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Let $A$ be a positive semi-definite (hence symmetric) $n\times n$ real matrix. We know that the eigenvalues of $A$ are non-negative real numbers. Let's denote them by $\lambda_i$, for $i=\{1,\dots,n\}$, and sort them in a non-decreasing order (i.e., $\lambda_i \leq \lambda_{i+1}$). Furthermore, as a real symmetric matrix, $A$ has the following eigenvalue decomposition: \begin{equation} A=\sum_{i=1}^n\lambda_iq_iq_i^T, \end{equation} where $q_i$ is the eigenvector corresponding to $\lambda_i$, $\langle q_i,q_i\rangle=1$, and $\langle q_i,q_j\rangle=0$, for any $i\neq j$.

Now assume that ${\text{trace}}(A)=n$, it implies that $\lambda_1 \leq 1$ and $\lambda_n \geq 1$. Also, by the Arithmetic mean-Geometric mean inequality, we have \begin{equation} \Bigg(\prod_{i=1}^{n}\lambda_i \Bigg)^{1/n}\leq \frac{\sum_{i=1}^n \lambda_i}{n}. \end{equation} Hence, $0\leq\det(A)\leq 1$.

Finally, assume that the main diagonal of $A$ consists of only ones, that is, ${\text{diag}(A)=\vec{1}}$. In some branches of math, this kind of matrix my be called the Correlation matrix. Now the question is: what properties does the matrix $A$ have other than the aforementioned properties, especially with respect to eigenvalues and eigenvectors?

  • Once you know the diagonal is all $1$'s, it implies the trace is $n$. You don't need that as an extra assumption. – Robert Israel Nov 12 '19 at 04:53
  • My intention of separating the two classes (i.e. ${\text{tr}}(A)=n$ and ${\text{diag}}(A)=\vec{1}$) was to stress that we are looking for the properties that matrices with ${\text{diag}}(A)=\vec(1)$ have but the matrices with ${\text{tr}}(A)=n$, as the more general class, don't. –  Nov 12 '19 at 05:14
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    You may add taking $2\times 2$ minors that $|a_{i,j}|\le 1$. – Toni Mhax Nov 12 '19 at 08:02
  • It is known that every matrix is similar to a matrix with constant diagonal entries. If we could additionally say that symmetric matrices are unitarily similar to a matrix with constant diagonal entries, then we'd be able to conclude that there are no differences in the spectral properties between correlation matrices and trace-$n$ positive semidefinite matrices. – Ben Grossmann Nov 12 '19 at 11:30

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Let $A$ be an arbitrary real positive semidefinite matrix with size $n$ and trace $n$. By the construction in this post (this is a post on the complex case by the same author for those curious), there exists an orthogonal matrix $U$ such that $UAU^T$ is positive definite with $1$'s on the diagonal. In other words, $UAU^T$ is a correlation matrix.

In other words, every positive semidefinite matrix with size $n$ and trace $n$ is similar to a correlation matrix. So, there is no difference between the eigenvalue properties of positive semidefinite matrices with size $n$ and trace $n$ and those of correlation matrices.

Ben Grossmann
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