On positive definite matrices and eigenvalues

Question

If I have symmetric matrix $A,$ then why is proving that all of the eigenvalues of $A$ are positive sufficient to show that $A$ is positive definite? Is this also true for non-symmetric matrices?

Answer 1

By definition, a symmetric matrix $A$ is positive-definite whenever all of the eigenvalues of $A$ are positive. On the other hand, if $A$ is not symmetric, then $A$ cannot be positive-definite.

Edit: By the Spectral Theorem, a symmetric matrix is orthogonally diagonalizable, i.e., there exists an orthogonal matrix $P$ such that $A = P^T DP,$ where $D$ is the diagonal matrix whose nonzero entries are the eigenvalues of $A.$ By hypothesis that the eigenvalues of $A$ are positive, it follows that $D$ is a positive-definite matrix. Considering that $P$ is orthogonal (and hence invertible), the linear transformation determined by $f(x) = Px$ is bijective. Particularly, we have that $x^T A x > 0$ if and only if $y^T D y > 0,$ where $y = Px.$ Consequently, the matrix $A$ is positive-definite.