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A First Course in Linear Algebra, 2017a

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288 L<strong>in</strong>ear Transformations<br />

Def<strong>in</strong>ition 5.29: One to One<br />

Suppose ⃗x 1 and ⃗x 2 are vectors <strong>in</strong> R n . A l<strong>in</strong>ear transformation T : R n ↦→ R m is called one to one<br />

(often written as 1 − 1) if whenever⃗x 1 ≠⃗x 2 it follows that :<br />

T (⃗x 1 ) ≠ T (⃗x 2 )<br />

Equivalently, if T (⃗x 1 )=T (⃗x 2 ), then ⃗x 1 =⃗x 2 . Thus, T is one to one if it never takes two different<br />

vectors to the same vector.<br />

The second important characterization is called onto.<br />

Def<strong>in</strong>ition 5.30: Onto<br />

Let T : R n ↦→ R m be a l<strong>in</strong>ear transformation. Then T is called onto if whenever⃗x 2 ∈ R m there exists<br />

⃗x 1 ∈ R n such that T (⃗x 1 )=⃗x 2 .<br />

We often call a l<strong>in</strong>ear transformation which is one-to-one an <strong>in</strong>jection. Similarly, a l<strong>in</strong>ear transformation<br />

which is onto is often called a surjection.<br />

The follow<strong>in</strong>g proposition is an important result.<br />

Proposition 5.31: One to One<br />

Let T : R n ↦→ R m be a l<strong>in</strong>ear transformation. Then T is one to one if and only if T (⃗x)=⃗0 implies<br />

⃗x =⃗0.<br />

Proof. We need to prove two th<strong>in</strong>gs here. <strong>First</strong>, we will prove that if T is one to one, then T (⃗x)=⃗0 implies<br />

that ⃗x =⃗0. Second, we will show that if T (⃗x)=⃗0 implies that ⃗x =⃗0, then it follows that T is one to one.<br />

Recall that a l<strong>in</strong>ear transformation has the property that T (⃗0)=⃗0.<br />

Suppose first that T is one to one and consider T (⃗0).<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. If T (⃗x)=⃗0 itmustbe<br />

thecasethat⃗x =⃗0 because it was just shown that T (⃗0)=⃗0 andT is assumed to be one to one.<br />

Now assume that if T (⃗x)=⃗0, then it follows that⃗x =⃗0. If T (⃗v)=T (⃗u), then<br />

T (⃗v) − T(⃗u)=T (⃗v −⃗u)=⃗0<br />

which shows that⃗v −⃗u = 0. In other words,⃗v =⃗u, andT is one to one.<br />

♠<br />

Note that this proposition says that if A = [ A 1 ··· A n<br />

]<br />

then A is one to one if and only if whenever<br />

it follows that each scalar c k = 0.<br />

0 =<br />

n<br />

∑ c k A k<br />

k=1

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