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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).

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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.

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