I understand that the definition of (semi-)definiteness of matrix $A$ is $$\forall z_{\neq0}\in\mathbb R^k: z^TAz>0$$
I also know that this doesn't mean that all elements of a negative definite matrix $A$ are negative (in fact, they may all be positive or 0, such as with a 180 degree rotation in $\mathbb R^2$).
Nevertheless, I'm wondering if there is a way to recognize, just by looking at the matrix, whether it is likely going to be a positive definite matrix? Is there a way to see this just from the matrix itself, or does it always require some form of computation first?
Best Answer
There are some things you can test...
Full Rank
If the matrix is not of full rank, it has a non-trivial kernel. Hence there exists a vector $v$ such that $Av=0$ and thus $v^\top Av=0$. And therefore $A$ is not positive definite.
An easy way to test this, is linear dependence of the rows / columns.
Eigenvalues
If $A$ is symmetric/hermitian and all eigenvalues are positive, then the matrix is positive definite.
Main Diagonal Elements
Because of $a_{ii}=e_i^\top Ae_i>0$ all main diagonal entries have to be positive. If not the matrix is not positive definite.
Gerschgorin Circles
This is a cool criterion, that you usually don't learn in linear algebra. Consider the circle disks $$K^i:= K_{r_i}(a_{ii})⊂ℂ$$ with midpoint $a_{ii}$ and radius $r_i = \sum_{j\neq i}|a_{ij}|$. All eigenvalues of $A$ are in the union of these disks.
Again you can tell something about the eigenvalues. E.g. if $A$ is symmetric, $a_{ii}>0$ and $a_{ii}>r_i$ all circles are in the right half of the complex plane, and thus $A$ is positive definite. [keyword diagonal dominance]
Note, that you can also compute the circles for $A^\top$.
Hurwitz criterion
The north-west minors of a matrix $A$ are the determinants of the sub-matrices $$ H_1 = \pmatrix{a_{11}}, \qquad H_2=\pmatrix{a_{11} & a_{12} \\ a_{21} & a_{22}}, \qquad H_3=\pmatrix{a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}} \qquad ...$$ If $\det(H_i)>0$ for all $i=1,…,n$ then $A$ is positive definite. If they have alternating sign $\det(H_i)=(-1)^ia$, with $a>0$, then $A$ is negative definite.
Note that this is very expensive to compute.
Addition of Transpose
It is $v^TAv=(v^TAv)^T=v^TA^Tv$. And therefore $$2 v^\top A v = v^\top Av + v^\top A^\top v = v^\top(A+A^\top)v,$$ So, $A$ is positive definite if, and only if, $A+A^\top$ is positive definite. Since $A+A^\top$ is symmetric it might be easier to check for positive definiteness.