I'm having trouble with some maths regarding the expression of the matrix quadratic form (i.e. $x^TAx$) and the proof that, where the eigenvalues $\lambda_1,\lambda_2,...,\lambda_n$ are all positive, the quadratic form is positive definite.
My understanding is that the definition of positive-definiteness is when $x^TAx>0$ for all x where at least one element of $x \neq 0$.
My textbook produces the following proof, but I don't understand the last line:
Where $s_i$ = a normalized eigenvector of A, $\lambda_i$ = the corresponding eigenvalue, and $C = [s_1|s_2|...|s_n]$, $C^{-1}AC=D=$ the "diagonalization" of A.
Since the eigenvectors of A are orthonormal, $C^{-1} = C^T$.
Suppose we define a transformation $x=CX$. Then the equation becomes: $x^TAx = (CX)^{T}ACX = X^TDX = \lambda_1X_1^2 + \lambda_2X_2^2+...+\lambda_nX_n^2$.
It follows from this that a quadratic form is positive-definite if and only if all its eigenvalues are positive.
So, in summary, I don't understand why the following derivation true:
$x^TAx = \lambda_1X_1^2 + \lambda_2X_2^2+...+\lambda_nX_n^2$ therefore, for $x^TAx > 0$ for all x where at least one element of $x_i \neq 0$ to be true, $\lambda_1, \lambda_2, ..., \lambda_n > 0$.
Can someone please help with the derivation of the last step?a
Thanks in advance for your help.