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A First Course in Linear Algebra, 2017a

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9.7. Isomorphisms 499<br />

9.7 Isomorphisms<br />

Outcomes<br />

A. Apply the concepts of one to one and onto to transformations of vector spaces.<br />

B. Determ<strong>in</strong>e if a l<strong>in</strong>ear transformation of vector spaces is an isomorphism.<br />

C. Determ<strong>in</strong>e if two vector spaces are isomorphic.<br />

9.7.1 One to One and Onto Transformations<br />

Recall the follow<strong>in</strong>g def<strong>in</strong>itions, given here <strong>in</strong> terms of vector spaces.<br />

Def<strong>in</strong>ition 9.62: One to One<br />

Let V ,W be vector spaces with⃗v 1 ,⃗v 2 vectors <strong>in</strong> V . Then a l<strong>in</strong>ear transformation T : V ↦→ W is called<br />

one to one if whenever⃗v 1 ≠⃗v 2 it follows that<br />

T (⃗v 1 ) ≠ T (⃗v 2 )<br />

Def<strong>in</strong>ition 9.63: Onto<br />

Let V ,W be vector spaces. Then a l<strong>in</strong>ear transformation T : V ↦→ W is called onto if for all ⃗w ∈ ⃗W<br />

there exists⃗v ∈ V such that T (⃗v)=⃗w.<br />

Recall that every l<strong>in</strong>ear transformation T has the property that T (⃗0) =⃗0. This will be necessary to<br />

prove the follow<strong>in</strong>g useful lemma.<br />

Lemma 9.64: One to One<br />

The assertion that a l<strong>in</strong>ear transformation T is one to one is equivalent to say<strong>in</strong>g that if T (⃗v)=⃗0,<br />

then⃗v = 0.<br />

Proof. Suppose first that T is one to one.<br />

( )<br />

T (⃗0)=T ⃗ 0 +⃗0 = T (⃗0)+T (⃗0)<br />

and so, add<strong>in</strong>g the additive <strong>in</strong>verse of T (⃗0) to both sides, one sees that T (⃗0)=⃗0. Therefore, if T (⃗v)=⃗0,<br />

itmustbethecasethat⃗v =⃗0 because it was just shown that T (⃗0)=⃗0.<br />

Now suppose that if T (⃗v) =⃗0, then ⃗v = 0. If T (⃗v) =T (⃗u), thenT (⃗v) − T (⃗u) =T (⃗v −⃗u) =⃗0 which<br />

shows that⃗v −⃗u = 0or<strong>in</strong>otherwords,⃗v =⃗u.<br />

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