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Suppose A is a covariance matrix whose diagonal elements are same, i.e. $A_{1,1}=A_{2,2}=\cdots=A_{N,N}$, can we conclude that A is full rank? Suppose the absolute values of the off-diagonal elements in A are all smaller than the diagonal elements in A. Thanks!

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The answer to your first question is: no, you can't conclude that, since a square matrix in which every entry is the same positive number is a rank-$1$ covariance matrix. –  Michael Hardy Jun 26 '13 at 21:10
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closed as off-topic by Mikael de la Salle, Yemon Choi, Benjamin Steinberg, Andy Putman, Andrey Rekalo Jun 27 '13 at 6:12

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No. Let $a,b,c,d$ be complex numbers of modulus $1$, chosen so that $a\neq c$, $b\neq d$, and $ab^*\neq cd^*$. Form the column vectors $v=(1,a,b)^T$ and $w=(1,c,d)^T$. Then the matrix $$ vv^* +ww^*=\begin{pmatrix} 2 & a^*+c^* & b^*+d^* \\ a+c & 2 & ab^*+cd^* \\ b+d & a^*b+c^*d & 2 \end{pmatrix} $$ has rank two, but all the off-diagonal entries have modulus strictly less than $2$.

EDIT: A simple example would be $$ \begin{pmatrix} 1 & \frac35 & \frac45 \\ \frac35 & 1 & 0 \\ \frac45 & 0 & 1\end{pmatrix} $$ where one can replace $(\frac35, \frac45)$ with any pair $(a,b)$ such that $|a|^2+|b|^2=1$.

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Thanks! It is interesting. How do you see the matrix has rank 2 without calculation? How did you construct such matrix? Is there any theory behind this construction? –  user2008790 Jun 27 '13 at 0:45
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I've added an easy example. The original example is rank 2 because it is constructed as a sum of two rank 1 matrices. (We are probably at the point that some people would say this is more appropriate for math.stackexchange.) –  Mike Jury Jun 27 '13 at 2:59
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