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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.

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    $\begingroup$ What do you mean by $A[-i,-i]$? $\endgroup$ – tomasz Mar 17 '13 at 3:19
  • $\begingroup$ And what do you mean by that you can assume that they are specified? If you mean that if it is at all possible to choose other entries so as to make the matrix positive-definite, then it is also possible for some specific values on the diagonal, then it is true, but rather trivial... $\endgroup$ – tomasz Mar 17 '13 at 3:22
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As a consequence of Gershgorin's Theorem, we know that the eigenvalues of a matrix $A$ live in balls $B(a_{ii},\sum_{j\neq i} |a_{ij}|)$ (centered at $a_{ii}$ of radius $\sum_{j\neq i} a_{ij}$). So if your matrix has real entries, but you have the freedom to choose the diagonal entries, then choosing each diagonal entry to be greater than the sum of the absolute values of the other entries in the same row will immediately imply that all of the eigenvalues of $A$ are positive, and therefore that $A$ is positive definite.

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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 $z\ge|x|+2$.

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Let suppose C is non positive definite correlation matrix $$C=Q\Lambda Q^*=Q (\Lambda_+ -\Lambda_-)Q^*$$ Where $\Lambda$ is diagonal matrix of Eigen values. Replace all negative eigen values with zero. Then we use the Symmetric , non negative definite matrix $\rho^2C$ with suitable value of $\rho$. Two choices of $\rho$ are $$\rho_1=tr(\Lambda)/tr(\Lambda_+) \space\space\space\space\space \rho_1=\sqrt{tr(\Lambda)/tr(\Lambda_+)}$$ User defined $\rho$ is also allowed. For +ve definite matrix $\Lambda=\Lambda_+$ and $\rho=1$ .Another Way is to add $C*I$ to your corr-matrix, where is C is a constant and I is an identity matrix.

Chan, Grace; Wood, Andrew T.A., An algorithm for simulating stationary Gaussian random fields, J. R. Stat. Soc., Ser. C 46, No.1, 171-181 (1997). ZBL0913.65142.

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