# Monthly Archives: February 2014

## The Rank Nullity Theorem

One of the important results in linear algebra is the rank nullity theorem. Here I am going to present a proof of it which is slightly less known. The reason I like this proof is because it ties together many concepts and results quite nicely, and also because I independently thought of it.

The theorem (as is well known) says that if $V,W$ are vector spaces with $n=\dim V<\infty$ and $T:V\to W$ a linear map then $\text{rank} (T)+\text{nullity}(T)=n$.

In this proof I will further assume that $W$ is finite dimensional with dimension $m$. A more general proof can be found on wikipedia.

We start by fixing two bases of $V$ and $W$ and obtain a $m\times n$ matrix $A=\begin{pmatrix} r_1\\ r_2\\ \vdots\\ r_m \end{pmatrix}$ of $T$ relative to these bases. (Each $r_i$ is a $1\times n$ row matrix). Then our theorem basically translates to $\text{rank} (A)+\text{nullity}(A)=n$. We let $\text{Row Space} (A)=R,\text{Null Space} (A)=N$ and claim that $R^\perp=N$.

Clearly if $x\in N$ then $Ax=0$ and so $\begin{pmatrix} r_1\\ r_2\\ \vdots\\ r_m \end{pmatrix}x=\begin{pmatrix} r_1x\\ r_2x\\ \vdots\\ r_mx \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ \vdots\\ 0 \end{pmatrix}$ so that each $r_i$ is orthogonal to $x$. Hence $x\in R^\perp$. Conversely if $x\in R^\perp$ then $x^Tr_i^T=0$ so that $x^TA^T=0$, i.e. $Ax=0$ following which $x\in N$.

Now it only remains to invoke the result $\dim U+\dim U^\perp=\dim V$ for any subspace $U$ of an inner product space $V$ to conclude that $\dim R+\dim N=\dim \mathbb{R}^n$. In other words $\text{rank} (A)+\text{nullity}(A)=n$.$\Box$