Assume the existence of an $A$-orthonormal system $\{u^{(1)},u^{(2)},..., u^{(n)}\}$ and prove that $A$ is symmetric and positive definite.

Question

Let $A$ be an $n\times n$ matrix, not assumed to be symmetric or positive definite. Assume the existence of an $A$-orthonormal system $\{u^{(1)},u^{(2)},..., u^{(n)}\}$ and prove that $A$ is symmetric and positive definite.

I know that $U^{T}AU=I$ where $U$ is the $n\times n$ matrix whose columns are $u^{(1)}, u^{(2)},...,u^{(n)}$, with which $(U^{T}AU)^{T}=I^{T}$ and so $U^{T}A^{T}U=I$ so $U^{T}AU=U^{T}A^{T}U$ so $A=A^{T}$. Besides, let $x\in \mathbb{R}^n$ such that $x\neq 0$, then there are $c_1,c_2,...,c_n$ such that $x=U\begin{bmatrix}c_1\\. \\. \\c_n\end{bmatrix}$ and so $x^{T}Ax=[c_1...c_n]U^{T}AU\begin{bmatrix}c_1\\. \\. \\c_n\end{bmatrix}=c_1^2+c_2^2+...+c_n^2>0$, then $A$ is positive definite.

Is this reasoning correct? Thank you very much.

Answer 1

Yes, it is correct. Short, sweet, very elegant. Nice job!

I can say no more because there is no more to be said!

Answer 2

As a basis matrix, $U$ is invertible. Thus $$ A=U^{-T}U^{-1} $$ which is automatically symmetric and positive semi-definite by construction. As $U$ has full rank, it is regular, so does $A$ and you get positive definite.