Question: Assume $U,V,W$ are finite dimensional vector spaces and $T:U\rightarrow V, S:V\rightarrow W$ are linear maps. Prove or disprove that $$ \dim (\operatorname{Null}(ST)) = \dim (\operatorname{Null}(S)) + \dim (\operatorname{Null} (T))$$

Here is my approach. I know that $$ \dim (\operatorname{Null}(ST)) \le \dim (\operatorname{Null}(S)) + \dim (\operatorname{Null}(T))$$ Thus, I want to show the counter example that $$ \dim (\operatorname{Null}(ST)) \ne \dim (\operatorname{Null}(S)) + \dim (\operatorname{Null}(T))$$

(But I don't really have a good counter example) However, here are scratch ideas:

$T:U\rightarrow V, S:V\rightarrow W$, then $ST:U\rightarrow W $

Suppose that $T:(x_1,x_2) \rightarrow (x_1,0)$, then $\operatorname{Null}(T) = {(0,x_2): x_2 \in \mathbb{R}^2}.$

Then $\dim(\operatorname{Null}(T)) = 1$

Suppose that $S:(x_1,x_2) \rightarrow (x_1,0)$, then $\operatorname{Null}(S) = {(0,x_2): x_2 \in \mathbb{R}^2}.$

Then $\dim(\operatorname{Null}(S)) = 1$

Suppose that $ST:(x_1,x_2) \rightarrow (0,x_2)$, then $\operatorname{Null}(ST) = {(x_1,0): x_1 \in \mathbb{R}^2}.$

Then $\dim(\operatorname{Null}(ST)) = 1$

Then, $$ \dim (\operatorname{Null}(ST)) = \dim (\operatorname{Null}(S)) + \dim (\operatorname{Null}(T))$$ but $1\ne 1+1=2$. Hence, $$ \dim (\operatorname{Null}(ST)) \ne \dim (\operatorname{Null}(S)) + \dim (\operatorname{Null}(T))$$

Please help me improve my answer! Thanks in advance!

  • 1
    $\begingroup$ Should it be $T:U\rightarrow V$? $\endgroup$
    – user84413
    Commented Oct 11, 2014 at 23:17
  • $\begingroup$ Right, my bad! thanks for that. $\endgroup$
    – needhelp
    Commented Oct 11, 2014 at 23:18

1 Answer 1


You have the right idea but $ST$ is defined as the map obtained by multiplying the matrices of $S$ and $T$ (in that order), or equivalently by taking the composition of the linear maps. So in your example you take $U=V=W=\mathbb{R}^2$, and then define $S=T$ to be the linear map obtained by projection onto the first coordinate, i.e. $$S:\mathbb{R}^2\to \mathbb{R}^2, S(x_1,x_2)=(x_1,0) $$ As you have argued, the Kernel of $S$ is $1$-dimensional, i.e. $dim(Null(S))=1=dim(Null(T))$, to be specific, the null space of $S(=T)$ is the span of $(0,1)=e_2$.

Now let's look at $ST=S^2$. We have: $S^2(x_1,x_2)=S(x_1,0)=(x_1,0)$ for all $(x_1,x_2)\in \mathbb{R}^2$. So $S^2=S$, and thus $dim(Null(S^2))=dim(Null(S))=1$. Thus:

$$dim(Null(ST))=dim(Null(S))=1<dim(Null(S))+\dim(Null(T))=2. $$

This gives a counter example to the claim that if $U,V,W$ are vector spaces and $T:U\to V, S:V\to W$ linear maps then: $$dim(Null(ST))=dim(Null(S))+\dim(Null(T)). $$

  • $\begingroup$ So do you claim that $S=T$ since they have same dimension? Thanks for your answer though. $\endgroup$
    – needhelp
    Commented Oct 11, 2014 at 23:26
  • $\begingroup$ No, I do not. I defined $S$ and $T$ to be the same, both the projection maps onto the first coordinate. In other words I defined $S$ and then said "Let $T$ be this same map". $\endgroup$
    – CWsl2
    Commented Oct 11, 2014 at 23:31
  • $\begingroup$ Thanks! That clears my doubt. I was stuck when it comes to $ST$. Thanks again! $\endgroup$
    – needhelp
    Commented Oct 11, 2014 at 23:34
  • $\begingroup$ No problem, that seemed to be the only difficulty you were having, the rest was great. $\endgroup$
    – CWsl2
    Commented Oct 11, 2014 at 23:35

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