Find the coordinate vector of $1+x+x^{2}+x^{3}$ relative to the following basis of $P_3$:
$$B=\{1-x,1+x,x^{2}-x^{3},x^{2}+x^{3}\}$$
I have never to deal with something more than the standard basis and am really just confused on the approach to this problem. Can't understand my professor and my textbook is not the biggest help either. Any help is appreciated.
$\endgroup$ 22 Answers
$\begingroup$HINT
You have to find $$a,b,c,d\in \mathbb{R}$$ such that:
$$a(1-x)+b(1+x)+c(x^2-x^3)+d(x^2+x^3)=1+x+x^2+x^3$$
$\endgroup$ 3 $\begingroup$Say you have
b1 = 1-x
b2 = 1+x
b3 = x2-x3
b4 = x2+x3
The you need c1,c2,c3,c4 such that c1b1+c2b2+c3b3+c4b4 = 1+x+x2+x3
Note that you can rewrite B as
b1 = 1-1x+0x2-0x3
b2 = 1+1x+0x2-0x3
b3 = 0+0x+1x2-1x3
b4 = 0+0+1x21+x3
You also have the target vector
v=1+x+x2+x3
Each of these can be written in column vector form by taking the coefficients, e.g.
b1 = $\begin{bmatrix}1\\-1\\0\\0\end{bmatrix}$
If we write all of the b vectors as column vectors and arrange them left to right, we can get a matrix:
$\begin{bmatrix}1&1&0&0\\-1&1&0&0\\0&0&1&1\\0&0&-1&1\end{bmatrix}$
We can then use that matrix to represent B. I'll use bold to indicate that it's a matrix, i.e. B (bolded) represents the basis B (unbolded).
So now if we have a set of coefficients
c1b1+c2b2+c3b3+c4b4
Then we can represent all of the c's with another vecotr:
c = $\begin{bmatrix}c_1\\c_2\\c_3\\c_4\end{bmatrix}$
Now the problem can be stated succinctly as:
Find c such that Bc = v
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