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.


2 Answers 2

  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.

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

I hope this was helpful!


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