Let $A$ be a real symmetric matrix with rank $1$ , then can all the diagonal entries of $A$ be $0$ ?

Question

Let $A$ be a square real symmetric matrix with rank $1$ , then can all the diagonal entries of $A$ be $0$ ? I know that real symmetric matrices are diagonalizable . Also if all the diagonal entries be $0$ then sum of all the eigenvalues will be $0$ . But so what ? Please help . Thanks in advance

Answer 1

The matrix is diagonalizable for the spectral's theorem. Indeed the dimension of the eigenspace for the eigenvalue of $0$ is $n-1$. Note that the eigenspace for the eigenvalue of $0$ is the $\ker$ of function.

Answer 2

The quick answer to your question: note that the only diagonalizable matrix whose eigenvalues are all $0$ is the zero-matrix, and that a rank $1$ matrix can have at most one non-zero eigenvalue.

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Another approach:

Note that any rank $1$ matrix can be written in the form $uv^T$ for column vectors $u,v$, and that a rank-$1$ matrix will be symmetric if and only if it can be written in the form $uu^T$ for some vector $u$.

Now, if $A = uu^T$, then the diagonal entries are given by $A_{ii} = u_i^2$.