If the rank of a real symmetric matrix be $1$, show that the diagonal elements of the matrix can not be all zero.
Since the rank is $1$, the determinant of the entire matrix is $0$, so it is singular, it is not row-equivalent to $I$, so it can't be reduced to identity matrix of the same order (though it is not mentioned whether it is a square matrix or not, but if not, then we can't really get a diagonal).
If all the diagonal elements were zero, then, applying some Gaussian eliminations, we could get the the first element non-zero, and all the elements zero, which follows the conclusion of rank $1$.
I really don't understand what reasoning causes some of diagonal element become non-zero. Am I missing something serious?
Best Answer
A (square) matrix $A$ of rank $1$ can be written in the form $u v^T$ where $u$ and $v$ are nonzero vectors in the column spaces of $A$ and $A^T$ respectively. If $A$ is symmetric, those column spaces are equal, so $v = c u$ for some scalar $c \ne 0$. Then $A_{ii} = c u_i^2$. Since $u \ne 0$, some $u_i \ne 0$, so that $A_{ii} \ne 0$.
Note, btw, that "real" is not needed here: this works over any field.