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Chris L T521 · #2 November 14th, 2008, 05:41 PM

Quote:

Originally Posted by Apprentice123

The set ${1+t,t-t^2,1+2t-t^2}$ is a base of P2?

Why?

I think you mean a basis...

First off, in order for this set to be a basis of $P_2$ , it must have span $P_2$

This means that $a_1 (1+t)+a_2(t-t^2)+a_3(1+2t-t^2)=b_1t^2+b_2t+b_3$ $\implies (-a_2-a_3)t^2+(a_1+a_2+3a_3)t+a_1+a_3=b_1t^2+b_2t+b_3$

We see that the corresponding augmented matrix is $\left[\begin{array}{ccc|c}0&-1&-1&b_1\\1&1&1&b_2\\1&0&1&b_3\end{array}\right]$

Getting it into RREF (I leave this for you to do), we see that $\left[\begin{array}{ccc|c}1&0&0&b_1+b_2\\0&1&0&b_2-b_3\\0&0&1&-b_1-b_2+b_3\end{array}\right]$

Since the system has solutions (primarily $a_1=b_1+b_2,~a_2=b_2-b_3,$ and $a_3=-b_1-b_2+b_3$ ), we see that $\left\{1+t,t-t^2,1+2t-t^2\right\}$ spans $P_2$ .

Now, we need to show that these elements are linearly independent.

$a_1 (1+t)+a_2(t-t^2)+a_3(1+2t-t^2)=0$

This can be rewritten as $(-a_2-a_3)t^2+(a_1+a_2+3a_3)t+a_1+a_3=0$

Thus, an augmented matrix can be constructed:

$\left[\begin{array}{ccc|c}0&-1&-1&0\\1&1&1&0\\1&0&1&0\end{array}\right]$

To test for linear independence, the easiest way is to evaluate $\det\left(\left[\begin{array}{ccc}0&-1&-1\\1&1&1\\1&0&1\end{array}\right]\right)$

This equals $\left|\begin{matrix}-1&-1\\1&1\end{matrix}\right|+\left|\begin{matrix}0&-1\\1&1\end{matrix}\right|=0+1=1\neq0$

Thus, they are linearly independent. Since this set spans $P_2$ and are linearly independent, they form a basis for $P_2$

Does this make sense?

--Chris

The following users thank Chris L T521 for this useful post:
Apprentice123
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