# Is it true that $\text{null} \, T \cap \text{range} \, T \cap \text{null} \, S = \{ 0 \}$ for arbitrary linear maps $S, T$?

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?

• I think by null you mean ker where null = dim ker Oct 7, 2019 at 7:08
• What does it mean for you the intersection, say, of $null\,T$ and $range\, T$ if they belong to completely different spaces, $U$ and $V$? Oct 7, 2019 at 7:08
• @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 $U=V=W$? Oct 7, 2019 at 7:17
• @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 Oct 7, 2019 at 7:33
• 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? Oct 7, 2019 at 7:47

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 $$n+m$$.