Dim null ST <= dim null S + dim null T

[fishturtle1]

Homework Statement

Suppose $U$ and $V$ are finite dimensional vector spaces and $S \in \mathcal{L}(V, W)$ and $T \in \mathcal{L}(U, V)$. Prove that $\dim \operatorname{null} ST \le \dim \operatorname{null} S + \dim \operatorname{null} T$.

Relevant Equations

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For a linear map $T \in \mathcal{L}(U, V)$, $\text{null} T = \lbrace u \in U : Tu = 0 \rbrace$

Fundamental theorem of linear maps: $\dim U = \dim \text{null} T + \dim T(U)$
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Proof: Since $U, V$ are finite dimensional, we have that $\operatorname{null} S, \operatorname{null} T$ are finite dimensional. Let $v_1, \dots v_m$ be a basis of $\operatorname{null}S$ and $u_1, \dots, u_n$ be a basis of $\operatorname{null} T$. It is enough to show there are $m + n$ vectors that span $\operatorname{null} ST$. Let $x \in \operatorname{null}ST$. Then $x \in \operatorname{null}T$ or $Tx \in \operatorname{null}S$. So $u_1, \dots, u_n$ is in our list of $m+n$ vectors.

If $T$ maps onto $\operatorname{null}S$, then for each $v_i$ there is $y_i \in U$ such that $Ty_i = v_i$. So $Ty_1, \dots, Ty_m$ spans $\operatorname{null}S$. This implies $\operatorname{span} \lbrace y_1, \dots, y_m, u_1, \dots, u_n \rbrace\supseteq \operatorname{null}ST$. But I don't think I can say $T$ maps onto $\operatorname{null}S$. Can I have a hint, please?

 

[Infrared]

The approach of finding a spanning set of at most $m+n$ should work, and the $u_i$ should be included. Instead of taking these vectors $y_i$, how about taking a basis for $\text{Range}(T)\cap\text{Nul}(S)$?

Edit: To clarify, if $Tx_1,\ldots, Tx_n$ is such a basis, I mean to use the vectors $x_1,\ldots, x_n.$

 

[member 587159]

Hint: $\operatorname{null}(ST) = T^{-1}(\operatorname{null}(S))$ and rank-nullity theorem.

 

[fishturtle1]

Thank you, I think I got it using what you said:

Proof: Since $U, V$ are finite dimensional, we have $\operatorname{null} S$ and $\operatorname{null}T$ are finite dimensional. Let $u_1, \dots, u_n$ be a basis for $\operatorname{null} T$ and $Tx_1, \dots, Tx_k$ be a basis for $\operatorname{Range}(T) \cap \operatorname{null} S$. We note that $k \le \dim \operatorname{null} S$. We'll show the statement $\operatorname{span} \lbrace u_1, \dots, u_n, x_1, \dots, x_k \rbrace = \operatorname{null} ST$ is true.

$(\subseteq)$ Let $a_1u_1 + \dots a_nu_n + b_1x_1 + \dots + b_kx_k \in \operatorname{span} \lbrace u_1, \dots, u_n, x_1, \dots, x_k \rbrace$. Then

$\begin{align*} ST(a_1u_1 + \dots a_nu_n + b_1x_1 + \dots + b_kx_k) &= S(a_1T(u_1) + \dots + a_n T(u_n) + b_1T(x_1) + \dots + b_kT(x_k)) \\ & = S(0 + \dots + 0 + b_1T(x_1) + \dots + b_kT(x_k)) \\ &= S(b_1T(x_1) + \dots + b_kT(x_k)) \\ &= 0 \end{align*}$
This shows $\operatorname{span} \lbrace u_1, \dots, u_n, x_1, \dots, x_k \rbrace \subseteq \operatorname{null} ST$.

$(\supseteq)$ If $x \in \operatorname{null}ST$, then $x \in \operatorname{null}T$ or $Tx \in \operatorname{null} S$. I.e., $x \in \operatorname{span} \lbrace u_1, \dots, u_n \rbrace$. Otherwise, $Tx \in \operatorname{Range}(T) \cap \operatorname{null}S$ and since $x_1, \dots, x_k$ is linearly independent, we have $x \in \operatorname{span} \lbrace x_1, \dots, x_k \rbrace$. This shows $\operatorname{span} \lbrace u_1, \dots, u_n, x_1, \dots, x_k \rbrace \supseteq \operatorname{null} ST$.

We can conclude $\dim \operatorname{null}ST \le n + k \le n + m = \dim \operatorname{null}T + \dim \operatorname{null}S$. []

 

[fishturtle1]

I see using post #3:

Proof: First we show $\operatorname{null}(ST) = T^{-1}(\operatorname{null}(S))$.

$(\subseteq)$ Let $x \in \operatorname{null}(ST)$. Then $S(Tx) = 0$ which means $Tx \in \operatorname{null}(S)$ i.e. $x \in T^{-1}(\operatorname{null}(S))$.

$(\supseteq)$ Let $x \in T^{-1}(\operatorname{null}(S))$. Then $Tx \in \operatorname{null}(S)$ so $S(Tx) = 0$ i.e. $x \in \operatorname{null}(ST)$.

We can conclude $\operatorname{null}(ST) = T^{-1}(\operatorname{null}(S))$.

By rank nullity theorem: $\dim \operatorname{null}(ST) = \dim T^{-1}(\operatorname{null}(S)) = \dim \operatorname{null}(T) + \dim T(T^{-1}(\operatorname{null}(S))) \le \operatorname{null}(T) + \operatorname{null}(S)$. []

 

[WWGD]

Or you can consider that $KerT \subset KerST$. How much larger can it be?