Determinant of a positive semi-definite matrix If $M$ is a Hermitian matrix, then $M$ is positive definite if and only if its leading principal minors have positive determinant, i.e the following matrices have positive determinant:


*

*The upper-left 1-by-1 corner of M

*The upper-left 2-by-2 corner of M

*...

*M.


I was wondering if in the positive semi-definite case the equivalent condition on all the above matrices would be non-negative instead (determinant $\geq$ 0).
 A: For a Hermitian matrix $A$ to be positive semi-definite, it is necessary for its leading principal minors to be non-negative, but it is not sufficient as the following example shows:
Consider the matrix
$$
A =
\begin{bmatrix}
0 &  0 \\
0 & -1 \\
\end{bmatrix}
$$
with both leading principle minors
$$
M_{0} = \det(A) = 0*(-1) - 0*0 = 0,
$$
$$
M_{1} =
\det(A_{2,2}) =
\det(\begin{bmatrix} 0 & \square \\ \square & \square \\ \end{bmatrix}) =
\det(\begin{bmatrix} 0 \end{bmatrix}) = 0,
$$
non-negative, but $A$ is not positive semi-definite as it has a negative eigenvalue $-1$.
To check if a Hermitian matrix $A$ is positive semi-definite one has to test if all principal minors (not only the leading principal minors) are non-negative. (proof)
If we look at the example above, the principal minors are
$$
M_{0,0} = \det(A) = M_0 = 0,
$$
$$
M_{1,1} =
\det(A_{1,1}) = 
\det(\begin{bmatrix} \square & \square \\ \square & -1 \\ \end{bmatrix}) =
\det(\begin{bmatrix} -1 \end{bmatrix}) = -1,
$$
$$
M_{2,2} = \det(A_{2,2}) = M_1 = 0.
$$
We see that $M_{1,1}$ is negative, so the matrix is not positive semi-definite.
