Change in eigenvalues by changing only one entry of a square matrix Consider following square matrix $A$ of order $n$ 
$A=\begin{bmatrix}
0 & a_{12} & a_{13} & a_{14} & \cdots  & a_{1n}  \\ 
a_{21} & 0 & a_{23} & a_{24} & \cdots & a_{2n} \\ 
a_{31} & a_{32} & 0 & a_{34} & \cdots & a_{3n}\\
a_{41} & a_{42} & a_{43} & 0 & \cdots & a_{4n}\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & a_{n3} & a_{n4} & \cdots & 0\\
\end{bmatrix}_n$
Let $A$ is symmetric matrix with eigenvalues $\lambda_i,i=1,2\cdots,n$.
Now consider two matrices $B$ and $C$ with eigenvalues $\mu_i$ and $\gamma_i,i=1,2,\cdots,n$,which are obtained from matrix $A$ by replacing $a_{11}$ entry to $1$ and $-1$ respectively
i.e 
$B=\begin{bmatrix}
1 & a_{12} & a_{13} & a_{14} & \cdots  & a_{1n}  \\ 
a_{21} & 0 & a_{23} & a_{24} & \cdots & a_{2n} \\ 
a_{31} & a_{32} & 0 & a_{34} & \cdots & a_{3n}\\
a_{41} & a_{42} & a_{43} & 0 & \cdots & a_{4n}\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & a_{n3} & a_{n4} & \cdots & 0\\
\end{bmatrix}_n$
$C=\begin{bmatrix}
-1 & a_{12} & a_{13} & a_{14} & \cdots  & a_{1n}  \\ 
a_{21} & 0 & a_{23} & a_{24} & \cdots & a_{2n} \\ 
a_{31} & a_{32} & 0 & a_{34} & \cdots & a_{3n}\\
a_{41} & a_{42} & a_{43} & 0 & \cdots & a_{4n}\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & a_{n3} & a_{n4} & \cdots & 0\\
\end{bmatrix}_n$
My questions are
$1.$Can we find any relation between $\lambda_i$ and $\mu_i$?
$2.$Can we find any relation between  $\lambda_i$ and $\gamma_i$?
$3.$Can we find any $\sum_{i=1}^{n} \left|\mu_i\right|$ in terms of $\lambda_i$?
$4.$Can we find any $\sum_{i=1}^{n} \left|\gamma_i\right|$ in terms of $\lambda_i$?
 A: As a result, the relation exists between each eigenvalues. At first, let replace $B$ and $C$ with $A'$ matrix and define its eigenvalues $\lambda'$. The matrix $A'$ can be described as follows:
$$A'=A\pm\mathbf{vv^{\text{T}}}$$
where vector $\mathbf{v}$ has all zero except first element as one:
$$
\mathbf{v}=
\begin{bmatrix}
1 & 0 & 0 & \cdots & 0
\end{bmatrix}
^{\text{T}}
$$
Then, the determinant can be calculated using this lemma:
$$
\begin{aligned}
\det{(A')}=&\det{(A\pm\mathbf{vv^{\text{T}}})} \\
=&(1\pm\mathbf{v^{\text{T}}}A^{-1}\mathbf{v})\det{(A)}
\end{aligned}
$$
Generally, arbitrary matrix determinant equals product of its all eigenvalues. Additionally, the quadratic form $\mathbf{v^{\text{T}}}A^{-1}\mathbf{v}$ includes only upper left element of inverse matrix $A^{-1}$. Hence, above determinant can be expressed following equation:
$$
\prod_{i=1}^{n}\lambda'_{i}=
\prod_{i=1}^{n}\lambda_{i}\pm\Delta_{11}
$$
where $\Delta_{11}$ is top left cofactor matrix. By the way, relation of eigenvalues can be also expressed using trace operation. Trace has an important property that the value equals summation of all eigenvalues. Therefore, the relation is:
$$
\begin{aligned}
\text{tr}(A')=&\text{tr}(A\pm\mathbf{vv^{\text{T}}}) \\
\sum_{i=1}^{n} \lambda_{i}'=&\sum_{i=1}^{n} \lambda_{i} \pm 1
\end{aligned}
$$
However, actual trace of matrix $\text{tr}(A)$ is zero so that above equation can be divided below cases:
$$
\begin{cases}
\displaystyle
\sum_{i=1}^{n} \lambda_{i}=0 \\
\displaystyle
\sum_{i=1}^{n} \lambda_{i}'=\pm 1
\end{cases}
$$
