## Positive eigenvalues implies positive definite

Can someone just tell me if this proof works? As a quick word of explanation since most of my posts will probably be like this - I'm studying math on my own and am without a professor or anyone else to check things over. It would be nice to think that I'd know when a proof doesn't work, but I know I miss things sometimes.

"Let $\displaystyle T$ be a self-adjoint linear operator on an n-dimensional vector space $\displaystyle V$, and let $\displaystyle A=[T]_\beta$, where $\displaystyle \beta$ is an orthonormal basis for $\displaystyle V$. Prove: $\displaystyle T$ is positive definite if and only if all of its eigenvalues are positive."

I know I got the forward direction, so here's the other direction.

Since T is self-adjoint, there exists an orthonormal basis (call it $\displaystyle \beta$) for $\displaystyle V$ consisting of eigenvectors of $\displaystyle T$, which implies that $\displaystyle T$ is diagonalizable - that is, there exists an invertible matrix $\displaystyle P$ such that $\displaystyle D=P^{-1}[T]_\beta P$ is a diagonal matrix. In particular, the entries of $\displaystyle D$ are the eigenvalues of $\displaystyle T$. Letting $\displaystyle d_{ii}$ be the entries of $\displaystyle D$ and $\displaystyle x_i$the $\displaystyle i$-th component of $\displaystyle x$, we have $\displaystyle \displaystyle\sum_i d_{ii}x_i^2 > 0$, and so $\displaystyle \langle T(x), x \rangle > 0$ for all nonzero $\displaystyle x$ and $\displaystyle T$ is positive definite.