# Matrix with no negative elements = Positive Semi Definite?

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?

No. The matrix $$A=\begin{pmatrix}0&1\\1&0\end{pmatrix}$$ is not psd, as you can check by seeing that $$(1,-1)A(1,-1)^T=-2$$.
$$\left( \begin{array}{cc} 1 & 153 \\ 153 & 1 \end{array} \right)$$
In general no. One way of defining positive definiteness is through the leading principal minors of a matrix. The $$k^{th}$$ leading minor if found by computing the deturminant of the matrix after deleting the last $$n-k$$ colomns and rows in an $$n \times n$$ matrix. It is quite common to see a matrix with all positive entries that has a negative deturminnant, this therefore means this matrix would not be positive definite. For example, If you look at the leading principal minors of the following matrix $$A \in \mathbb{R}^{n\times n}$$:
$$A = \left( \begin{array}{cc} 1 & 1 \\ 2 & 1 \end{array}\right),$$
for $$A$$ to be positive definite $$det(1)>0$$ and $$det(A)>0$$. This is clearly not the case as $$det(A)=-1$$. In fact this particular matrix is indefinite.