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