$\mathbf {The \ Problem \ is}:$ Given that $\operatorname A$ is an $m×m$ orthogonal matrix such that $\operatorname A$ is of the form :
$$ \begin{pmatrix}
1/{\sqrt m} & 1/{\sqrt m} \cdots 1/{\sqrt m} \\
\operatorname P
\end{pmatrix}
$$
where $\operatorname P$ is an $(m-1)×m$ rectangular matrix . And let $\operatorname B$ be an $m×m$ symmetric matrix with rank $(m-1)$ and $\operatorname B1_m=0$ where ${1_m}^t=(1,1,....1)^t$ .
Then, show that : $1)$ $\operatorname P^TP= \operatorname {I_m} - 1/m({1_m}{1_m}^t)$
$2)$ rank of $\operatorname { PBP^T} = m-1$ where $\operatorname I_m$ is the $m×m$ identity matrix .
$\mathbf {My \ approach} :$ Actually, I tried upto the portion that as $\operatorname A$ is orthogonal, then for any of its two columns $v_i$ and $v_j$, we have ${v_i}^tv_j = \delta_{ij}$ and then for any row vector ${p_i}^t$ of $\operatorname P$ ;
$$ {p_i}^tp_j = \begin{cases} (1-1/m), & \text {for $i=j$} \\ -1/m , & \text{for $i \neq j$} \end{cases} $$ hence first part is done.
Now, as $\operatorname B$ is symmetric and $0$ is an eigenvalue appearing once with eigenvector ${1_m}$ and so $\operatorname { BP^TP} = B$ with $\operatorname { PBP^T}$ being orthogonally diagonalizable but I can't help myself to show that none of it's eigenvalues are $0 .$
Obs. : Sum of the entries of each row of $\operatorname B$ is $0 .$
A small hint is warmly appreciated .
1) Let $\mathbf{v}= \frac{1}{\sqrt{m}}\mathbf{1}_m$ and \begin{align} A = \begin{pmatrix} \mathbf{v}^T\\ P \end{pmatrix}. \end{align} Since $A$ is orthogonal, then by block multiplication \begin{align} I=A^TA= \begin{pmatrix} \mathbf{v} & P^T \end{pmatrix} \begin{pmatrix} \mathbf{v}^T\\ P \end{pmatrix} = \mathbf{v}\mathbf{v}^T+P^TP. \end{align}
For 2), you will use the fact that \begin{align} \operatorname{rank}(AB)\leq \min\{\operatorname{rank}(A), \operatorname{rank}(B)\}. \end{align} Hence we see that \begin{align} \operatorname{rank}(PBP^T) \le \operatorname{rank}(B) =m-1. \end{align} Moreover, we have that \begin{align} \operatorname{rank}(P^TPBP^TP)\le \operatorname{rank}(PBP^T). \end{align} Since $P^TP$ is invertible from part (1), then it follows \begin{align} \operatorname{rank}(P^TPBP^TP)=\operatorname{rank}(B)=m-1. \end{align}