How to make a matrix positive semidefinite We have a symmetric matrix $A$, with some entries specified and others not. We are trying to find the values of the unspecified entries so that the matrix $A$ becomes positive semidefinite. How can I prove that I can assume that the diagonal entries of $A$ are specified?
Intuitively, if we do not specify a  diagonal entry, say $i$th entry, we can take it to infinity. Then the positive definiteness of $A$ is equal to the positive definiteness of the new matrix $A[-i,-i]$ where we remove the $i$th column and row. I do not know how to show this mathematically.

Your question is vague about the meaning of "one can assume that the diagonal entries of $A$ are specified". If you mean to first set the unspecified diagonal entries to some large numbers, then determine the rest to make $A$ positive semidefinite, you will not always succeed. For example, consider
$$
A=\begin{pmatrix}1&0&x\\0&1&2\\x&2&z\end{pmatrix}.
$$
The first two leading principal minors of $A$ are clearly positive. So $A$ is positive semidefinite iff its determinant is nonnegative, i.e. iff $z\ge x^2+4$. Surely, when $z$ is sufficiently large (in this example we need $z\ge4$), you can always pick a suitable $x$ that makes $A$ positive semidefinite, but the caveat is that in general, it is hard to know how large is large enough.
Note that simple tools like Gerschgorin disc theorem may not get you anywhere: in the above example, $a_{22}=1$ is never a dominant diagonal entry in the first place; so you cannot guarantee that $A$ is positive semidefinite by keeping the other two Gerschgorin discs on the right half plane. In fact, if you merely try to make the other two discs disjoint from $0$, $A$ may fail to be positive semidefinite, because you cannot force $z\ge x^2+4$ from the two inequalities $|x|\le 1$ and $z\ge|x|+2$.
