The Rank-Nullity Theorem

Rank-Nullity Theorem:

Let T:V\to W be a linear transformation between vector spaces V and W. Then

\dim V=\dim\mbox{Im }T+\dim\ker T.


Take a basis \{u_1,\dots, u_m\} for \ker T, and extend it to a basis \{u_1,\dots,u_m,v_1,\dots,v_n\} for V. Now it suffices to show that \{Tv_1,\dots, Tv_n\} form a basis for \mbox{Im }T (since then we would have \dim V=m+n,  \dim\ker T=m, and \dim\mbox{Im}T=n ).

First we will check that \{Tv_1,\dots, Tv_n\} are linearly independent. Suppose c_1 Tv_1+\cdots+c_n Tv_n=0. Then, T(c_1v_1+\cdots+c_nv_n)=0, so c_1v_1+\cdots+c_nv_n\in\ker T. This means it can be represented in terms of the \{u_1,\dots, u_m\}. Write

c_1v_1+\cdots+c_nv_n=b_1u_1+\cdots +b_mu_m

c_1v_1+\cdots+c_nv_n-b_1u_1-\cdots -b_mu_m=0.

But this is a linear combination of basis vectors of V which sum to zero, so it must be that each of the c_i=0 and b_i=0 (we don’t really care about the b_i, but it is true anyway). The interesting part is that c_i=0 for every i, exactly the condition needed to say that \{Tv_1,\dots, Tv_n\} is linearly independent.

Now we need only show that \mbox{span}(\{Tv_1,\dots, Tv_n\})=\mbox{Im }T. Let w\in\mbox{Im }T. There is some vector v which has Tv=w. Let’s write v=\lambda_1u_1+\cdots+\lambda_mu_m+\mu_1v_1+\cdots+\mu_nv_n. When we apply T, we get



But since, for each i, Tu_i=0,


Therefore, \{Tv_1,\dots,Tv_n\} form a basis for \mbox{Im }T, proving the theorem.



Notice the distinct lack of assumptions. This is true for any linear transformation over any vector spaces over any field. That’s cool.

Another thing that is cool is that this theorem generalizes wildly. This is the vector space version of what is known as Noether’s first isomorphism theorem. It’s true in the context of many other algebraic structures. We’ll probably see this result soon, at least in the context of groups.

This is the first real theorem about dimension. I wanted you to see this so that you at least understand the flavor of proof. We’ll have another similar proof when we get to Galois Theory.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s