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Theorem. A $M_{n \times n}$ matrix is positive (negative) definite iff the determinate of all upper-left sub matrices are positive (negative).

However, consider this matrix:

\begin{pmatrix} 3 & 1 & 2 \\ 2 & 4 & 3 \\ -1 & -2 & 1 \\ \end{pmatrix}

If I am not mistaken I take the upper-left submatrices and their respective determinant to be

$$\begin{pmatrix} 3 \\ \end{pmatrix} \Longrightarrow 3$$

$$\begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix} \Longrightarrow 10$$

$$\begin{pmatrix} 3 & 1 & 2 \\ 2 & 4 & 3 \\ -1 & -2 & 1 \\ \end{pmatrix} \Longrightarrow 25$$

But this matrix isn't strictly positive definite of negative definite. It you find the eigenvalues (I'd suggest using some type of software - can also try to plug in for the Cholesky Decomposition and should given an error) you get eigenvalues with real and complex parts.

The real parts are all $> 0$, but the complex parts are not. Does this mean the matrix is positive definite in the reals, but not strictly positive definite?

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    $\begingroup$ Your matrix isn't symmetric $\endgroup$ Apr 22, 2017 at 22:28
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    $\begingroup$ Yes, the result you quote is true of symmetric matrices, not general ones. $\endgroup$
    – Chappers
    Apr 22, 2017 at 22:29
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    $\begingroup$ Is there no extension for matrices that are not symmetric? $\endgroup$
    – EzioBosso
    Apr 22, 2017 at 22:32
  • $\begingroup$ @user334916 a non-symmetric matrix $A$ is positive definite iff the symmetric matrix $A + A^T$ is positive definite. $\endgroup$ Apr 22, 2017 at 22:33
  • $\begingroup$ Just found on the wiki page that "some authors choose to say that a complex matrix M is positive definite if Re(z*Mz) > 0 for all non-zero complex vectors z. This weaker definition encompasses some non-Hermitian complex matrices, including some non-symmetric real ones, such as ((1,1),(-1,1))" $\endgroup$
    – EzioBosso
    Apr 22, 2017 at 22:34

1 Answer 1

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As was stated in comments, the Sylvester's criterion requires the matrix to be symmetric. A simpler example with a non-symmetric matrix would be $$ A = \begin{pmatrix} 1 & 4 \\ 0 & 1\end{pmatrix} $$ which makes it clear that none of the upper-left determinants are influenced by the entry $2$; thus they do not detect its effect on the signature of the matrix.

To determine positive-definiteness of a non-symmetric matrix one can apply Sylvester's criterion to $(A+A^T)/2$, which generates exactly the same quadratic form as $A$ itself.

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