If I know the eigenvalues of matrix $A$, is there any way to calculate eigenvalues of $B$ using some tricks? Here $A$ is an $m\times m$, $O$ is the zero matrix and $I$ denotes $m \times m$ identity matrix. I want to calculate the eigenvalues of the matrix $B$.
$B = \left( \begin{array}{ccc} A & I-kI \\ I & O \end{array} \right) $.
Here $k$ is any positive integer. I tried to calculate the eigenvalues of the matrix $B$ by taking $k=1$. But I could not anything relevant. Is it possible to find any relation between eigenvalues of $B$ and eigenvalues of $A$? What can be said if $A$ is a symmetric matrix? Thanks in advance
The eigenvalues of B can be found by the expression:
$$ \det \left( \left( \begin{array}{cc} A & I-kI \\ I & O \end{array} \right) - \left( \begin{array}{cc} \lambda I & O \\ O & \lambda I \end{array} \right) \right) = \det \left( \begin{array}{cc} A-\lambda I & I-kI \\ I & -\lambda I \end{array} \right) = 0$$
There are relations for calculating the determinant of a block matrix:
$$\det\begin{pmatrix}A& B\\ C& D\end{pmatrix} = \det(A)\times \det\left(D - C A^{-1} B\right) = \det(D)\times \det\left(A - B D^{-1} C\right)$$
$$ \det \left( \begin{array}{cc} A-\lambda I & I-kI \\ I & -\lambda I \end{array} \right) = \det(A-\lambda I)\times \det\left(-\lambda I - (A-\lambda I)^{-1} (I-kI)\right) = \det(-\lambda I)\times \det\left( (A-\lambda I) - (I-kI) (-\lambda I)^{-1} \right) = 0$$
Looking into the first equation, when $\lambda$ is also an eigenvalue of $A$, $\det(A-\lambda I)$ is equal to zero which means that $(A-\lambda I)$ is not invertible and this equation can't be used.
On the other hand, the second equation is only valid if $\lambda$ is different from zero. So if all the eigenvalues of $A$ are different from zero and $k=1$, the eigenvalues of $A$ are also eigenvalues of $B$.