Rank–nullity theorem

Proofs

Here we provide two proofs. The first operates in the general case, using linear maps. The second proof looks at the homogeneous system ${\displaystyle \mathbf{A}\mathbf{x}=\mathbf{0},}$ where ${\displaystyle \mathbf{A}}$ is a ${\displaystyle m\times n}$ with rank ${\displaystyle r,}$ and shows explicitly that there exists a set of ${\displaystyle n-r}$ linearly independent solutions that span the null space of ${\displaystyle \mathbf{A}}$.

While the theorem requires that the domain of the linear map be finite-dimensional, there is no such assumption on the codomain. This means that there are linear maps not given by matrices for which the theorem applies. Despite this, the first proof is not actually more general than the second: since the image of the linear map is finite-dimensional, we can represent the map from its domain to its image by a matrix, prove the theorem for that matrix, then compose with the inclusion of the image into the full codomain.

First proof

Let ${\displaystyle V,W}$ be vector spaces over some field ${\displaystyle F,}$ and ${\displaystyle T}$ defined as in the statement of the theorem with ${\displaystyle \dim V=n}$.

As ${\displaystyle \mathrm{Ker}T\subset V}$ is a subspace, there exists a basis for it. Suppose ${\displaystyle \dim \mathrm{Ker}T=k}$ and let $${\displaystyle \mathcal{K}:=\{v_1,\dots ,v_k\}\subset \mathrm{Ker}(T)}$$ be such a basis.

By the Steinitz exchange lemma, there is a linearly independent set ${\displaystyle \mathcal{S}=\{w_1,\dots ,w_{n-k}\}}$ of ${\displaystyle n-k}$ vectors such that the union ${\displaystyle \mathcal{B}=\mathcal{K}\cup \mathcal{S}=\{v_1,\dots ,v_k,w_1,\dots ,w_{n-k}\}}$ is a basis of ${\displaystyle V}$. From this, it follows $${\displaystyle \begin{array}{cc}\mathrm{Im}T& =\mathrm{Span}T(\mathcal{B})=\mathrm{Span}\{T(v_1),\dots ,T(v_k),T(w_1),\dots ,T(w_{n-k})\}\\ & =\mathrm{Span}\{T(w_1),\dots ,T(w_{n-k})\}=\mathrm{Span}T(\mathcal{S}).\end{array}}$$ We now claim that ${\displaystyle T(\mathcal{S})}$ is a basis for ${\displaystyle \mathrm{Im}T}$. The above equality already states that ${\displaystyle T(\mathcal{S})}$ is a generating set for ${\displaystyle \mathrm{Im}T}$; it remains to be shown that it is also linearly independent to conclude that it is a basis.

To this end, consider a linear combination $${\displaystyle \sum _{j=1}^{n-k}{\alpha }_{j}T(w_j)=0_W}$$ for some ${\displaystyle {\alpha }_j\in F}$. We must show that all the coefficients ${\displaystyle {\alpha }_j}$ are equal to ${\displaystyle 0}$. Owing to the linearity of ${\displaystyle T}$, it follows that $${\displaystyle T\left(\sum _{j=1}^{n-k}{\alpha }_j{w}_j\right)=0_W,}$$ and therefore that $${\displaystyle \left(\sum _{j=1}^{n-k}{\alpha }_j{w}_j\right)\in \mathrm{Ker}T=\mathrm{Span}\mathcal{K}.}$$ Therefore, there are coefficents ${\displaystyle {\beta }_1,\dots ,{\beta }_k}$ such that ${\displaystyle \sum _{j=1}^{n-k}{\alpha }_j{w}_j=\sum _{i=1}^k{\beta }_i{v}_i}$, and so that ${\displaystyle \sum _{j=1}^{n-k}{\alpha }_j{w}_j-\sum _{i=1}^k{\beta }_i{v}_i=0_V}$. This last relation expresses ${\displaystyle 0_V}$ as a linear combination of the basis ${\displaystyle \mathcal{B}}$; since ${\displaystyle \mathcal{B}}$ is a basis, all the coefficients ${\displaystyle {\alpha }_j}$ are equal to zero. This shows that ${\displaystyle T(\mathcal{S})}$ is linearly independent, and more specifically that it is a basis for ${\displaystyle \mathrm{Im}T}$.

