Can a matrix be positive semidefinite, even though it has negative leading principle minors?

Question

(Disclaimer. I don't remember the tag, but this is a task I need/want to solve as part of a problem set at uni.)

Question: For which values of $a$ is \begin{matrix} 1 & a & 1 \\a & 0 & 0 \\1 & 0 & 1 \end{matrix} is the matrix positive semidefinite or indefinite using the principal minor methodology?

I first looked at the leading principal minors.

$|A_1| = 1$ $|A_2| = 1*0 - a*a = -a^2$ $|A_3| = -a^2$

I learned if all principal minors are $\geq 0$, then the matrix is positive semidefinite, if all principal minors of odd order are $\leq 0$ and of even order $\geq 0$, then it is negative semidefinite.

So I determine all other pincipal minors that are not leading ones.

Of order one: I get one $=0$ and one $=1$, thus all prinicpal minors of order one (odd) are $\geq 0$.

Of order two: I get $1$ and $0$. Thus, all of even order are $\geq 0$

So to the question of whether it is positive semideifnite or not?

From the leading principal minors it follows for $a\neq 0$ the matrix is indefinite and for $a = 0$ the matrix is definite. (that is correct acc. to solution)

however, i don't get how the solution says that forall $a\geq 0$ the matrix is positive semidefinite as the third and second leading principal minors are already negative.

Answer 1

Zero is nonnegative, and for the correct value of $a$ we can have $-a^2=0$. Therefore, a real value of $a$ exists for which the matrix is positive semi definite.

By contrast, a positive definite matrix would require the leading principal minors to be actually positive rather than possibly zero, thus is not possible for any real $a$ in this case.