Which of the following matrices are positive? \[ \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \qquad \begin{bmatrix} 1 & 1 + i \\ 1 - i & 3 \end{bmatrix}, \qquad \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 1 \\ 3 & -1 & 1 \end{bmatrix}, \qquad \begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \end{bmatrix}. \]
Solution
Call the four matrices \(A_1,A_2,A_3,A_4\) in the order displayed.
1. \(A_1\) is not positive.
We have \[ A_1= \begin{pmatrix}1&2\\3&4\end{pmatrix}. \] A positive matrix must be Hermitian, but \[ A_1^t= \begin{pmatrix}1&3\\2&4\end{pmatrix} \ne A_1. \] Hence \(A_1\) is not positive.
2. \(A_2\) is positive.
The matrix \[ A_2= \begin{pmatrix} 1&1+i\\ 1-i&3 \end{pmatrix} \] is Hermitian. Its leading principal minors are \[ \Delta_1=1>0 \] and \[ \Delta_2 =3-|1+i|^2 =3-2 =1>0. \] Therefore Sylvester’s criterion shows that \(A_2\) is positive.
3. \(A_3\) is not positive.
The matrix \[ A_3= \begin{pmatrix} 1&-1&1\\ 2&-1&1\\ 3&-1&1 \end{pmatrix} \] is not symmetric, since its \((1,2)\) and \((2,1)\) entries are \(-1\) and \(2\). Thus it is not Hermitian and hence not positive.
4. \(A_4\) is positive.
The matrix \[ A_4= \begin{pmatrix} 1&1/2&1/3\\ 1/2&1/3&1/4\\ 1/3&1/4&1/5 \end{pmatrix} \] is real symmetric. Its leading principal minors are \[ \Delta_1=1, \qquad \Delta_2=\frac13-\frac14=\frac1{12}, \] and \[ \Delta_3=\det A_4=\frac1{2160}. \] All are positive, so Sylvester’s criterion gives that \(A_4\) is positive.
Equivalently, for \(x=(x_1,x_2,x_3)^t\ne0\), \[ x^tA_4x =\int_0^1(x_1+x_2t+x_3t^2)^2\,dt>0, \] which gives an independent check of positivity.
Therefore the positive matrices are exactly \[ \boxed{A_2\text{ and }A_4}. \]