What will be eigenvalues of this supermatrix Let $A$ be a symmetric matrix whose eigenvalues are known. We get supermatrix $B$ by adding a new row $r^T$ having only one nonzero entry, and a new column $r$ to $A$.
What is relation between eigenvalues of $A,B$?
 A: Let $A$ be a $n \times n$ matrix, $v_1, ..., v_n$ are eigenvectors (orthogonal) of $A$ and $\lambda_1,...,\lambda_n$ are eigenvalues of $A$ respectfully (counted with multiplicity). 
First, assume that $r^T = (\vec{0}^T,1)$, where $\vec{0}^T = (\overbrace{0,..,0}^{n })$. Hence,
$$B = \begin{bmatrix} A && \vec{0} \\
                       \vec{0}^T && 1\end{bmatrix},$$
consider $v_i' = \begin{bmatrix} v_i  \\
                       0 \end{bmatrix},$ therefore
$$Bv_i' = \lambda_iv_i',$$
so, in this case all eigenvalues of $A$ are eigenvalues of $B$.
Second, assume that $r^T = (q^T, 0)$, where $q \in \mathbb{R^n}$. Hence,
$$B = \begin{bmatrix} A && q \\
                       q^T && 0\end{bmatrix},$$
Now, let's assume that $\exists k:q=v_k$. Consider $v_i' = \begin{bmatrix} v_i  \\
                       0 \end{bmatrix}$, where $i\neq k$, therefore 
$$Bv_i' = \lambda_iv_i',$$
so, in this case all eigenvalues of $A$ except one are eigenvalues of $B$. 
This result is good, but let's go a bit further. Let $\gamma \in \mathbb{R}$ and $\gamma$ is not an eigenvalue of $A$. We have 
$$\det(B-\gamma I) = det\begin{bmatrix} A - \gamma I && v_k \\
                       v_k^T && -\gamma\end{bmatrix},$$
since matrix $(A-\gamma I)$ is invertible we can apply the following formula
$$\det \begin{bmatrix} A && B \\
                       C && D \end{bmatrix} = \det(A)\det(D - CA^{-1}B).$$
Hence,
$$det\begin{bmatrix} A - \gamma I && v_k \\
                       v_k^T && -\gamma\end{bmatrix} = \det(A-\gamma I)(-\gamma-v_k^T(A-\gamma I)^{-1}v_k) = 0,$$
and we obtained the equation for eigenvalue $\gamma$ of $B$
$$-\gamma-v_k^T(A-\gamma I)^{-1}v_k = 0$$
Now let's have a look a the equation, using the fact that $A=U\Lambda U^T$, where $UU^T=U^TU=I$ and $\Lambda = diag(\lambda_1,...,\lambda_n)$ we rewrite the equation as it follows
$$-\gamma = v_k^TU^T(\Lambda-\gamma I)^{-1}Uv_k.$$
Since columns of $U$ are eigenvectors of $A$ and $v_i^Tv_k=\delta_{ik}$, we have
$$\gamma = \frac{1}{\gamma-\lambda_k},$$
Basically this is a second order polynom that gives you two additional eigenvalues.
You can extend this result for the case when $q = \sum_{i=1}^k\alpha_i v_i.$
Best.