To summarize, we have ${\displaystyle \mathcal{K}}$, a basis for ${\displaystyle \mathrm{Ker}T}$, and ${\displaystyle T(\mathcal{S})}$, a basis for ${\displaystyle \mathrm{Im}T}$.

Finally we may state that $${\displaystyle \mathrm{Rank}(T)+\mathrm{Nullity}(T)=\dim \mathrm{Im}T+\dim \mathrm{Ker}T}$$

${\displaystyle =|T(\mathcal{S})|+|\mathcal{K}|=(n-k)+k=n=\dim V.}$

This concludes our proof.

Second proof

Let ${\displaystyle \mathbf{A}}$ be an ${\displaystyle m\times n}$ matrix with ${\displaystyle r}$ linearly independent columns (i.e. ${\displaystyle \mathrm{Rank}(\mathbf{A})=r}$). We will show that:

1. There exists a set of ${\displaystyle n-r}$ linearly independent solutions to the homogeneous system ${\displaystyle \mathbf{A}\mathbf{x}=\mathbf{0}}$.
2. That every other solution is a linear combination of these ${\displaystyle n-r}$ solutions.

To do this, we will produce an ${\displaystyle n\times (n-r)}$ matrix ${\displaystyle \mathbf{X}}$ whose columns form a basis of the null space of ${\displaystyle \mathbf{A}}$.

Without loss of generality, assume that the first ${\displaystyle r}$ columns of ${\displaystyle \mathbf{A}}$ are linearly independent. So, we can write $${\displaystyle \mathbf{A}=\left(\begin{array}{cc}{\mathbf{A}}_1& {\mathbf{A}}_2\end{array}\right),}$$ where

• ${\displaystyle {\mathbf{A}}_1}$ is an ${\displaystyle m\times r}$ matrix with ${\displaystyle r}$ linearly independent column vectors, and
• ${\displaystyle {\mathbf{A}}_2}$ is an ${\displaystyle m\times (n-r)}$ matrix such that each of its ${\displaystyle n-r}$ columns is linear combinations of the columns of ${\displaystyle {\mathbf{A}}_1}$.

This means that ${\displaystyle {\mathbf{A}}_2={\mathbf{A}}_1\mathbf{B}}$ for some ${\displaystyle r\times (n-r)}$ matrix ${\displaystyle \mathbf{B}}$ (see rank factorization) and, hence, $${\displaystyle \mathbf{A}=\left(\begin{array}{cc}{\mathbf{A}}_1& {\mathbf{A}}_1\mathbf{B}\end{array}\right).}$$

Let $${\displaystyle \mathbf{X}=\left(\begin{array}{c}-\mathbf{B}\\ {\mathbf{I}}_{n-r}\end{array}\right),}$$ where ${\displaystyle {\mathbf{I}}_{n-r}}$ is the ${\displaystyle (n-r)\times (n-r)}$ identity matrix. So, ${\displaystyle \mathbf{X}}$ is an ${\displaystyle n\times (n-r)}$ matrix such that $${\displaystyle \mathbf{A}\mathbf{X}=\left(\begin{array}{cc}{\mathbf{A}}_1& {\mathbf{A}}_1\mathbf{B}\end{array}\right)\left(\begin{array}{c}-\mathbf{B}\\ {\mathbf{I}}_{n-r}\end{array}\right)=-{\mathbf{A}}_1\mathbf{B}+{\mathbf{A}}_1\mathbf{B}={\mathbf{0}}_{m\times (n-r)}.}$$

Therefore, each of the ${\displaystyle n-r}$ columns of ${\displaystyle \mathbf{X}}$ are particular solutions of ${\displaystyle \mathbf{A}\mathbf{x}=0_{F^m}}$.

