I'm looking for nice applications of Rank–nullity theorem to show to my students for a 45 minutes class. I'm going to begin this class showing the demonstration of the theorem and I don't know what to do in rest of the time, any suggestions?

Let $V$ and $W$ be finite-dimensional vector spaces and let $T:V\to W$ be a linear transformation from $V$ into $W$. Then:

$$\dim V=\dim \text{Im}(T)+\dim\ker(T)$$

Remark: The only nice applications I remember is the fact the row and column ranks of a matrix are equal and a linear transformation $T:V\to W$ with $\dim V=\dim W$ is injective iff it's surjective.

  • $\begingroup$ You could perhaps show an example calculation or two using actual numbers and matrices. In fact, if you pick the numbers nicely, you could potentially give the argument of the proof only in a special case (say, $V=W=\mathbb{R}^{3}$ and $T$ has some judiciously chosen matrix in the standard basis so that all the gory details appear) and leave the generalization to the students as an exercise. $\endgroup$
    – Will R
    Nov 27, 2017 at 12:01
  • $\begingroup$ @WillR I must prove the theorem in its full generalization. The problem is I don't know what to do afterwards. $\endgroup$
    – user42912
    Nov 27, 2017 at 12:07
  • $\begingroup$ Perhaps try this en.wikipedia.org/wiki/Fundamental_theorem_of_linear_algebra ? $\endgroup$
    – lhf
    Nov 27, 2017 at 13:37

1 Answer 1


A bit too long for a comment ...

Your memory is faulty. It’s easy to prove that $T$ is injective iff $\ker(T)=\{0\}$.

Now the theorem reveals that if $\dim(V)>\dim(W)$ the map $T$ can’t be injective since the dimension of the image of $T$ is at most the dimension of $W$.

OTOH if $\dim(V)<\dim(W)$ then $T$ can’t be surjective for the same reason.

Now furthermore if the dimensions are equal your statement is right.

  • $\begingroup$ Thank you. You're right, the dimensions must be equal. Do you think I can use this theorem in my calculus classes with functions $f:\mathbb R\to \mathbb R$, since $\mathbb R$ is itself a vector space? $\endgroup$
    – user42912
    Nov 27, 2017 at 13:33
  • 1
    $\begingroup$ The theorem demands $f$ to be linear, in your case we must have $f(x)=ax$, which is pretty boring. $\endgroup$ Nov 27, 2017 at 14:07
  • $\begingroup$ It's true, you're right! $\endgroup$
    – user42912
    Nov 27, 2017 at 14:11

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