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This was a question in one of my online assignments that I got wrong but have no idea why. Unfortunately I do not have the option to re-enter other attempts to see if they are correct, hence stack exchange.

Question:

Let p $=2x^2+6x+7$ .Find the coordinate vector of p relative to the following basis for $P_2$

$p_1 = 1$

$p_2 = 1+x$

$p_3 = 1+x+x^2$

My attempt:

Plugging in for simplicity $x=1$ for p , $p_1$, $p_2$, and $p_3$ results in the following vectors.

p = $(2,6,7)$

$p_1= $ $(1,0,0)$

$p_2= $ $(1,1,0)$

$p_3= $ $(1,1,1)$

Transposed of $p_n$ have:

$\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}$ *$\begin{bmatrix}7\\6\\2\end{bmatrix} = $$\begin{bmatrix}15\\8\\2\end{bmatrix}$

I reversed p because it was reversed relative to $p_n$

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This is incorrect because you have simply taken the image of your polynomial under the matrix whose columns are the basis vectors.

Instead, you want to augment that same matrix with the vector you were given: $\left (\begin{array}{rrr|r}1&1&1&7\\0&1&1&6\\0&0&1&2\end{array}\right)$.

Now you can read off the solution. For starters, $z=2$.

You are looking for the linear combination of the basis vectors which will give you $(7,6,2)^t$.

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  • $\begingroup$ Ah ah ah, I see now. Thank you! $\endgroup$ – Mark Nov 28 '18 at 1:59
  • $\begingroup$ But I was correct in changing (2,6,7) to (7,6,2) right? $\endgroup$ – Mark Nov 28 '18 at 2:00
  • $\begingroup$ Yes. As the standard basis is$\{1,x,x^2\}$. $\endgroup$ – user403337 Nov 28 '18 at 2:08
  • $\begingroup$ In other words, you need to multiply the vector by the inverse of this matrix. $\endgroup$ – amd Nov 28 '18 at 3:01
  • $\begingroup$ Yes that would do it. $\endgroup$ – user403337 Nov 28 '18 at 3:16

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