Woodbury formula problem Hello guys i have this linear algebra problem and i want some help on how i should proceed with it. i think the aim is to use the Woodbury formula
Suppose $u,v \in \mathbb{R}^n, \quad A \in \mathbb{R}^{n*n}$ is invertible and consider for scalars $\alpha , \beta \in \mathbb{R} : \\$ $B(\alpha,\beta)= A +\alpha(uu^T +vv^T ) +\beta(uv^T + vu^T ) \\$ 
 give conditions on $\alpha$ and $\beta$ in terms of $m_{ij}$ where : 
$\begin{bmatrix} m_{11} & m_{12} \\ m_{21} & m_{22}\end{bmatrix}= \begin{bmatrix} u^T \\ v^T \end{bmatrix} A^{-1} \begin{bmatrix} u & v \end{bmatrix} \\$ 
That describes precisely when $B(\alpha,\beta)x=0$ has a nontrivial solution $x \neq 0$.
Thank you.
 A: We start with recalling two standard results:


*

*If $A$ is invertible then $\begin{pmatrix} A & B \\ C & D \end{pmatrix}$ is invertible iff $D-CA^{-1}B$ is invertible. This follows directly from $ \begin{pmatrix} I & 0 \\ -CA^{-1} & I \end{pmatrix} \begin{pmatrix} A & B \\ C & D \end{pmatrix} \begin{pmatrix} I & -A^{-1}B \\ 0 & I \end{pmatrix} = \begin{pmatrix} A & 0 \\ 0 & D - CA^{-1}B \end{pmatrix}.$

*If $A$ is invertible $A+uv^T$ is invertible iff $1+ v^TA^{-1}u \neq 0.$
Now to this problem.
Note $B(\alpha,\beta) = A + \begin{pmatrix} u & v \end{pmatrix} \begin{pmatrix} \alpha & \beta \\ \beta & \alpha \end{pmatrix} \begin{pmatrix} u^T \\ v^T \end{pmatrix}.$
First we find $\alpha,\beta$ such that $B(\alpha,\beta)$ is singular and $\alpha^2 \neq \beta^2.$
If $\alpha^2 \neq \beta^2$ then $\begin{pmatrix} \alpha & \beta \\ \beta & \alpha \end{pmatrix}$ is invertible and its inverse is $\dfrac{1}{\alpha^2 - \beta^2} \begin{pmatrix} \alpha & -\beta \\ -\beta & \alpha \end{pmatrix}.$
Consider the matrix $$ D = \begin{pmatrix}  \dfrac{\alpha}{\alpha^2 - \beta^2} & \dfrac{-\beta}{\alpha^2 - \beta^2} & u^T \\ \dfrac{-\beta}{\alpha^2-\beta^2} & \dfrac{\alpha}{\alpha^2 - \beta^2} & v^T \\
-u & -v & A \end{pmatrix}.$$
Since $\dfrac{1}{\alpha^2 - \beta^2} \begin{pmatrix} \alpha & -\beta \\ -\beta & \alpha \end{pmatrix}$ is the inverse of $\begin{pmatrix} \alpha & \beta \\ \beta & \alpha \end{pmatrix}$ and is hence invertible, $D$ is invertible iff $A + \begin{pmatrix} u & v \end{pmatrix} \begin{pmatrix} \alpha & \beta \\ \beta & \alpha \end{pmatrix} \begin{pmatrix} u^T \\ v^T \end{pmatrix}$ i,e., $B(\alpha,\beta)$ is invertible.
Since $\begin{pmatrix} A & -u & -v \\  u^T & \dfrac{\alpha}{\alpha^2 - \beta^2} & \dfrac{-\beta}{\alpha^2-\beta^2} \\ v^T & \dfrac{-\beta}{\alpha^2-\beta^2} & \dfrac{\alpha}{\alpha^2-\beta^2} \end{pmatrix} = \begin{pmatrix} 0 & 0 & I \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} D \begin{pmatrix} 0 & 1 & 0  \\ 0 &0&1\\I & 0 &0\end{pmatrix}$ can be obtained form permuting rows and columns and $A$ is invertible, $D$ is also invertible iff  $\dfrac{1}{\alpha^2 - \beta^2} \begin{pmatrix} \alpha & -\beta \\ -\beta & \alpha \end{pmatrix} + \begin{pmatrix} u^T \\ v^T\end{pmatrix}A^{-1}\begin{pmatrix}u & v\end{pmatrix} =\begin{pmatrix} \alpha & \beta \\ \beta & \alpha \end{pmatrix}^{-1} + M$ is invertible.
So if if $\alpha^2 \neq \beta^2$ then a necessary and sufficient condition for $B(\alpha,\beta)$ to be singular is that $\texttt{det}( \begin{pmatrix} \alpha & \beta \\ \beta & \alpha \end{pmatrix}^{-1} + M ) = 0$ or equivalently $\texttt{det}( I + \begin{pmatrix} \alpha & \beta \\ \beta & \alpha \end{pmatrix}M ) = 0.$
If $\alpha^2 = \beta^2$ then we we have $\alpha = \pm \beta.$
First consider the case $\alpha = \beta.$ In this case $B(\alpha,\beta) = A + \alpha (u+v) (u + v)^T$ and so $B(\alpha,\beta)$ is singular iff $\alpha (u+v)^T A^{-1} (u + v) + 1 = 0$ i.e. $\alpha \mathbf{1}^T M \mathbf{1}  + 1 = 0$ where $\mathbf{1}$ is a $ 2 \times 1$ vector of 1's. 
If $\alpha = -\beta$ then $B(\alpha,\beta) = A + \alpha(u-v)(u-v)^T$ which leads to then condition that $B(\alpha,\beta)$ is singular iff $\alpha \begin{pmatrix}1 -1 \end{pmatrix} M \begin{pmatrix}1 \\ -1 \end{pmatrix}  + 1 = 0.$
