How to get a symmetric positive definite 5x5 matrix? \begin{pmatrix}
1&0&0&0&0\\
0&1&0&0&0\\
0&0&1&0&0\\
0&0&0&1&0\\
0&0&0&0&1\end{pmatrix}
is an example.
But I don't find another one. There also should be entries other than 0 or 1.
Is there a systematic way to do this?
 A: Take any invertible matrix $A$ (so just about any matrix unless you're unlucky) and compute $A^TA$.
It is easy to check that $(A^TA)^T=A^T(A^T)^T=A^TA$ (symetric), and for all $x$
$x^T(A^TA)x = (x^T A^T)(Ax)= (Ax)^T(Ax) = (Ax) \cdot (Ax) = \|Ax\|^2\geq 0$ (positive).
Plus since $A$ is invertible, this is equal to $0$ if and only if $x=0$. (definite)
Therefore $A^TA$ is symmetric definite positive. 
A: Yes. Choose any 5 strictly positive numbers and form a diagonal matrix $\Lambda$.
Now take any orthogonal $5 \times 5$ matrix $U$, and form the product $A=U \Lambda U^T$. It is straightforward to check that $A>0$ and $A^T = A$.
(Note that for any $A$ satisfying  $A>0$ and $A=A^T$, then it can be written in the form $A=U \Lambda U^T$.)
A: Replace all non-diagonal entries symmetrically and randomly with numbers whose absolute values are smaller than $\frac14$. By Gershgorin circle theorem, all eigenvalues of the result random matrix will fall inside a circle of radius smaller than $1$ centered at $1$. Since the eigenvalues of a real symmetric matrix are real, the eigenvalues will be all positive. i.e. Any random matrix generated in this manner will be positive definite.
A: Yet another way, particularly useful for generating SPD matrices with constant diagonal entries, is to take $A=\alpha I+B$, where $\alpha > 0$ and $B$ is any symmetric matrix with $\rho(B)=\|B\|<\alpha$ (can be done simply by taking an arbitrary symmetric matrix and scaling it such that the inequality holds).
