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A First Course in Linear Algebra, 2017a

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9.3. L<strong>in</strong>ear Independence 469<br />

9.3 L<strong>in</strong>ear Independence<br />

Outcomes<br />

A. Determ<strong>in</strong>e if a set is l<strong>in</strong>early <strong>in</strong>dependent.<br />

In this section, we will aga<strong>in</strong> explore concepts <strong>in</strong>troduced earlier <strong>in</strong> terms of R n and extend them to<br />

apply to abstract vector spaces.<br />

Def<strong>in</strong>ition 9.17: L<strong>in</strong>ear Independence<br />

Let V be a vector space. If {⃗v 1 ,···,⃗v n }⊆V, then it is l<strong>in</strong>early <strong>in</strong>dependent if<br />

where the a i are real numbers.<br />

n<br />

∑<br />

i=1<br />

a i ⃗v i =⃗0 implies a 1 = ···= a n = 0<br />

The set of vectors is called l<strong>in</strong>early dependent if it is not l<strong>in</strong>early <strong>in</strong>dependent.<br />

Example 9.18: L<strong>in</strong>ear Independence<br />

Let S ⊆ P 2 be a set of polynomials given by<br />

S = { x 2 + 2x − 1,2x 2 − x + 3 }<br />

Determ<strong>in</strong>e if S is l<strong>in</strong>early <strong>in</strong>dependent.<br />

Solution. To determ<strong>in</strong>e if this set S is l<strong>in</strong>early <strong>in</strong>dependent, we write<br />

a(x 2 + 2x − 1)+b(2x 2 − x + 3)=0x 2 + 0x + 0<br />

If it is l<strong>in</strong>early <strong>in</strong>dependent, then a = b = 0 will be the only solution. We proceed as follows.<br />

a(x 2 + 2x − 1)+b(2x 2 − x + 3) = 0x 2 + 0x + 0<br />

ax 2 + 2ax − a + 2bx 2 − bx + 3b = 0x 2 + 0x + 0<br />

(a + 2b)x 2 +(2a − b)x − a + 3b = 0x 2 + 0x + 0<br />

It follows that<br />

a + 2b = 0<br />

2a − b = 0<br />

−a + 3b = 0

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