# Positive Definite iff the determinant of all upper-left submatrices $> 0$ - Always true?

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?

• Your matrix isn't symmetric – carmichael561 Apr 22 '17 at 22:28
• Yes, the result you quote is true of symmetric matrices, not general ones. – Chappers Apr 22 '17 at 22:29
• Is there no extension for matrices that are not symmetric? – user334916 Apr 22 '17 at 22:32
• @user334916 a non-symmetric matrix $A$ is positive definite iff the symmetric matrix $A + A^T$ is positive definite. – Ben Grossmann Apr 22 '17 at 22:33
• 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))" – user334916 Apr 22 '17 at 22:34

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.