Therank–nullity theorem is a theorem inlinear algebra, which asserts:

• the number of columns of a matrixM is the sum of therank ofM and thenullity ofM; and
• thedimension of thedomain of alinear transformationf is the sum of therank off (the dimension of theimage off) and the nullity off (the dimension of thekernel off).^[1][2][3][4]

It follows that for linear transformations ofvector spaces of equal finite dimension, eitherinjectivity orsurjectivity impliesbijectivity.

Let${\displaystyle T:V\to W}$ be a linear transformation between two vector spaces where${\displaystyle T}$'s domain${\displaystyle V}$ is finite dimensional. Then$${\displaystyle \mathrm{rank}(T)+\mathrm{nullity}(T)=\dim V,}$$where$\mathrm{rank}(T)$ is therank of${\displaystyle T}$ (thedimension of itsimage) and${\displaystyle \mathrm{nullity}(T)}$ is thenullity of${\displaystyle T}$ (the dimension of itskernel). In other words,$${\displaystyle \dim (\mathrm{Im}T)+\dim (\mathrm{Ker}T)=\dim (\mathrm{Domain}(T)).}$$This theorem can be refined via thesplitting lemma to be a statement about anisomorphism of spaces, not just dimensions. Explicitly, since${\displaystyle T}$ induces an isomorphism from${\displaystyle V/\mathrm{Ker}(T)}$ to${\displaystyle \mathrm{Im}(T),}$ the existence of a basis for${\displaystyle V}$ that extends any given basis of${\displaystyle \mathrm{Ker}(T)}$ implies, via the splitting lemma, that${\displaystyle \mathrm{Im}(T)\oplus \mathrm{Ker}(T)\cong V.}$ Taking dimensions, the rank–nullity theorem follows.

Linear maps can be represented withmatrices. More precisely, an${\displaystyle m\times n}$ matrixM represents a linear map${\displaystyle f:F^n\to F^m,}$ where${\displaystyle F}$ is the underlyingfield.^[5] So, the dimension of the domain of${\displaystyle f}$ isn, the number of columns ofM, and the rank–nullity theorem for an${\displaystyle m\times n}$ matrixM is$${\displaystyle \mathrm{rank}(M)+\mathrm{nullity}(M)=n.}$$

Here we provide two proofs. The first^[2] operates in the general case, using linear maps. The second proof^[6] looks at the homogeneous system${\displaystyle \mathbf{A}\mathbf{x}=\mathbf{0},}$ where${\displaystyle \mathbf{A}}$ is a${\displaystyle m\times n}$ withrank${\displaystyle r,}$ 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 thecodomain. 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.

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 asubspace, 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.

We may now, by theSteinitz exchange lemma, extend${\displaystyle \mathcal{K}}$ with${\displaystyle n-k}$ linearly independent vectors${\displaystyle w_1,\dots ,w_{n-k}}$ to form a full basis of${\displaystyle V}$.

Let$${\displaystyle \mathcal{S}:=\{w_1,\dots ,w_{n-k}\}\subset V\setminus \mathrm{Ker}(T)}$$such that$${\displaystyle \mathcal{B}:=\mathcal{K}\cup \mathcal{S}=\{v_1,\dots ,v_k,w_1,\dots ,w_{n-k}\}\subset V}$$is a basis for${\displaystyle V}$.From this, we know that$${\displaystyle \mathrm{Im}T=\mathrm{Span}T(\mathcal{B})=\mathrm{Span}\{T(v_1),\dots ,T(v_k),T(w_1),\dots ,T(w_{n-k})\}}$$

${\displaystyle =\mathrm{Span}\{T(w_1),\dots ,T(w_{n-k})\}=\mathrm{Span}T(\mathcal{S}).}$

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.

Suppose${\displaystyle T(\mathcal{S})}$ is not linearly independent, and let$${\displaystyle \sum _{j=1}^{n-k}{\alpha }_{j}T(w_j)=0_W}$$for some${\displaystyle {\alpha }_j\in F}$.

Thus, owing to the linearity of${\displaystyle T}$, it follows that$${\displaystyle T\left(\sum _{j=1}^{n-k}{\alpha }_j{w}_j\right)=0_W⟹\left(\sum _{j=1}^{n-k}{\alpha }_j{w}_j\right)\in \mathrm{Ker}T=\mathrm{Span}\mathcal{K}\subset V.}$$This is a contradiction to${\displaystyle \mathcal{B}}$ being a basis, unless all${\displaystyle {\alpha }_j}$ 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.

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:

To do this, we will produce an${\displaystyle n\times (n-r)}$ matrix${\displaystyle \mathbf{X}}$ whose columns form abasis 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 writewhere

• ${\displaystyle {\mathbf{A}}_1}$ 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${\displaystyle \mathbf{B}}$ (seerank factorization) and, hence,

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

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${\displaystyle \mathbf{X}}$ arelinearly 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 thatany solution of${\displaystyle \mathbf{A}\mathbf{x}={\mathbf{0}}_{F^m}}$ must be alinear 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,$${\displaystyle ⟹\mathbf{u}=\left(\begin{array}{c}{\mathbf{u}}_1\\ {\mathbf{u}}_2\end{array}\right)=\left(\begin{array}{c}-\mathbf{B}\\ {\mathbf{I}}_{n-r}\end{array}\right){\mathbf{u}}_2=\mathbf{X}{\mathbf{u}}_2.}$$

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 thenull space of${\displaystyle \mathbf{A}}$. Therefore, thenullity 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.

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 thenull space of${\displaystyle M}$, which represents thekernel of${\displaystyle T}$. In some texts, a third fundamental subspace associated to${\displaystyle T}$ is considered alongside its image and kernel: thecokernel of${\displaystyle T}$ is thequotient 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 thefundamental theorem of linear algebra.^[7][8]

This theorem is a statement of thefirst isomorphism theorem of algebra for the case of vector spaces; it generalizes to thesplitting 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 ashort 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 anexact sequence of finite-dimensional vector spaces, then^[9]$${\displaystyle \sum _{i=1}^r(-1{)}^i\dim (V_i)=0.}$$The rank–nullity theorem for finite-dimensional vector spaces may also be formulated in terms of theindex 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: theAtiyah–Singer index theorem states that the index of certain differential operators can be read off the geometry of the involved spaces.

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