Let V be the vector space of real valued polynomials with degree smaller equal 3 and let $$(p,q) = \int_{-1}^{1} p(x)q(x) \,dx - 100p(0)q(0)$$
Is the bilinear form degenerate and find a basis for V in which the matrix of the bilinear form is diagonal and only has 1, -1, 0 on the diagonal.
Let $W = span \{x\}$, then the restriction of the bilinear form to W is positive definite and $ W^\perp = span\{1, x^2\}$. Now since $W \oplus W^\perp \neq V $ it would appear that the bilinear form is degenerate, this felt wrong to me, since I couldn't work out a vector apart from the 0 vector that would be in the radical, so I worked out the matrix for the bilinear form and got $$\begin{pmatrix} -98 & 0 & \frac{2}{3} & 0\\ 0 & \frac{2}{3} & 0 & \frac{2}{5}\\ \frac{2}{3} & 0 & \frac{2}{5} & 0\\ 0 & \frac{2}{5} & 0 & \frac{2}{7} \end{pmatrix}$$ which is non-degenerate, so maybe I made an error somewhere, if anyone can spot it, that would be great. I feel like I should be able to solve this without the matrix.
using your matrix, $W^\perp$ is the collection of all polynomials $$ a + bx + c x^2 + d x^3 $$ for which $$ 5b + 3d = 0.$$ No restriction on $a,c.$ Span of $1, x^2, 3x-5x^3 .$ Only trivial intersection with $W$
Indeed, the inner product of $x$ and $3x - 5 x^3$ has zeroes for constant terms. Then the integral of $3x^2 - 5 x^4,$ antiderivative $x^3 - x^5 = x^3 (1-x^2)$ which evaluates to zero at each endpoint.