Let [tex] W=\lbrace p(x) \in P_{2} : p(2)=0\rbrace [/tex](adsbygoogle = window.adsbygoogle || []).push({});

Find a basis for W.

Since a basis must be elements of the set W we know that p(2)=0.

So if [tex]p(x)=ax^2+bx+c[/tex], then [tex]p(x) = 4a+2b+c=0[/tex].

Let c=t, b=s and s,t are real scalars. Then p(x) can be written as

[tex]

t(-\frac{1}{4} x^2+1)+s(-\frac{1}{2}x^2+x)

[/tex]

so the basis would be [tex]\{ (-\frac{1}{4} x^2+1), (-\frac{1}{2}x^2+x)\rbrace [/tex].

Does anyone see a problem with this?

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# Homework Help: Finding a basis for a subspace of P_2

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