Furthermore, the ${\displaystyle n-r}$ columns of ${\displaystyle \mathbf{X}}$ are linearly independent because ${\displaystyle \mathbf{X}\mathbf{u}={\mathbf{0}}_{F^n}}$ will imply ${\displaystyle \mathbf{u}={\mathbf{0}}_{F^{n-r}}}$ for ${\displaystyle \mathbf{u}\in F^{n-r}}$: $${\displaystyle \mathbf{X}\mathbf{u}={\mathbf{0}}_{F^n}⟹\left(\begin{array}{c}-\mathbf{B}\\ {\mathbf{I}}_{n-r}\end{array}\right)\mathbf{u}={\mathbf{0}}_{F^n}⟹\left(\begin{array}{c}-\mathbf{B}\mathbf{u}\\ \mathbf{u}\end{array}\right)=\left(\begin{array}{c}{\mathbf{0}}_{F^r}\\ {\mathbf{0}}_{F^{n-r}}\end{array}\right)⟹\mathbf{u}={\mathbf{0}}_{F^{n-r}}.}$$ Therefore, the column vectors of ${\displaystyle \mathbf{X}}$ constitute a set of ${\displaystyle n-r}$ linearly independent solutions for ${\displaystyle \mathbf{A}\mathbf{x}={\mathbf{0}}_{{\mathbb{F}}^m}}$.

We next prove that any solution of ${\displaystyle \mathbf{A}\mathbf{x}={\mathbf{0}}_{F^m}}$ must be a linear combination of the columns of ${\displaystyle \mathbf{X}}$.

For this, let $${\displaystyle \mathbf{u}=\left(\begin{array}{c}{\mathbf{u}}_1\\ {\mathbf{u}}_2\end{array}\right)\in F^n}$$

be any vector such that ${\displaystyle \mathbf{A}\mathbf{u}={\mathbf{0}}_{F^m}}$. Since the columns of ${\displaystyle {\mathbf{A}}_1}$ are linearly independent, ${\displaystyle {\mathbf{A}}_1\mathbf{x}={\mathbf{0}}_{F^m}}$ implies ${\displaystyle \mathbf{x}={\mathbf{0}}_{F^r}}$.

Therefore, $${\displaystyle \begin{array}{rcl}\mathbf{A}\mathbf{u}& =& {\mathbf{0}}_{F^m}\\ ⟹\left(\begin{array}{cc}{\mathbf{A}}_1& {\mathbf{A}}_1\mathbf{B}\end{array}\right)\left(\begin{array}{c}{\mathbf{u}}_1\\ {\mathbf{u}}_2\end{array}\right)& =& {\mathbf{A}}_1{\mathbf{u}}_1+{\mathbf{A}}_1\mathbf{B}{\mathbf{u}}_2& =& {\mathbf{A}}_1({\mathbf{u}}_1+\mathbf{B}{\mathbf{u}}_2)& =& {\mathbf{0}}_{{\mathbb{F}}^m}\\ ⟹{\mathbf{u}}_1+\mathbf{B}{\mathbf{u}}_2& =& {\mathbf{0}}_{F^r}\\ ⟹{\mathbf{u}}_1& =& -\mathbf{B}{\mathbf{u}}_2\end{array}}$$ $${\displaystyle ⟹\mathbf{u}=\left(\begin{array}{c}{\mathbf{u}}_1\\ {\mathbf{u}}_2\end{array}\right)=\left(\begin{array}{c}-\mathbf{B}\\ {\mathbf{I}}_{n-r}\end{array}\right){\mathbf{u}}_2=\mathbf{X}{\mathbf{u}}_2.}$$

This proves that any vector ${\displaystyle \mathbf{u}}$ that is a solution of ${\displaystyle \mathbf{A}\mathbf{x}=\mathbf{0}}$ must be a linear combination of the ${\displaystyle n-r}$ special solutions given by the columns of ${\displaystyle \mathbf{X}}$. And we have already seen that the columns of ${\displaystyle \mathbf{X}}$ are linearly independent. Hence, the columns of ${\displaystyle \mathbf{X}}$ constitute a basis for the null space of ${\displaystyle \mathbf{A}}$. Therefore, the nullity of ${\displaystyle \mathbf{A}}$ is ${\displaystyle n-r}$. Since ${\displaystyle r}$ equals rank of ${\displaystyle \mathbf{A}}$, it follows that ${\displaystyle \mathrm{Rank}(\mathbf{A})+\mathrm{Nullity}(\mathbf{A})=n}$. This concludes our proof.

