symmetric matrix , an orthonormal basis for V is $\{XD^{^{\frac{-1}{2}}}e_{1},...,XD^{^{\frac{-1}{2}}}e_{n}\}$?

Question

Let $V=\mathbb{R}^{n}$ vector space, $Q$ is symmetric matrix and positive definitive, it decomposes $Q=XDX^{T}$ yields an orthonormal basis for $v$ given by the scaled columns of $X$, namely, $\{XD^{^{\frac{-1}{2}}}e_{1},...,XD^{^{\frac{-1}{2}}}e_{n}\}$.

My attempt: I know if $Q$ is symmetric, then is diagonalizable, there exists a basis consisting of eigenvectors of $Q$, but I don't know how it looks like that$\big(\{XD^{^{\frac{-1}{2}}}e_{1},...,XD^{^{\frac{-1}{2}}}e_{n}\}\big)$, it's very general. Any help to proof, thanks in advance.

Answer 1

It seems that the question you are trying to answer is something like the following:

Let $V$ be the vector space $\mathbb{R}^{n}$, let $Q$ be a symmetric and positive definite matrix . Noting that $Q=XDX^{T}$ for some diagonal $D$ and orthogonal $X$, show that $\{XD^{^{\frac{-1}{2}}}e_{1},...,XD^{^{\frac{-1}{2}}}e_{n}\}$ is an orthonormal basis for $V$ (consisting of scaled columns of $X$) relative to the inner product induced by $Q$.

(The bold portions indicate information missing from the original question)

With this in mind, we need to verify the following: for $1 \leq i < j \leq n$, we have $$ \langle XD^{-1/2}e_i,XD^{-1/2}e_i \rangle_Q := (XD^{-1/2}e_i)^T QX(D^{-1/2}e_i) = 1\\ \langle XD^{-1/2}e_i,XD^{-1/2}e_j \rangle_Q := (XD^{-1/2}e_i)^T QX(D^{-1/2}e_j) = 0. $$ It is straightforward to show that the above holds if we use our decomposition $Q = XDX^T$ and the fact that $X^TX = I$.