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

(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.

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.