I understand that the eigenvalues of a positive definite matrix must be positive, but if the matrix isn't diagonalizable, does this still hold? Also, it doesn't seem like the reverse would be true but I can't think of a counterexample, so that would also be helpful.
Thank you!
On
Hint (necessity): $$ \begin{bmatrix} 0&0&1&0 \end{bmatrix} \begin{bmatrix} a&\color{#A0A0A0}{b}&\color{#A0A0A0}{c}&\color{#A0A0A0}{d}\\ \color{#A0A0A0}{e}&f&\color{#A0A0A0}{g}&\color{#A0A0A0}{h}\\ \color{#A0A0A0}{j}&\color{#A0A0A0}{k}&m&\color{#A0A0A0}{n}\\ \color{#A0A0A0}{p}&\color{#A0A0A0}{q}&\color{#A0A0A0}{r}&s \end{bmatrix} \begin{bmatrix} 0\\0\\1\\0 \end{bmatrix}=m $$ Counterexample (insufficiency): $$ \begin{bmatrix} 1&1 \end{bmatrix} \begin{bmatrix} 1&-2\\ -2&1 \end{bmatrix} \begin{bmatrix} 1\\1 \end{bmatrix}=-2 $$
A linear transformation $P: V \to V$ is positive definite if and only if $\langle P(v),v \rangle >0$ for each non-zero vector $v,$ where $\langle, \rangle$ is a positive definite inner product on $V$. In particular if $\{v_{1},v_{2},\ldots,v_{n} \}$ is an orthonormal basis with respect to $\langle, \rangle$, then we have $\langle P(v_{i}), v_{i} \rangle >0$ for each $i.$ Hence $v_{i}$ appears with real positive coefficient when $P(v_{i})$ is expressed as a linear combination of $\{v_{1},v_{2}, \ldots, v_{n} \}.$ This means precisely that the matrix for $P$ with respect to the orthonormal basis $\{v_{1},v_{2},\ldots,v_{n} \}$ has positive entries on the diagonal. In particular, this applies when $\langle, \rangle$ is the standard inner product, and $\{v_{1},\ldots,v_{n} \}$ is the standard basis.