A matrix $A$ is positive semi-definite IFF $x^TAx\geq 0$ for all non-zero $x\in\mathbb{R}^d$. If all elements of $A$ are non-negative, does this guarantee that $A$ is positive semi-definite?
Matrix with no negative elements = Positive Semi Definite
positive-semidefinite
Related Solutions
There are two approaches. One is to write the (symmetric) matrix associated the quadratic form, the other is to "complete the square", as the a deleted answer notes.
For example: $$ q(x_1,x_2) = x_1^2 - 4x_1x_2 + x_2^2 = x_1^2 - 2x_1x_2 - 2x_2x_1 + x_2^2 $$ The matrix associated with this quadratic form is $$ M = \pmatrix{1&-2\\-2&1} $$ and we wish to determine its definiteness. Applying Sylvester's criterion to both $M$ and $-M$ allows us to check if $M$ is positive/negative definite. Since the $2 \times 2$ determinant $\det(M)$ is negative, we can conclude that $M$ is neither positive nor negative definite.
Now, $\det(M) \neq 0$, so $M$ is invertible. Since $M$ is invertible without being positive/negative definite, it is invertible.
If $M$ were singular, we'd have to try something else. In practice (with pencil and paper), one would find the eigenvalues of $M$. If all non-zero eigenvalues have the same sign, then the singular $M$ is postive/negative semidefinite. Otherwise, $M$ would be indefinite.
Another approach to apply (especially for large matrices) is to attempt a Cholesky factorization. If $M$ has a Cholesky factorization, it is positive semidefinite. If $-M$ has a Cholesky factorization, it is negative semidefinite. Otherwise, $M$ is neither.
In this case, we find $M$ and $-M$ have no Cholesky factorization
Equivalent (in a sense) to Cholesky factorization is the approach of completing the square. We find $$ q(x_1,x_2) = x_1^2 - 4x_1x_2 + x_2^2 = (x_1^2 - 4x_1x_2 + 4x_2^2) - 3x_2^2 = \\ (x_1 - 2x_2)^2 - 3x_2^2 $$ Correspondingly, we see that $$ M = \pmatrix{1&0\\-2&1}^T\pmatrix{1&0\\0&-3}\pmatrix{1&0\\-2&1} $$
For a Hermitian matrix $A$ to be positive semi-definite, it is necessary for its leading principal minors to be non-negative, but it is not sufficient as the following example shows:
Consider the matrix $$ A = \begin{bmatrix} 0 & 0 \\ 0 & -1 \\ \end{bmatrix} $$ with both leading principle minors $$ M_{0} = \det(A) = 0*(-1) - 0*0 = 0, $$
$$ M_{1} = \det(A_{2,2}) = \det(\begin{bmatrix} 0 & \square \\ \square & \square \\ \end{bmatrix}) = \det(\begin{bmatrix} 0 \end{bmatrix}) = 0, $$ non-negative, but $A$ is not positive semi-definite as it has a negative eigenvalue $-1$.
To check if a Hermitian matrix $A$ is positive semi-definite one has to test if all principal minors (not only the leading principal minors) are non-negative. (proof)
If we look at the example above, the principal minors are $$ M_{0,0} = \det(A) = M_0 = 0, $$
$$ M_{1,1} = \det(A_{1,1}) = \det(\begin{bmatrix} \square & \square \\ \square & -1 \\ \end{bmatrix}) = \det(\begin{bmatrix} -1 \end{bmatrix}) = -1, $$
$$ M_{2,2} = \det(A_{2,2}) = M_1 = 0. $$
We see that $M_{1,1}$ is negative, so the matrix is not positive semi-definite.
Best Answer
No. The matrix $A=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ is not psd, as you can check by seeing that $(1,-1)A(1,-1)^T=-2$.