# A symmetric real matrix with all diagonal entries unspecified can be completed to be positive semidefinite

In my matrix theory studies I recently came across the following problem:

Let $A$ be a square real partial matrix (some entries are unspecified), with all diagonal entries unspecified, and we have the following symmetry $a_{ij} = a_{ji}$ for all indices $1\leq i\neq j \leq n$ whenever both entries are specified, or that both entries are unspecified. We are asked to show that this matrix can be completed (meaning that we can find real numbers to substitute) for the unspecified entries such that $A$ is real symmetric positive semidefinite.

I really have no idea how to solve this, I thought about using an alternative characterization for positive semidefinite matrices like such a matrix being the Gram matrix of some n vectors in an inner product space, but I really have no idea how to proceed with this and I certainly need help.

• 1. A symmetric matrix is positive semidefinite if and only if it has nonnegative eigenvalues. 2. Gerschgorin's theorem. – Ian Oct 17 '16 at 20:27

## 1 Answer

Let $A_0$ be the matrix $A$ in the case where all the missing entries are all equal to zero. Since $A_0$ is symmetric and real, its eigenvalues are real. Let $\lambda_1\leq\lambda_2\leq\cdots\lambda_n$ be the eigenvalues of $A_0$. If we know let $$A=A_0+(1-\lambda_1)I,$$ it is a symmetric real matrix with eigenvalues $$1\leq1+\lambda_2-\lambda_1\leq1+\lambda_3-\lambda_1\leq\cdots\leq1+\lambda_n-\lambda_1.$$ As all the eigenvalues of $A$ are positive, and it is real and symmetric, we have that $A$ is positive-definite.