Principal minors and semi-definiteness

Question

And a positive semidefinite matrix has all principal minors $\geq 0$.
We were asked in class to come up with an example of a square, symmetric matrix that has all leading principal minors $\geq 0$, but is NOT positive semi-definite.
I was trying to come up with a $2 \times 2$ matrix, but I have tried for a while and can't do it. Is it that hard or am I missing something?

Answer 1

I was able to get another answer, so I thought I'd add that as well: $ \pmatrix{0&0&1\\ 0&0&1\\ 1&1&0}. $

Answer 2

Consider $\pmatrix{0&0\\ 0&-1}$ for instance. In general, for any $n\ge2$, every diagonal matrix $\operatorname{diag}(0,d_2,\ldots,d_n)$ with some $d_i<0$ can serve as a counterexample.

There are even counterexamples without any zero entry. For every $n\ge3$, let $A$ be the all-one matrix. Now modify the bottom right entry of $A$ to $-1$. E.g. when $n=3$, the modified $A$ is $$ \pmatrix{1&1&1\\ 1&1&1\\ 1&1&-1}. $$ Then all leading principal minors of $A$ of size $2$ or larger are zero (hence nonnegative), because the first two rows of $A$ are identical. The leading principal $1\times1$ minor ($=1$) is also clearly nonnegative. But $A$ isn't positive semidefinite because it has a negative (though not leading principal) minor $-1$.

This counterexample can be modified to become entrywise positive when $n\ge4$. E.g. consider $$ \pmatrix{1&1&1&1\\ 1&1&1&1\\ 1&1&1&2\\ 1&1&2&1}. $$