Rank–Nullity (Linear Maps)

Let $T:V\to W$ be a linear map between vector spaces over $\mathbb{R}$, with $V$ finite-dimensional. Two subspaces come along for free.

• The kernel (or null space) is everything the map sends to zero: $\ker T=\{\,v\in V : T(v)=\mathbf{0}_W\,\}.$ It lives inside the domain $V$ and is a subspace of $V$. Intuitively it measures how much information the map collapses.

• The image (or range) is everything the map actually hits: $\operatorname{im} T=\{\,T(v) : v\in V\,\}=\{\,w\in W:\exists\,v\in V,\;T(v)=w\,\}.$ It lives inside the codomain $W$ and is a subspace of $W$. It measures how much of $W$ the map reaches.

We give their dimensions short names:

• nullity $=\dim(\ker T)$ — the size of what gets squashed to $\mathbf 0$.
• rank $=\dim(\operatorname{im} T)$ — the size of what gets produced.

Here $\dim(\cdot)$ is the dimension (number of vectors in a basis) and $\mathbf 0_W$ is the zero vector of $W$. Keep straight that the kernel is measured inside the domain and the image inside the codomain.

The central fact is a kind of conservation law: the domain's dimension splits cleanly into what the map kills and what it keeps.

$\dim(\ker T)+\dim(\operatorname{im} T)=\dim V,$

or in words,

$\text{nullity}+\text{rank}=\dim(\text{domain}).$

Why it is believable. Pick a basis $\{v_1,\dots,v_k\}$ of $\ker T$ (so $k=$ nullity) and extend it to a basis $\{v_1,\dots,v_k,v_{k+1},\dots,v_n\}$ of all of $V$ (so $n=\dim V$). One shows that $\{T(v_{k+1}),\dots,T(v_n)\}$ is a basis of $\operatorname{im} T$. That image basis has $n-k$ vectors, so rank $=n-k$, giving $k+(n-k)=n$. The directions inside the kernel contribute nothing new to the image; the remaining $n-k$ directions are exactly what survives.

How you actually use it. The equation has three slots, and knowing any two gives the third: $\text{nullity}=\dim V-\text{rank},\qquad \text{rank}=\dim V-\text{nullity}.$ For a map given by an $m\times n$ matrix $A$ (a map $\mathbb{R}^n\to\mathbb{R}^m$), the rank is the number of pivots after row reduction and the nullity is the number of free columns, and these always sum to $n$, the number of columns.

The theorem turns two qualitative questions — is $T$ one-to-one? onto? — into arithmetic.

Injective $\iff$ trivial kernel. A linear map is injective (one-to-one) exactly when $\ker T=\{\mathbf 0\}$, i.e. when $\text{nullity}=0.$ (Reason: if $T(u)=T(v)$ then $T(u-v)=\mathbf 0$, so $u-v\in\ker T$; the kernel being trivial forces $u=v$.)

Surjective $\iff$ full image. A linear map is surjective (onto) exactly when $\operatorname{im} T=W$, i.e. when $\text{rank}=\dim W.$

Now combine these with Rank–Nullity. Because rank $\le\dim W$ always, and rank $=\dim V-\text{nullity}\le\dim V$:

• If $\dim V>\dim W$, then nullity $=\dim V-\text{rank}\ge \dim V-\dim W>0$, so $T$ cannot be injective. (Too big a domain must collapse something.)
• If $\dim V<\dim W$, then rank $\le\dim V<\dim W$, so $T$ cannot be surjective. (Too small a domain cannot fill the codomain.)

These are counting arguments only — no formulas for $T$ needed.

The two bounds above produce the classic quick verdicts.

• $T:\mathbb{R}^3\to\mathbb{R}^2$ can never be injective. Here $\dim V=3>2=\dim W$, so rank $\le 2$ and nullity $=3-\text{rank}\ge 1>0$. The kernel is nontrivial, so some nonzero vector maps to $\mathbf 0$.
• $T:\mathbb{R}^2\to\mathbb{R}^3$ can never be surjective. Here rank $\le\dim V=2<3=\dim W$, so the image is a proper subspace (at most a plane) of $\mathbb{R}^3$.

The square case $T:V\to V$ with $\dim V=n$. Rank–Nullity reads rank $+$ nullity $=n$. So $T\text{ injective}\iff\text{nullity}=0\iff\text{rank}=n\iff T\text{ surjective}.$ For a linear operator on a finite-dimensional space, injective, surjective, and bijective are all the same condition. (This fails for infinite-dimensional spaces: the shift map $(x_1,x_2,\dots)\mapsto(0,x_1,x_2,\dots)$ is injective but not surjective.)