing through the origin" (a line, plane, hyperplane...). If you can scale and add and never leave, you have a complete vector world with no edges. A line not through the origin fails rule 1, so it is not a subspace.

2. Building the four subspaces from scratch

Let $A$ be $m \times n$ . Write its columns and rows: $A = \begin{bmatrix} | & & | \\ a_1 & \cdots & a_n \\ | & & | \end{bmatrix}, \qquad A \in \mathbb{R}^{m\times n}.$

(a) Column space $C(A)$ — reachable outputs

WHY: $Ax = x_1 a_1 + \dots + x_n a_n$ is literally a combination of the columns. So the set of all possible outputs $b=Ax$ is exactly the span of the columns. HOW to use: $Ax=b$ is solvable $\iff b \in C(A)$ .

(b) Null space $N(A)$ — inputs crushed to zero

WHY it's a subspace: if $Ax=0$ and $Ay=0$ then $A(cx+dy)=cAx+dAy=0$ . Closed. WHAT it means: these are the inputs the machine completely destroys.

(c) Row space $C(A^T)$ — and (d) Left null space $N(A^T)$

The other two are just the column space and null space of the transpose: $C(A^T) \subseteq \mathbb{R}^n \quad(\text{span of the rows}),\qquad N(A^T)\subseteq \mathbb{R}^m \quad(\{y: A^Ty=0\}).$

Figure — Subspaces — four fundamental subspaces of a matrix

3. Dimensions: the rank theorem (derive it)

Let $r = \text{rank}(A) =$ number of pivots after row reduction.

Derivation. Row reduce $A$ to RREF.

Pivot columns are linearly independent and span $C(A)$ $\Rightarrow \dim C(A) = r$ .
Row operations don't change the row space, and the $r$ nonzero rows of RREF are independent $\Rightarrow \dim C(A^T) = r$ . So row rank = column rank. This is why "rank" is one number.
Free variables = $n - r$ . Each free variable gives one independent special solution to $Ax=0$ $\Rightarrow \dim N(A) = n - r$ .
By the same argument on $A^T$ (size $n\times m$ , also rank $r$ ): $\dim N(A^T) = m - r$ .

4. Orthogonality: the deep pairing

Derivation of $N(A)\perp C(A^T)$ . If $x\in N(A)$ then $Ax=0$ . Row $i$ of $A$ is $r_i$ , and $(Ax)_i = r_i\cdot x = 0$ . So $x$ is orthogonal to every row, hence to their whole span $C(A^T)$ . Since $\dim N(A)+\dim C(A^T)=n$ and they only share $\mathbf 0$ , together they fill $\mathbb{R}^n$ . ∎

5. Worked examples

Worked example Find all four subspaces of

$A=\begin{bmatrix}1&2&3\\2&4&6\end{bmatrix}$ Step 1 — Row reduce. $R_2\to R_2-2R_1$ : $\begin{bmatrix}1&2&3\\0&0&0\end{bmatrix}$ . Why? RREF exposes pivots → rank. Here $r=1$ ( $m=2,n=3$ ). Step 2 — Column space. Pivot is column 1, so $C(A)=\text{span}\{(1,2)\}$ , a line in $\mathbb{R}^2$ . Why? Pivot columns span the column space; column 2,3 are multiples of column 1. Step 3 — Row space. Nonzero RREF row: $C(A^T)=\text{span}\{(1,2,3)\}$ , a line in $\mathbb{R}^3$ . Step 4 — Null space. $x_1+2x_2+3x_3=0$ . Free: $x_2,x_3$ . Special solutions: set $x_2=1,x_3=0\Rightarrow(-2,1,0)$ ; set $x_2=0,x_3=1\Rightarrow(-3,0,1)$ . So $N(A)=\text{span}\{(-2,1,0),(-3,0,1)\}$ , dim $=n-r=2$ . Why? one equation in 3 unknowns leaves 2 free directions. Step 5 — Left null space. Solve $A^Ty=0$ , i.e. $y_1(1,2,3)+y_2(2,4,6)=0$ . Row 2 $=2\times$ row 1 forces $y_1+2y_2=0\Rightarrow y=(−2,1)$ . $N(A^T)=\text{span}\{(-2,1)\}$ , dim $=m-r=1$ . Check orthogonality: $(1,2)\cdot(-2,1)=0$ ✓ (column $\perp$ left-null).

6. Forecast-then-Verify drill

Before reducing, predict the four dimensions, then check. $A$ is $4\times 6$ with rank $3$ . Forecast: $\dim C(A)=3$ , $\dim C(A^T)=3$ , $\dim N(A)=6-3=3$ , $\dim N(A^T)=4-3=1$ . Sums: $3+3=6=n$ ✓, $3+1=4=m$ ✓. If your forecast doesn't sum correctly, your rank guess is wrong.

