Why $\langle Ax,x\rangle = \langle Dx,x\rangle$ holds for a diagonalisable matrix $A = XDX^*$?

Question

Let $A$ be selfadjoint with $spec(A) \subset (0, \infty) $. Show that $\langle Ax, x\rangle > 0$ for $x \ne 0 $

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Since $A$ is selfadjoint there is a matrix $X$ and $D$ such that $A=XDX^*$. So: $$\langle Ax,x\rangle = \langle XDX^*x, x\rangle = \langle Dx,x\rangle = \sum_{i=1}^n \lambda_i x_i^2$$

Since all eigenvalues are positive the term is greater than zero for $x \ne 0$ but how can we reduce $\langle XDX^*x,x\rangle$ to $\langle Dx,x\rangle$? Can someone explain what is happening between these two terms?

Answer 1

This is because a self adjoint matrix is diagonalizable in an orthonormal basis. We get the result as the inner product is kept between orthonormal basis

$$\langle Ax,x\rangle = \langle XDX^*x, x\rangle=\langle XDy, Xy\rangle = \langle Dy,y\rangle = \sum_{i=1}^n \lambda_i y_i^2$$

where $y=X^*x$.