Suppose $U$ is a finite-dimensional vector space, that $S ∈ L(V, W )$, and that $T ∈ L(U, V )$. I'm trying to prove that $\operatorname{dim} \operatorname{null}(ST ) = \operatorname{dim} \operatorname{null}(T ) + \operatorname{dim} (\operatorname{range}(T ) ∩ \operatorname{null}(S))$.

First I realized that for a vector $x$ to be in $\operatorname{null}(ST)$, it can

(1) be in $\operatorname{null} T$, in which case $Sx=0$ (since $S$ is a linear map) and thus $x \in \operatorname{null}(ST)$

(2) get mapped to an element of $\operatorname{range}T$ that happens to be in the null space of $S$.

Hence $\operatorname{null}T$ and $\operatorname{range}(T ) ∩ \operatorname{null}(S)$ combine to make $\operatorname{null}(ST)$, i.e. $$\operatorname{null}T + \operatorname{range}(T ) ∩ \operatorname{null}(S) = \operatorname{null}(ST).$$

Now, taking $\text{dim}$s, I get

$$\operatorname{dim}\operatorname{null}T + \operatorname{dim}(\operatorname{range}(T ) ∩ \operatorname{null}(S)) - \operatorname{dim}(\text{null} \, T \cap \text{range} \, T \cap \text{null} \, S) = \operatorname{dim}\operatorname{null}(ST).$$

It seems like for the theorem to hold, we must have that $\text{dim}(\text{null} \, T \cap \text{range} \, T \cap \text{null} \, S) = 0$, i.e. that $\text{null} \, T \cap \text{range} \, T \cap \text{null} \, S = \{ 0 \}$ for arbitrary linear maps $S, T$. Could someone please point me in the right direction as to how go about conceptualizing and proving this?

  • $\begingroup$ I think by null you mean ker where null = dim ker $\endgroup$ Oct 7, 2019 at 7:08
  • $\begingroup$ What does it mean for you the intersection, say, of $null\,T$ and $range\, T$ if they belong to completely different spaces, $U$ and $V$? $\endgroup$
    – GReyes
    Oct 7, 2019 at 7:08
  • $\begingroup$ @GReyes That's what I was thinking about; if we can guarantee that $U$ and $V$ are disjoint (with the exception of the $0$ element), then we're done. But what if there is overlap, or if $U=V=W$? $\endgroup$
    – actinidia
    Oct 7, 2019 at 7:17
  • $\begingroup$ @Tiwa I think what GReyes is pointing out is that range(T) belongs to a different space, so you cannot eg combine it (or a subspace of it) with the kernel of T, at least not with vector addition of elements which is what I presume '+' means in this context $\endgroup$ Oct 7, 2019 at 7:33
  • 2
    $\begingroup$ The question is not phrased with $U=V=W$, so what gives you the ability to write $ker T + range T\cap ker S$, in the general case? $\endgroup$ Oct 7, 2019 at 7:47

1 Answer 1


Say $u_1,\dots,u_n \in U$ is a basis of the kernel of $T$ and $v_1,\dots,v_m \in V$ is a basis of the space formed by intersecting the range of $T$ with the kernel of $S$. As the $v_i$s are in the range of $T$, they have preimages in $U$. Pick one preimage for each $v_i$; you can group them with the earlier $u_i$s to get a collection of linearly independent (why?) vectors $u_1,\dots,u_{n+m}\in U$. This implies that the nullity of $ST$ is at least $n+m$.

Conversely let $u\in U$ be an arbitrary vector in the kernel of $ST$. As you noted, it is either in the kernel of $T$ or is mapped by $T$ into the kernel of $S$. Thus it is in the span of $u_1,\dots,u_{n+m}$ (does it matter that we chose specific preimages?), and so the nullity is at most $n+m$.


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