I tried to solve this problem, but, I can't. I need help.

Let $U$ and $V$ a finite-dimensional vector spaces, $T:U\rightarrow V$ and $S:V\rightarrow W$ linear transformations. Proof that $Nul(S\circ T)\leq Nul(S)+Nul(T)$

($Nul(T)$ means the dimension of the Kernel of $T$ and $Ran(T)$ means the dimension of the image of $T$)

My incomplete proof:

We know that $S\circ T: U\rightarrow W$, then

$Nul(S\circ T)+Ran(S\circ T)=Dim(U)=Nul(T)+Ran(T)$

Now, $Nul(T)+Ran(T)\leq Nul(T)+dim(V)$ because the image of $T$ is a subespace of $V$.


Hence, $Nul(S\circ T)+Ran(S\circ T)\leq Nul(T)+Nul(S)+Rang(T)$...

But I do not know how to conclude. I think that if $Ran(S\circ T)\geq Ran(T)$, the proof it's ok because in the last inequality we obtain $Nul(S\circ T)\leq Nul(T)+Nul(S)$, but, it is true?

  • $\begingroup$ Can you show that $\ker(T)\subset\ker (S\circ T)$? But perhaps $T$ doesn't hit all the kernel of $S$. Then what? $\endgroup$ Nov 28, 2016 at 3:05

1 Answer 1


We have to look at the dimension of $\ker S\circ T=T^{-1}(\ker S)$.

Consider the restriction $T|_{T^{-1}\ker S}:T^{-1}\ker S\to\ker S$. By the Dimension theorem, $$\dim(T^{-1}\ker S)=\dim(\ker T|_{T^{-1}\ker S})+\dim T(T^{-1}\ker S)\leq\dim\ker T+\dim\ker S$$

  • 1
    $\begingroup$ Then, it's true that $Ker(S\circ T)=T^{-1}(Ker(S))$? I think that I can prove this with the next: $x\in Ker(S\circ T)$ if and only if $(S\circ T)(x)=0$ if and only if $S(T(x))=0$ if and only if $T(x)\in Ker(S)$ if and only if $x\in T^{-1}(Ker(S))$. It is correct? $\endgroup$ Nov 28, 2016 at 3:24
  • $\begingroup$ @CarlosJiménez Yes, that is correct. It's very good that you can manage to find points which need further explanation, and find the explanation by yourself. $\endgroup$ Nov 28, 2016 at 3:29

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