Linear Algebra · Lesson 04

The rank–nullity theorem

A linear map cannot create dimensions and cannot destroy them without a trace. Every dimension of the domain either survives into the image or is lost in the kernel, so the two sizes always add up to the size of the domain.

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dim ker (T) + dim im (T) = dim (V)

ker (T)

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Intuition

A linear map cannot create dimensions and cannot destroy them without a trace. Every dimension of the domain either survives into the image or is lost in the kernel, so the two sizes always add up to the size of the domain.

A budget split between spending and saving: whatever is not spent is saved, and the two amounts add up to the whole. The dimensions of the domain are split in the same way between the image and the kernel.

Rank and nullity

For a linear map $T:V\to W$ with of finite dimension, the rank of is and its nullity is . The theorem below says they add up to . For the map of an matrix it is the count of the chapter on elimination: pivots plus free variables make .

im (T)

rank (T)

V

V

T

\dim\operatorname{im}T

\dim\ker T

\dim V

m\times n

n

\dim V=\dim\ker T+\dim\operatorname{im}T

Consequences

For an $m\times n$ matrix: $n=\dim\operatorname{Nul}A+\operatorname{rank}A$ .
If $\dim V>\dim W$ , then $T$ is not one-to-one: $\dim\ker T\ge\dim V-\dim W>0$ .
If $\dim V<\dim W$ , then $T$ is not onto: $\dim\operatorname{im}T\le\dim V<\dim W$ .
If $\dim V=\dim W$ , then $T$ is one-to-one exactly when it is onto: $\ker T=\{\mathbf{0}\}$ exactly when $\dim im T =$ .
The proof uses one more fact: in a space of finite dimension, every independent list extends to a basis. While it does not span, add a vector outside its span, which keeps it independent; it cannot grow past $\dim V$ .

The rank–nullity theorem

\dim V=\dim\ker T+\dim\operatorname{im}T

Take a basis $\mathbf{u}_1,\ldots,\mathbf{u}_k$ of the kernel and extend it to a basis of $V$ by vectors $\mathbf{v}_1,\ldots,\mathbf{v}_r$ . The images $T\mathbf{v}_1,\ldots,T\mathbf{v}_r$ span the image, because every $T\mathbf{u}_i$ is $\mathbf{0}$ . They are independent, because a combination of them that is zero comes from a combination of the $\mathbf{v}_j$ lying in the kernel, and that can only be the trivial one. So the image has dimension $r$ , and $\dim V=k+r$ .

Proof steps

\mathbf{u}_1,\ldots,\mathbf{u}_k\ \text{a basis of }\ker T

Start from a basis of the kernel, a subspace of the domain.

\mathbf{u}_1,\ldots,\mathbf{u}_k,\mathbf{v}_1,\ldots,\mathbf{v}_r\ \text{a basis of }V

Extend it to a basis of the whole domain; then $\dim V=k+r$ .

T\big(\textstyle\sum a_i\mathbf{u}_i+\sum c_j\mathbf{v}_j\big)=\sum c_jT\mathbf{v}_j

Each $T\mathbf{u}_i$ is $\mathbf{0}$ , so every image is a combination of the $T\mathbf{v}_j$ : they span $\operatorname{im}T$ .

\textstyle\sum c_jT\mathbf{v}_j=\mathbf{0}\implies\sum c_j\mathbf{v}_j\in\ker T

If a combination of the $T\mathbf{v}_j$ is $\mathbf{0}$ , the same combination of the $\mathbf{v}_j$ is in the kernel, so it is a combination of the $\mathbf{u}_i$ . Independence of the whole basis forces every $c_j=0$ .

\dim\operatorname{im}T=r=\dim V-\dim\ker T

So $T\mathbf{v}_1,\ldots,T\mathbf{v}_r$ is a basis of the image, and counting gives the theorem.

Applications

Practice

Counting Both Sides

Every dimension of the domain is either lost in the kernel or survives into the image. Know two of the three numbers, and the third follows.

T:\mathbb{R}^5\to\mathbb{R}^3,\ \dim\operatorname{im}T=3\implies\dim\ker T=2

Try it

A linear map $T:\mathbb{R}^7\to\mathbb{R}^4$ is onto. What is $\dim\ker T$ ?