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Find the change-of-basis-matrix of the following ordered basis of $Pol_2(\Bbb R)$

a) $B=(x^2,x,1), B'=(a_2x^2+a_1x+a_0,b_2x^2+b_1x+b_0, c_2x^2+c_1x+c_0)$
b) $B=(1,x,x^2), B'=(a_2x^2+a_1x+a_0,b_2x^2+b_1x+b_0, c_2x^2+c_1x+c_0)$
c) $B=(x^2-x,x^2+1,x-1), B'=(5x^2-2x-1,-x^2+4x+2, 2x^2-5x-3)$

So for a) i have \begin{bmatrix}a_2&b_2&c_2\\a_1&b_1&c_1\\a_0&b_0&c_0\end{bmatrix}

for b) \begin{bmatrix}a_2x^2&b_2x^2&c_2x^2\\a_1&b_1&c_1\\0&0&0\end{bmatrix}

I will leave the third for now, (I got rubbish numbers) probably wrong....

Jerry Holmes
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  • Your answer for b) is incorrect, simply because your matrix is supposed to consist of scalars, not vectors (i.e. polynomials) like $a_2x^2$. This is a very common misstep for linear algebra students, especially when they have not properly understood what a change of basis matrix actually is/does. I would suggest having a look back at your notes, as at some point, the change of basis matrix will have been introduced. Check out my and ennar's answers here for some insights. – Theo Bendit Nov 08 '21 at 07:19
  • @TheoBendit Hi thanks i read it over. I think i understood it better. However, i have a question, in my case, can i now take any $3x3$ matrix and set it into the equation to get my transformation matrix i need? or is there a way to directly calculate what the transformation matrix should be? – Jerry Holmes Nov 08 '21 at 08:44
  • There is a direct method for calculation, yes. The change of basis matrix, from $B'$ to $B$ (I'm assuming based on part a); order is definitely important) is the matrix for the identity transformation from $B'$ to $B$. That means, you should take, in order, the basis vectors from $B'$, apply the identity map to them (which does nothing; I include for completeness), then write each (transformed) vector as a linear combination of basis vectors from $B$. The coordinate column vectors (column vectors containing the coefficients of these linear combinations) are the columns of the desired matrix. – Theo Bendit Nov 08 '21 at 20:36
  • In part (c), for example, calculation reveals that$$5x^2-2x-1=4(x^2-x)+1(x^2+1)+2(x-1).$$This means that the first column of the matrix in (c) is$$\begin{bmatrix}4\1\2\end{bmatrix}.$$ – Theo Bendit Nov 08 '21 at 20:44
  • @TheoBendit Thanks! I worked out the matrix for b) and c) could you have a look and tell me if correct? for b) $\begin{bmatrix}a_0&b_0&c_0\a_1&b_1&c_1\a_2&b_2&c_2\end{bmatrix}$ for c) i have \begin{bmatrix}4&-3.5&-2\1&2.5&4\2&0.5&-3\end{bmatrix} – Jerry Holmes Nov 09 '21 at 07:00
  • The answer for b) is correct. The answer for c) is almost correct, though I believe your third column is not correct. If you double-check your calculations of the third column, I'm sure you'll find the error. – Theo Bendit Nov 09 '21 at 07:11
  • @TheoBendit oh yeah right sorry it should be (5,-3,0) Please post something as answer so i can give you the correct answer thing – Jerry Holmes Nov 09 '21 at 07:18
  • I could, but I'd prefer if you wrote the answer, given you did all the calculation, then I give you a +1 on both question and answer. You can even accept your own answer! – Theo Bendit Nov 09 '21 at 07:19

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