Matrix with different dimensions and same eigenvalues (multiplicity)

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Suppose we have two positive definite symmetric matrices $A\in\mathbb{R}^{m\times m}$ and $B\in\mathbb{R}^{(m+1)\times (m+1)}$. In particular, $B\equiv \left( \begin{matrix} a_{11} & ... & a_{1n} & b \\ a_{21} & a_{22} & ... & a_{1n} \\ a_{31} & a_{32} & ... & a_{1(n-1)} \\ ... & ... & ... & ... \\ b & a_{1n} & a_{1(n-1)} & ... \end{matrix} \right)$

that is the north-west part of $B$ is $A$, while the last column and row are constructed with adding the two $b$ and stacking the last column and row of $A$ in order to reach the $m+1$ rank. If all the eigenvalues are the same, in particular the last eigenvalue of $B$ has algebraic multiplicity 2 (so $\lambda_m det(A)= \prod^m \lambda_i \lambda_m = \lambda_m^2 \prod^{m-1}\lambda_i = det(B)$), can we say something about the value of the entry $b$? Is it zero?

Plus, is it trivial to conclude that $A$ and $B$ share the same eigenvalues put to $m$ (and with the additional assumption $\lambda_{m+1}=\lambda_m$ they have the same eigenvalues and determinants)? Is $b=0$ a sufficient condition to guarantee such properties?