Recall Feynman: explain to a 12-year-old

Imagine a fancy juice mixer (the matrix). You pour in fruits (input vectors).

Some fruit combos give nothing — the mixer cancels them out. That collection of "wasted" recipes is the null space.
All the different juices you can possibly make is the column space.
The mixer can only make so many distinct juices — that count is the rank.
And there are some juices you can never make no matter what — that "impossible juice" zone is the left null space. Two of these talk about your recipes (inputs), two about your juices (outputs), and recipe-leftovers are always at perfect right angles to the recipes that matter.

Connections

Rank of a Matrix — the single number $r$ controlling all four dimensions.
Rank–Nullity Theorem — $\dim C(A^T)+\dim N(A)=n$ .
Orthogonal Complements — why the pairs are perpendicular.
Solving Ax=b — solvability ( $b\in C(A)$ ) and general solution (particular $+ N(A)$ ).
RREF and Pivots — the computational engine.
Least Squares & Projections — uses $C(A)$ and $N(A^T)$ directly.

Flashcards

What are the four fundamental subspaces of an

m\times n

matrix

A

Column space

C(A)

, Row space

C(A^T)

, Null space

N(A)

, Left null space

N(A^T)

Which subspaces live in the input space

\mathbb{R}^n

Row space

C(A^T)

and Null space

N(A)

Which subspaces live in the output space

\mathbb{R}^m

Column space

C(A)

and Left null space

N(A^T)

\dim C(A)=?

r

(the rank).

\dim N(A)=?

n-r

\dim N(A^T)=?

m-r

Why is row rank = column rank?

RREF leaves

r

independent nonzero rows and

r

pivot columns; both equal the pivot count

r

Ax=b

is solvable iff?

b\in C(A)

(the column space).

Which two subspaces are orthogonal in

\mathbb{R}^n

N(A)\perp C(A^T)

, and they sum to

\mathbb{R}^n

Why is

N(A)\perp C(A^T)

Ax=0

means each row dotted with

x

is 0, so

x\perp

every row, hence

\perp

row space.

A matrix has trivial null space iff?

Its columns are linearly independent (

r=n

4\times 6

matrix of rank 3: dim of left null space?

m-r=4-3=1

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Socho ek matrix $A$ ek machine hai jo $\mathbb{R}^n$ se vector leke $\mathbb{R}^m$ mein vector deti hai. Is machine ke andar chhupe hue char khaas "subspaces" hote hain. Do input wali duniya $\mathbb{R}^n$ mein rehte hain — Row space $C(A^T)$ aur Null space $N(A)$ . Do output wali duniya $\mathbb{R}^m$ mein rehte hain — Column space $C(A)$ aur Left null space $N(A^T)$ .

Null space matlab woh saare input $x$ jinko machine bilkul zero bana deti hai ( $Ax=0$ ). Column space matlab woh saare output jo banana possible hai (saare columns ka span). Isliye $Ax=b$ tabhi solve hoga jab $b$ column space mein ho. Rank $r$ ek hi number hai jo sab kuch control karta hai: $\dim C(A)=r$ , $\dim C(A^T)=r$ , $\dim N(A)=n-r$ , $\dim N(A^T)=m-r$ . Yaad rakho — null space input space mein hota hai, isliye $n$ se count karo, $m$ se nahi.

Sabse khoobsurat baat: input space mein Row space aur Null space ek dusre ke perpendicular ( $\perp$ ) hote hain, aur milke poora $\mathbb{R}^n$ ban jaate hain. Iska reason simple hai — agar $Ax=0$ hai to har row ka $x$ ke saath dot product zero hai, matlab $x$ har row ke perpendicular hai. Yahi cheez least squares, projections aur solvability sab mein kaam aati hai. Exam tip: pehle rank guess karo, phir char dimensions predict karo, aur check karo ki $r+(n-r)=n$ aur $r+(m-r)=m$ sahi aa rahe hain.

Subspaces — four fundamental subspaces of a matrix

1. What is a subspace? (foundation first)

2. Building the four subspaces from scratch

(a) Column space $C(A)$ — reachable outputs

(b) Null space $N(A)$ — inputs crushed to zero

(c) Row space $C(A^T)$ — and (d) Left null space $N(A^T)$

3. Dimensions: the rank theorem (derive it)

4. Orthogonality: the deep pairing

5. Worked examples

6. Forecast-then-Verify drill

Connections

Flashcards

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Linear Algebra (Full)

Connections