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I have the following Matrix $A$ given as:

$A=\begin{pmatrix}10&0.1&1&0.9&0\\ 0.2&9&0.2&0.2&0.2\\ 0.3&-0.1&5+i&0&0.1\\ 0&0.6&0.1&6&-0.3\\ 0.3&-0.3&0.1&0&1\end{pmatrix}$

Applying Gerschgorin's Theorem (details not necessary), I know for sure that all Eigenvalues of this Matrix are positive. In my professor's notes it says and I quote:

If $\lambda_{i} > 0 $ for all $i=1,2,...,n$, then $M$ is called positive definite $<=>$ $M>0$

So I'd say, $A$ here is positive definite. Yet the solution tells me

... but having positive Eigenvalues is not sufficient for $A$ being positive definite. Since it is not symmetric, it will not be positive definite.

But there are for sure non-symmetric Matrices, that are positive definite. So I'm very confused here.

Any help would be greatly appreciated.

Rouben
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  • By definition, a positive definite matrix is symmetric and has all eigenvalues are positive. – the_candyman Nov 10 '21 at 18:04
  • @the_candyman This post says otherwise: https://math.stackexchange.com/questions/1954167/do-positive-semidefinite-matrices-have-to-be-symmetric#:~:text=No%2C%20they%20don%27t%2C%20but%20symmetric%20positive%20definite%20matrices,of%20a%20non-symmetric%20positive%20definite%20matrix%20is%20%24%24M%3Dpmatrix%7B2%260%5C2%262%7D.%24%24 – Rouben Nov 10 '21 at 18:05
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    There are two definitions of positive definite that are in common use, as is noted on the wiki page for "definiteness". It seems that the author of the solution in this case has a definition whereby positive definite matrices are necessarily symmetric. – Ben Grossmann Nov 10 '21 at 18:05
  • @Rouben See also this post – Ben Grossmann Nov 10 '21 at 18:06
  • @Rouben By the way: for a non-symmetric matrix, positivity of (the real part of) the eigenvalues is necessary but is not sufficient for positive definiteness. For example, the matrix $$ A = \pmatrix{1 & 6\0 & 2} $$ has positive (in this case, real) eigenvalues, but it is not positive definite. – Ben Grossmann Nov 10 '21 at 18:09
  • @Rouben Note also that writing $\lambda_i > 0$ typically implies that $\lambda_i$ is a real number, which is different from the way you use the term "positive". – Ben Grossmann Nov 10 '21 at 18:11
  • For many applications of a positive definite matrix you want the eigenvectors to be a complete set and to be mutually orthogonal, and this only is true if the matrix is symmetric as well. – Paul Nov 10 '21 at 18:11
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    General explanation, for those unfamiliar with the alternative definition. A (not necessarily symmetric) real matrix $A$ is sometimes said to be "positive definite" if it satisfies $x^TAx > 0$ for all non-zero $x$. This is equivalent to the condition that the "symmetric part" $\frac 12 (A + A^T)$ of $A$ is (symmetric and) positive definite. – Ben Grossmann Nov 10 '21 at 18:14
  • @Rouben my definition is good for the real case. In general, when moving to the $\mathcal{C}$ world, symmetric is substituted by "the Hermitian is equal to the starting matrix$ – the_candyman Nov 10 '21 at 18:51

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