$\textbf{Question:}$ Consider the inner product on $\mathbb{R}^4$ defined by $\langle \vec{u},\vec{v} \rangle = C \vec{u} \dot{}C\vec{v}$ where $$C=\begin{bmatrix}1&2&3&4\\2&3&4&5\\3&4&5&6\\4&5&6&7\end{bmatrix}$$Find an orthonormal basis for $\mathbb{R}^4$ with respect to this inner product.
$\textbf{My Attempt:}$ So, this question caught me a bit off guard and I am looking for some clarification/guidance. Do I take every column vector of this matrix as the basis, then use Gram-Schmidt process to make them orthonormal? Sorry if this is a really obvious/bad question, I've just never had a problem like this in my book so far.
For the particular $C$ you are giving, the product you define fails to be an inner product because $C$ is not invertible. For example $$ \left\langle\begin{bmatrix}1\\-2\\1\\0\end{bmatrix},\begin{bmatrix}1\\-2\\1\\0\end{bmatrix}\right\rangle=0. $$
If $C$ were (selfadjoint, positive, and) invertible and you don't have any additional information, the only resource you have is to perform Gram-Schmidt using the new inner product.
If you were to know the eigenvalues and eigenvectors of $C$, say $\lambda_1,\ldots,\lambda_4$ and $x_1,\ldots,x_4$, then $$ \langle x_k,x_j\rangle = Cx_k\cdot Cx_j=(\lambda_k\lambda_j)\,x_k\cdot x_j=0 $$ (eigenvectors corresponding to different eigenvalues are orthogonal because $C$ is selfadjoint). Thus $$ \frac{x_1}{\lambda_1},\frac{x_2}{\lambda_2},\frac{x_3}{\lambda_3},\frac{x_4}{\lambda_4} $$ is an orthonormal basis.