Null space (kernel) and column space (image) — basis, dimension

WHAT are these spaces?

HOW to find them (the algorithm)

Everything comes from row reduction to RREF. Reducing $A$ to $R$ does not change the null space (row operations don't change solution sets of $Ax=0$ ), but it does change the column space — so we read columns from the original $A$ .

Basis of the column space

WHY the original columns? Row reduction mixes rows, distorting column entries, so $\mathrm{Col}(R)\neq\mathrm{Col}(A)$ in general. But RREF correctly tells you which columns are independent — those dependence relations are exactly preserved (they live in the null space, which is unchanged). So we keep the indices, but take the columns from $A$ .

Basis of the null space

Solve $Rx=0$ . Express pivot variables in terms of free variables; set one free variable to $1$ and the rest to $0$ at a time. Each free variable gives one special solution.

The Rank–Nullity Theorem (derived)

Figure — Null space (kernel) and column space (image) — basis, dimension

Worked Examples

Worked example Example 1 — a 3×4 matrix

$A=\begin{bmatrix}1&2&2&4\\3&6&6&12\\1&2&3&6\end{bmatrix}$ Step 1: Row reduce. $R_2\to R_2-3R_1$ , $R_3\to R_3-R_1$ :

\begin{bmatrix}1&2&0&0\\0&0&1&2\\0&0&0&0\end{bmatrix}=R.$$ *Why this step?* RREF exposes pivots cleanly; clearing above pivots gives free-variable formulas directly. **Step 2: Pivots** in columns 1 and 3 → $r=2$. **Col(A) basis** = columns 1 & 3 of *original $A$*: $\{(1,3,1)^\top,\ (2,6,3)^\top\}$. So $\dim\mathrm{Col}=2$. *Why original columns?* The mixed columns of $R$ no longer point in the right output directions. **Step 3: Null space.** Free variables $x_2,x_4$. From $R$: $x_1=-2x_2$, $x_3=-2x_4$. - $x_2=1,x_4=0$: $s_1=(-2,1,0,0)^\top$. - $x_2=0,x_4=1$: $s_2=(0,0,-2,1)^\top$. **Null(A) basis** $=\{s_1,s_2\}$, $\dim\mathrm{Null}=2$. **Check:** $r+\text{nullity}=2+2=4=n$. ✓ And indeed $A s_1 = -2(\text{col}_1)+(\text{col}_2)=0$. ✓

Common Mistakes (steel-manned)

Active Recall

Recall Quick self-test (cover the answers)

Dimension of Col(A)? → rank $r$ .
Dimension of Null(A)? → $n-r$ .
Which space lives in $\mathbb{R}^m$ ? → Col(A).
Where do you take Col(A) basis columns from? → original $A$ , at pivot positions.
How many special solutions? → one per free variable.

Recall Feynman: explain to a 12-year-old

Imagine a vending machine. You press buttons (that's your input $x$ ).

The column space is all the snacks the machine can ever give you — maybe it stocks only chips and soda, so you can't get chocolate no matter what you press.
The null space is button combos that give you nothing — press these and the machine just eats your effort and hands back empty. Counting rule: every button you press is either useful (gives a new snack) or wasted (does nothing new). Total buttons = useful + wasted. That's rank + nullity = $n$ .

Connections

Rank of a matrix — rank is $\dim\mathrm{Col}(A)$ .
Rank–Nullity Theorem — the counting identity used here.
Row reduction & RREF — the engine for both bases.
Linear independence and basis — pivots ⇔ independent columns.
Injective and surjective linear maps — trivial kernel ⇔ injective; full row rank ⇔ surjective.
Four fundamental subspaces — Col(A), Null(A), Row(A), Null(Aᵀ).
Solving Ax=b — solvable iff $b\in\mathrm{Col}(A)$ ; solution set = particular + Null(A).

Definition of Col(A)?

The span of the columns of

A

; the set

\{Ax\}

, living in output space

\mathbb{R}^m

Definition of Null(A)?

\{x : Ax=0\}

, the set of inputs crushed to zero, living in input space

\mathbb{R}^n

Basis of Col(A)?

The pivot columns taken from the ORIGINAL matrix

A

(pivot positions found via RREF).

Basis of Null(A)?

The special solutions — one per free variable, found by solving

Rx=0

Dimension of Col(A)?

The rank

r

= number of pivots.

Dimension of Null(A) (nullity)?

n-r

= number of free variables.

State Rank–Nullity.

\dim\mathrm{Col}(A)+\dim\mathrm{Null}(A)=n

, i.e.

r+(n-r)=n

Why derive Rank–Nullity?

Each of the

n

columns is either pivot (counts in rank) or free (counts in nullity); count columns two ways.

Why use original columns, not RREF columns, for Col(A)?

Row ops change column values; RREF only tells you WHICH columns are independent, not their actual output directions.

When is Null(A)={0}?

When there are no free variables, i.e. full column rank

r=n

; the map is injective.

Why is Null(A) a subspace?

Ax=0,Ay=0

then

A(\alpha x+\beta y)=0

; contains 0 and closed under linear combos.

Row reduction effect on the two spaces?

Preserves Null(A) (same solution set of

Ax=0

); changes Col(A).

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, ek matrix $A$ ko ek machine ki tarah socho jo input vector $x$ leke output $Ax$ deti hai. Ab do important sawaal hote hain. Pehla: machine kaunse outputs de sakti hai — yeh hai column space, yaani $A$ ke columns ka span, aur yeh output wali jagah $\mathbb{R}^m$ mein rehta hai. Doosra: kaunse inputs ko machine zero bana deti hai — yeh hai null space (kernel), jo $Ax=0$ ko satisfy karne wale saare $x$ ka set hai, aur input wali jagah $\mathbb{R}^n$ mein rehta hai.

Inko nikalne ka tareeka simple hai: matrix ko RREF tak row-reduce karo. Jitne pivots milte hain, utna rank $r$ — yahi $\dim$ Col(A). Important baat: column space ki basis ke liye pivot wale columns original $A$ se uthao, RREF se nahi, kyunki row operations columns ki values badal dete hain (sirf yeh batate hain kaunse columns independent hain). Null space ke liye free variables count karo; har free variable ek "special solution" deta hai, aur yeh sab milke null space ki basis bante hain. Nullity $=n-r$ .

Sabse pyaari baat hai Rank–Nullity: $r+(n-r)=n$ . Iska derivation rattana mat — bas socho, har column ya toh pivot hai (rank mein gina) ya free hai (nullity mein gina), koi column do baar nahi ginta. Toh total columns $n$ = rank + nullity. Bas counting hai, jaadu nahi!

Exam tip (80/20): agar koi obvious row/column dependence dikh jaaye (jaise ek row doosri ka multiple), turant rank kam ho jaata hai — full reduction ki zaroorat nahi. Aur yaad rakho: Null trivial ( $\{0\}$ ) tabhi hota hai jab full column rank ho, matlab map injective hai.

Null space (kernel) and column space (image) — basis, dimension

WHAT are these spaces?

HOW to find them (the algorithm)

Basis of the column space

Basis of the null space

The Rank–Nullity Theorem (derived)

Worked Examples

Common Mistakes (steel-manned)

Active Recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Linear Algebra (Full)

Connections