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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

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    $\begingroup$ $\det M > 0$ is definitely not enough -- just look at $-I_2$ $\endgroup$
    – user369210
    Mar 1, 2017 at 18:31
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    $\begingroup$ Your teacher is wrong. What is sufficient that every upper-left $j\times j$-matrix (called principle minor) has positive determinant. $\endgroup$
    – Peter
    Mar 1, 2017 at 18:31

2 Answers 2

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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:

$\Delta_3 = \det \begin{bmatrix}2&-1&0\\-1&2&-1 \\ 0 & -1 &2 \end{bmatrix} = 4 > 0$

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  • $\begingroup$ Only in special cases (for example diagonal-dominant-matrices) there are easier ways.I think as well that no easier general criterion exists. $\endgroup$
    – Peter
    Mar 1, 2017 at 18:55
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The matrix $\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$ is not positive definite.

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