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Question: If the rank of a real symmetric matrix be $1$, show that the diagonal elements of the matrix cannot be all zero.

Progress: Let $A=(a_{ij})_{n\times n}$ be a symmetric matrix with rank $1$. Then $a_{ij}=a_{ji}$ for $i,j=1,2,\dots, n$. If possible, let $a_{ii}=0$ for all $i=1,2,\dots, n$. How can I prove the result?

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Hint: Use $trac(A)=trac(P^{-1}AP)=trac(D)$

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If a matrix has rank 1 then all its rows are multiples of each other. Suppose $$a_{ij}=0$$ if $$i=j$$ and $$a_{ij}≠0$$ if $$i≠j$$ . Then R2 can't be expressed as a multiple of R1 since the first element $$a_{12}$$ of R2 can't be a multiple of the first element of R1 which is $$a_{11}$$since $$a_{12}$$ is non zero and $$a_{11}=0$$. Thus atleast one diagonal element must be non zero.

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