Linear Algebra- Quadratic form and change of basis

[zeion]

Homework Statement

Suppose that for each v = (x1, x1, ... xn) in Rn, q(v) = X^TAX for the given matrix A. For the given basis B of Rn, find the expression for q(v) in terms of the coordiantes yi of v relative to B.

a) A = [tex]
\begin{bmatrix}3/2&{\sqrt{2}}&-1/2\\{\sqrt{2}}&1&-{\sqrt{2}}\\-1/2&{\sqrt{2}}&-5/2\end{bmatrix}


B = {(1,0,1), (3, {\sqrt{2}}, 1), (3{\sqrt{2}}, -4, {\sqrt{2}}) [/tex]

Homework Equations

The Attempt at a Solution

So I read the theorem about A wrt to B is P^TAP where P is the change of basis matrix from B to E3.. so I can just get A wrt to B by doing that right? I transpose P and the multiply it out like that? And it should give me a diagonal matrix?

But when I do that I get something really wrong? Like nothing is diagonal.
Is it possible for the result to have a zero column and q(v) wrt to B will not have all the terms?

 

[mighty2000]

Yes, it is possible for the result to have a zero column and for q(v) with respect to B to not have all the terms. This would occur if the basis B is not a basis for Rn, meaning that the vectors in B are not linearly independent. In this case, the matrix P would not be invertible and the change of basis formula would not work.

To find q(v) with respect to B, you can use the formula q(v) = YT PTP Y, where Y is the coordinate vector of v with respect to B. This can be derived from the change of basis formula PTAP. The resulting matrix will not necessarily be diagonal, as it depends on the specific values of the coordinates yi and the matrix A.