Rank Nullity Theorem

The Rank Nullity Theorem is a result in linear algebra. To provide some background to understand the full claim, let $V$ and $W$ be vector spaces and let $\text{T}: V\to W$ be a linear transformation. We define $\text{nullity}(\text{T})=\dim(\text{N}(\text{T}))$ where $\text{N}(\text{T})=\{x\in V~|~\text{T}(x)=0\}$ . Additionally, we define $\text{rank}(\text{T})=\dim(\text{R}(\text{T}))$ where $\text{R}(\text{T})=\{\text{T}(x)~|~x\in V\}$ . In other words, rank is the dimension of the linear transformation's image, and nullity is the dimension of it's kernel (the word nullity comes from the term "null space" which is used to describe the physical subset of the domain of the linear transformation that maps to the zero vector). The theorem states the following:

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

Proof: Let $V$ and $W$ be vector spaces and let $\text{T}:V\to W$ be linear. We first claim that if $\beta=\{v_1,v_2,\ldots, v_n\}$ is a basis of $V$ , then

$\text{R}(\text{T})=\text{span}(\text{T}(\beta))=\text{span}(\{\text{T}(v_1),\text{T}(v_2),\ldots, \text{T}(v_n)\}).$

By definition we have $\text{T}(v_i)\in\text{R}(\text{T})$ . Because $\text{R}(\text{T})$ is a subspace (this is easy to verify), then the basis of $\text{R}(\text{T})$ is contained in $\text{R}(\text{T})$ or that it contains

$\text{span}(\{\text{T}(v_1),\text{T}(v_2),\ldots, \text{T}(v_n)\})=\text{span}(\text{T}(\beta))$ .

Now we must prove that $\text{span}(\text{T}(\beta))$ contains $\text{R}(\text{T})$ in order for $\text{R}(\text{T})=\text{span}(\text{T}(\beta))$ . To do this, let $w\in \text{R}(\text{T})$ . Then $w=\text{T}(v)$ for some $v\in V$ . Because $\beta$ is a basis for $V$ , we can use the linearity of $\text{T}$ to get

$v=\sum_{i=1}^na_iv_i\implies w=\text{T}(v)=\sum_{i=1}^na_i\text{T}(v_i)=\text{T}\left(\sum_{i=1}^na_iv_i\right)\in\text{span}(\text{T}(\beta))$

so $\text{R}(\text{T})$ is contained in its basis, which means that the two must be equal, done. With this corollary out of the way, we head to the main course. Obviously this is under the assumption that $\dim(T)$ is finite, so assume that $\dim(T)=n$ and $\text{nullity}(\text{T})=k$ such that $\{v_1,v_2,\ldots,v_k\}$ is a basis for $\text{N}(\text{T})$ . We can then extend this basis into $\beta=\{v_1,v_2,\ldots, v_n\}$ for $V$ . We seek to show that $\{\text{T}(v_{k+1}),\text{T}(v_{k+2}),\ldots, \text{T}(v_{n})\}$ is a basis for $\text{R}(\text{T})$ , because that would mean $\text{rank}(\text{T})=n-k$ which would prove what we want.

To show that $\{\text{T}(v_{k+1}),\text{T}(v_{k+2}),\ldots, \text{T}(v_{n})\}$ is a basis, we have two things to do. We need to show that $\{\text{T}(v_{k+1}),\text{T}(v_{k+2}),\ldots, \text{T}(v_{n})\}$ generates $\text{R}(\text{T})$ and that $\{\text{T}(v_{k+1}),\text{T}(v_{k+2}),\ldots, \text{T}(v_{n})\}$ is linearly independent, which would satisfy both criteria for a basis. The first part is easy, because since we have $\text{T}(v_i)=0$ for each $1\le i\le k$ we get

$\begin{align*} \text{R}(\text{T})&=\text{span}(\{\text{T}(v_1),\text{T}(v_2),\ldots, \text{T}(v_n)\})\\ &=\text{span}(\{\text{T}(v_{k+1}),\text{T}(v_{k+2})),\ldots, \text{T}(v_{n})\}\in \text{span}(\text{T}(\beta)). \end{align*}$

Now we have to show that $\{\text{T}(v_{k+1}),\text{T}(v_{k+2}),\ldots, \text{T}(v_{n})\}$ is linearly independent. Using the linearity of $\text{T}$ , we would have

$\sum_{i=k+1}^na_i\text{T}(v_i)=0\implies \text{T}\left(\sum_{i=k+1}^na_iv_i\right)=0$

so we can conclude that $\sum_{i=k+1}^na_iv_i\in\text{N}(\text{T})$ . Then by taking a set of scalars $\{b_1,b_2,\ldots, b_k\}\in \mathbb{F}$ we get

$\sum_{i=k+1}^na_iv_i=\sum_{i=1}^kb_iv_i\implies\sum_{i=k+1}^na_iv_i-\sum_{i=1}^kb_iv_i=0.$

Since $\beta$ is a basis of $V$ , we see that this implies that $\{\text{T}(v_{k+1}),\text{T}(v_{k+2}),\ldots, \text{T}(v_{n})\}$ is linearly independent, completing our work.

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Rank Nullity Theorem