Let $\mathbf{A}$ be a real, symmetric $n \times n$ with its spectral decomposition, $\mathbf{A} = \mathbf{X} {\rm diag}(\lambda_1,\lambda_2,\cdots,\lambda_n)\mathbf{X}^{\rm T}$. Is it true that if $\lambda_i = 0$, then ${\rm rank}(\mathbf{A}_{i,i})={\rm rank}(\mathbf{A})$, where $\mathbf{A}_{i,i}$ is a submatrix of $\mathbf{A}$ obtained by removing the $i$th row/column of $\mathbf{A}$?
I tried to analyze $\mathbf{A} \mathbf{x}_i = \mathbf{0}$, where $\mathbf{x}_i$ is the eigenvector assigned to $\lambda_i=0$. This connection yields
$\sum\limits_{j=1}^nx_{ji} \mathbf{a}_j=0$, (*)
where $\mathbf{a}_j$ is the $j$th column of $\mathbf{A}$. If $x_{ii} \ne 0$ was valid for this $\mathbf{x}_i$ vector, then $\mathbf{a}_i$ could be expressed from (*). However, I do not see anything which implies this assumption.
Best Answer
Let $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$. We have eigenvalues $\lambda_1 = 0$, $\lambda_2 = 1$ and orthonormal eigenvectors $v_1 = (0, 1)$, $v_2 = (1, 0)$. But $A_{1, 1} = \begin{pmatrix} 0 \end{pmatrix}$, which has rank $0$ not $1$.