# $T:U\rightarrow V$ and $S:V \rightarrow W$, then $\dim \ker ST \leq \dim \ker S + \dim \ker T$

I don't understand this proof, or rather I'm not convinced by some of the statements in it. However, I don't find the statement itself unreasonable. It is from lecture notes:

Assume $$U$$ and $$V$$ finite-dimensional and $$T:U\rightarrow V$$ and $$S:V\rightarrow W$$ linear operators. (They did not specify if W is finite dimensional)

Proof: Let $$U_0 = \ker ST,\; V_0=\ker S$$. $$U_0 \text{ and } V_0$$ are subspaces of $$U$$ and $$V$$. Furthermore, because $$ST=0$$ in $$U_0$$ it must be that $$T(U_0)\subset V_0$$. So we can consider T and S as transformations defined in $$U_0$$ and $$V_0$$. Now, because $$\operatorname{Ran}T$$ is a subspace of $$V_0$$ we have $$\dim \operatorname{Ran}T \leq \dim V_0 = \dim \ker S.$$ But according to the rank-nullity theorem, the following also holds $$\dim \ker T+\dim \operatorname{Ran}T = \dim U_0$$ This tells us that $$\dim \ker ST = \dim U_0 = \dim \ker T+\dim \operatorname{Ran}T \leq \dim \ker T+\dim \ker S$$

Some points that I'm uncertain about:

1. Why is $$T(U_0)\subset V_0$$?
2. What does "So we can consider T and S as transformations defined in $$U_0$$ and $$V_0$$." mean?
3. Why is the $$\operatorname{Ran}T$$ a subspace of $$V_0$$?
4. Does it matter whether $$W$$ is finite-dimensional or not?

I think if these are cleared up I might understand the proof and it might actually turn out to be convincing.

1. If $$x \in \ker(ST) = U_0$$ we have $$ST(x)=S(T(x))=0$$, consequently $$T(x)\in \ker(S)$$. In other words, $$T(U_0) \subset \ker(S) = V_0$$.

2. We can restrict the linear maps $$T$$ and $$S$$ to the subvectorspaces $$U_0$$ resp. $$V_0$$. Call these new maps $$T_0$$ and $$S_0$$, that is,

$$T_0 : U_0 \rightarrow V, x \mapsto T(x)$$ $$S_0 : V_0 \rightarrow W, x \mapsto S(x)$$

Note that we can only compose $$T_0$$ and $$S_0$$ if the image of $$T_0$$ is contained in $$V_0$$. This is true by 1., so it is actually possible to define the map $$S_0T_0: U_0 \rightarrow W$$.

1. What you call $$\operatorname{Ran}T$$ is now the image of $$T_0$$ defined above. Every element of it is by 1. an element of $$V_0$$. As the image of a linear map, it is therefore a subvectorspace of $$V_0$$.

2. No, it doesn't, because the image of $$S$$ will be finite dimensional and the rest of $$W$$ is irrelevant for the question.

• so the way I should read 1) $T(U_0)$ is like T(x), that is any x in $Ker(ST)$? Sep 16, 2021 at 10:20
• Any $x \in Ker(ST)$ must be mapped by $T$ to an element of $Ker(S)$, that is what 1. says. Sep 16, 2021 at 12:24

(1) Take some $$v \in T(U_0)$$, i.e. $$v = T(u)$$ for $$u \in U_0$$. Then we see that $$S(v)=S(T(u))=(ST)(u)=0$$, since $$u \in U_0$$, and hence $$v \in \ker(S)=V_0$$, so we have shown that $$T(U_0) \subseteq V_0$$.

(2) This means you can restrict the maps to those subspaces, i.e. instead of considering $$T : U \to V$$, you can consider $$T|_{U_0} : U_0 \to V_0$$, defined by $$T|_{U_0}(u) = T(u)$$. Essentially it is the same map as before, except that you only allow fewer values as inputs o the function.

(3) For arbitrary linear maps $$T : U \to V$$ we have that $$\text{Ran}(T)$$ is always a subspace of $$V$$ (showing this is a good exercise). I'm guessing in this case they meant $$\text{Ran}(T|_{U_0})$$. This is a subspace of $$V_0$$, as seen in (1).

(4) It does not matter, since all the spaces we're considering are subspaces of $$U$$ and $$V$$, and we're given that those are finite-dimensional.