Find a positive definite matrix with $a_{1,1}<0$

Question

Find a matrix $A=[a_{i,j}]$ such that $A$ is positive definite and $a_{1,1}<0$.

It is not easy. In my opinion, there is no such matrix if the matrix dimension is even.

Answer 1

If $x = \left(\begin{matrix} x_1 \\ \vdots \\ x_n \end{matrix}\right)$, then $$x^TAx = \sum\limits_{i,j=1}^{n}{a_{ij}x_ix_j}.$$ In particular, for $x = e_1 = \left(\begin{matrix} 1 \\ 0 \\ \vdots \\ 0 \end{matrix}\right)$, we have $$ e_1^TAe_1 = a_{11}.$$ If $A$ is positive definite, then $e_1^TAe_1$ is positive, and hence $a_{11}$ must be positive.

Answer 2

Using Sylvester's criterion, saying that $\mathrm A \succ \mathrm O$ is equivalent to saying that all the leading principal minors of $\mathrm A$ are positive. Note that $a_{11}$ is the $1 \times 1$ leading principal minor. Thus, $a_{11} > 0$.

Therefore, a matrix $\mathrm A$ with $a_{11} < 0$ cannot be positive definite.