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Exercise 3.B.22

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Proof. By the Fundamental Theorem of Linear Maps, we have

\begin{aligned} \dim null ST & = \dim U - \dim range ST \\ = \dim null T + \dim range T - \dim range ST \\ \leq \dim null T + \dim V - \dim range ST \\ = \dim null T + \dim V - \dim range S + \dim range S - \dim range ST \\ = \dim null T + \dim null S + \dim range S - \dim range ST \end{aligned}

Where the third line follows because $range T \subseteq V$ . We need only to prove that $\dim range S \geq \dim range ST$ . Suppose that $w \in range ST$ . Then there exists $u \in U$ such that $STu = w$ . Since $Tu \in V$ , it follows that for every vector $w \in range ST$ there is another vector in $V$ that $S$ maps to $w$ . Therefore $range ST \subseteq range S$ , which implies $\dim range S \geq \dim range ST$ . □

celiopassos

2017-10-06 00:00

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Define R: $null ST \to null S$
Where $\forall a \in null ST$ ,

Ra = Ta

By the fundamental theorem of linear maps(3.22),

\begin{matrix} \begin{aligned} dim null ST & = dim null R + dim range R \\ \leq dim null T + dim null S \end{aligned} \end{matrix}

(1)

Which is true since null R $\subset null T$ and $range R \subset null S □$

Arden_Shi

2025-04-04 00:58

Exercise 3.B.22

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