Is a positive semidefinite matrix always non-negative?

Can a positive semidefinite matrix contain negative elements? If by elements, you mean entries, then yes, $$\pmatrix{2&-1\\-1&2}$$ is positive definite.

The symbol $\succeq$ actually does not denote componentwise inequality for matrices in that book. If $A$ and $B$ are symmetric matrices then $A \succeq B$ means $A-B$ is positive semidefinite. Perhaps confusingly, if $x$ and $y$ are vectors rather than matrices, then in that case $x \succeq y$ does mean that each component of $x$ is greater than or equal to the corresponding component of $y$.

(As Lord Shark the Unknown mentioned, a positive semidefinite matrix can have some negative entries.)

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Here are a few more details. On p. 43, Boyd and Vandenberghe introduce the notation $x \preceq_K y$ (where $K$ is a proper cone) to mean that $y - x \in K$. If $K$ is the nonnegative orthant then $x \preceq_K y$ means that $y - x$ is in the nonnegative orthant, or in other words that $y_i \geq x_i$ for all $i$. If $K$ is the positive semidefinite cone then $A \preceq_K B$ means that $B - A$ belongs to the positive semidefinite cone, or in other words that $B - A$ is positive semidefinite.

The book goes on to state

The nonstrict and strict partial orderings associated with the nonnegative orthant arise so frequently that we drop the subscript $\mathbb R^n_+$; it is understood when the symbol $\preceq$ or $\prec$ appears between vectors.

Regarding the case where $K$ is the positive semidefinite cone, the book comments

Here, too, the partial ordering arises so frequently that we drop the subscript.