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



Filed under Algebra, Linear Algebra, Miscellaneous

2 responses to “The Rank Nullity Theorem

  1. Shiv

    you reduce the original problem to “our theorem basically translates to \text{rank} (A)+\text{nullity}(A)=n”. Is this step needs a proof?

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