let $M$ be a symetric $n$ by $n$ real matrix
on wikipedia it is said that all the eigenvalues of M are positve is equivalent to $M$ being positive definite. which means that if a matrix $M$ is positive definite then $det(M)>0$ but our teacher told us today that the reciprocal is also true ($M$ is said to be a symetric positive definite matrix if and only if $det(M)>0$) and I'm pretty sure this is wrong because $M$ could have some negative eigenvalues and at the same time a positive determinant.
please can someone shed some light on this and give me a solid counter exemple and tell when the reciprocal is true.
thank you
Best Answer
Determinant is just one of the leading principal minors. For $M$ to be positive definite all its leading principal minors have to be positive. I think, this is the easiest way to check positive definiteness. In the example above, the first leading principal minor is $-1$, thus this matrix is not positive definite.
Just as an example from Wikipedia:
$\begin{bmatrix}2&-1&0\\-1&2&-1 \\ 0 & -1 &2 \end{bmatrix}$.
First principal minor is just $\Delta_1 = 2 >0$. The second one is:
$\Delta_2 = \det\begin{bmatrix}2&-1\\-1 &2\end{bmatrix} = 4-1=3 > 0$
And finally, the last one is just the determinant of the whole matrix:
$\Delta_3 = \det \begin{bmatrix}2&-1&0\\-1&2&-1 \\ 0 & -1 &2 \end{bmatrix} = 4 > 0$