A third fundamental subspace

When ${\displaystyle T:V\to W}$ is a linear transformation between two finite-dimensional subspaces, with ${\displaystyle n=\dim (V)}$ and ${\displaystyle m=\dim (W)}$ (so can be represented by an ${\displaystyle m\times n}$ matrix ${\displaystyle M}$), the rank–nullity theorem asserts that if ${\displaystyle T}$ has rank ${\displaystyle r}$, then ${\displaystyle n-r}$ is the dimension of the null space of ${\displaystyle M}$, which represents the kernel of ${\displaystyle T}$. In some texts, a third fundamental subspace associated to ${\displaystyle T}$ is considered alongside its image and kernel: the cokernel of ${\displaystyle T}$ is the quotient space ${\displaystyle W/\mathrm{Im}(T)}$, and its dimension is ${\displaystyle m-r}$. This dimension formula (which might also be rendered ${\displaystyle \dim \mathrm{Im}(T)+\dim \mathrm{Coker}(T)=\dim (W)}$) together with the rank–nullity theorem is sometimes called the fundamental theorem of linear algebra.

Reformulations and generalizations

This theorem is a statement of the first isomorphism theorem of algebra for the case of vector spaces; it generalizes to the splitting lemma.

In more modern language, the theorem can also be phrased as saying that each short exact sequence of vector spaces splits. Explicitly, given that $${\displaystyle 0\to U\to V\stackrel{T}{\to }R\to 0}$$ is a short exact sequence of vector spaces, then ${\displaystyle U\oplus R\cong V}$, hence $${\displaystyle \dim (U)+\dim (R)=\dim (V).}$$ Here ${\displaystyle R}$ plays the role of ${\displaystyle \mathrm{Im}T}$ and ${\displaystyle U}$ is ${\displaystyle \mathrm{Ker}T}$, i.e. $${\displaystyle 0\to \ker T\hookrightarrow V\stackrel{T}{\to }\mathrm{im}T\to 0}$$

In the finite-dimensional case, this formulation is susceptible to a generalization: if $${\displaystyle 0\to V_1\to V_2\to \cdots \to V_r\to 0}$$ is an exact sequence of finite-dimensional vector spaces, then $${\displaystyle \sum _{i=1}^r(-1{)}^i\dim (V_i)=0.}$$ The rank–nullity theorem for finite-dimensional vector spaces may also be formulated in terms of the index of a linear map. The index of a linear map ${\displaystyle T\in \mathrm{Hom}(V,W)}$, where ${\displaystyle V}$ and ${\displaystyle W}$ are finite-dimensional, is defined by $${\displaystyle \mathrm{index}T=\dim \mathrm{Ker}(T)-\dim \mathrm{Coker}T.}$$

Intuitively, ${\displaystyle \dim \mathrm{Ker}T}$ is the number of independent solutions ${\displaystyle v}$ of the equation ${\displaystyle Tv=0}$, and ${\displaystyle \dim \mathrm{Coker}T}$ is the number of independent restrictions that have to be put on ${\displaystyle w}$ to make ${\displaystyle Tv=w}$ solvable. The rank–nullity theorem for finite-dimensional vector spaces is equivalent to the statement $${\displaystyle \mathrm{index}T=\dim V-\dim W.}$$

We see that we can easily read off the index of the linear map ${\displaystyle T}$ from the involved spaces, without any need to analyze ${\displaystyle T}$ in detail. This effect also occurs in a much deeper result: the Atiyah–Singer index theorem states that the index of certain differential operators can be read off the geometry of the involved spaces.