How can I prove $\operatorname{dim null}(ST)$ is less than or equal to $\operatorname{dim null}S + \operatorname{dim null}T$?

Question

Problem from linear algebra done right.

We are given that $U$ and $V$ are finite-dimensional and T $\in$ $\mathcal{L} (U,V)$ and $S \in \mathcal{L} (V,W)$. We want to show

$$\operatorname{dim null}(ST) \leq \operatorname{dim null} T + \operatorname{dim null} S.$$

I tried using the fundamental theorem of linear maps but wasn't able to get much process, so instead I am using bases to solve the problem.

This is my work so far:

If $u \in \operatorname{null} T$, then $(ST)u = S(Tu) = S(0) = 0$ which implies that $u \in \operatorname{null} ST$. Hence, $\operatorname{null} T$ is a subspace of $\operatorname{null} ST$.

Let $u_1,\ldots,u_n$ be a basis for $\operatorname{null} T$. Extend the list to a basis $u_1,\ldots,u_n, v_1, \ldots, v_m$ of $\operatorname{null}(ST)$. So,

$$\operatorname{dim null} (ST) = n + m.$$

If I want to complete my proof, I only need to show that $m \leq \operatorname{dim null} S $.

Now, notice that if $u \in \operatorname{null} ST$, then $u = b_1 u_1 +\ldots+ b_n u_n + a_1 v_1 + \ldots + a_m u_m$ where the $a$'s and $b$'s are constants. We also know that $Tu \in \operatorname{null} S$. Then,

$Tu = a_1Tv_1 + \ldots + a_mTv_m$ where the other $u$'s got deleted since they were in $\operatorname{null} T$.

But here is where I am not sure where to go. I don't know if this method actually will lead me anywhere or if it's just easier to use the linear map theorem.

Answer 1

I don't know how to make your attempt to work, but here is a (sketch of a) solution.

Let $R:\mathrm{null}(ST)\to V$ be the restriction of $T$ to $\mathrm{null}(ST)$, that is $$ Ru = Tu, \quad u\in \mathrm{null}(ST). $$ First prove that the image of $R$ is contained in the null space of $S$, that is $$ \mathrm{img}(R)\subseteq \mathrm{null}(S), $$ where (I write this just for make the notation clear) $$ \mathrm{img}(R) = \{Rx \; | \; x\in \mathrm{null}(ST)\}. $$ Now, you can easily verify that $\mathrm{null}(R) = \mathrm{null}(T)$ (here you must use that $\mathrm{null}(T)\subseteq \mathrm{null}(ST)$). Then by the rank-nullity theorem we have that (since the domain of $R$ is $\mathrm{null}(ST)$, $$ \dim\mathrm{null}(ST) = \dim\mathrm{null}(R) + \dim\mathrm{img}(R)\leq \dim\mathrm{null}(T) + \dim\mathrm{null}(S). $$

Answer 2

You worked with basis. Without considering basis or elements: $$\begin{align}\operatorname{dim null}(ST)&=\dim (ST)^{-1}(0)\\ &=\dim T^{-1}(S^{-1}(0))\\ &=\dim T^{-1}(\operatorname{Im} T\cap \operatorname{null}S)\\ &=\dim T^{-1}(W)\\ &=\dim T^{-1}(0)+\dim W \end{align}$$ where $\dim( W)\leq\operatorname{dim null}S$ Therefore, $$\operatorname{dim null}(ST) \leq \operatorname{dim null} T + \operatorname{dim null} S.$$

Answer 3

Your approach will work.

Note that $\ker T \subset \ker ST$. Let $u_1,...,u_n$ be a basis for $\ker T$, extend with $v_1,...v_m$ to form a basis for $\ker ST$.

The relevant fact is that $Tv_1,...,Tv_m$ are linearly independent and $STv_k = 0$ for all $k$, so $\operatorname{sp} \{ T v_k \} \subset \ker S$ and hence $ m \le \dim \ker S$.

Then $\dim \ker ST = n+m \le \dim \ker T + \dim \ker S$.

Elaboration: Suppose $\sum_{i=1}^m \alpha_i T v_i = 0$, then $\sum_{i=1}^m \alpha_i v_i \in \ker T$ and so $\sum_{i=1}^m \alpha_i v_i = \sum_{j=1}^n \beta_j u_j$ from which we get $\alpha_i = 0, \beta_j = 0$. Hence the $T v_i$ are linearly independent.