Show that the diagonal elements are not all $0$

Question

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?

Answer 1

If the matrix size is $1\times 1$, then it has only one entry, it is in the main diagonal and it is not zero if and only if the rank is $1$. So from now, let's suppose that the size of the matrix is $2\times 2$ or greater.

Since the rank is $1$, is not the zero matrix. But every $2\times2$ minor is zero.

Since it is not the zero matrix, it has one entry that it is not zero. Suppose that $a_{ij}\neq0$. Since the matrix is symmetric, $a_{ji}$ is not zero either.

If $i=j$, the statement is proved. So, suppose that $i\neq j$. Under our assumption, the minor $$\left|\begin{matrix}a_{ii}&a_{ij}\\a_{ji}&a_{jj}\end{matrix}\right|=-a_{ij}^2+a_{ii}a_{jj}$$ is zero. This implies $a_{ii}a_{jj}\neq 0$

Answer 2

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.

Answer 3

Suppose there is a matrix $A=(a_{ij})_{i,j}\in \text{Mat}(n\times n)$ symmetric with rank $1$ and $a_{ii}=0$ for all $1\leq i\leq n$.$n=1$ is obvious, so suppose $n>1$.

Because we have rank 1, there is an entry which isn't $0$. Say $a_{kl}\neq 0$ for a $k\neq l$. Because of rank $1$ the $k$-th column is linear dependent from from the $l$-th, so the $k$-th column must be $0$ because $a_{kk}=0$. But by symmetry we must have $a_{lk}=a_{kl}\neq 0$, a contradiction.

Answer 4

Since the matrix is real and symmetric, it is diagonalizable. Since its rank is $1$, we have an eigenvalue $\lambda\not=0$. The diagonalized matrix has $\lambda$ on one component of the diagonal. All the other components of the diagonalized matrix are $0$. So its trace is exaclty $\lambda$. By the invariance of the trace we have also that the trace of the starting matrix is $\lambda$, so not all the components of the diagonal are $0$.