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How to find eigenvalues of following matrix?

$\begin{bmatrix} mkI-A & -A & -A & \cdots & -A\\ -A & kI-A & O & \cdots & O\\ -A & O & kI-A & \cdots & O\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ -A & O & O & \cdots & kI-A\\ \end{bmatrix}_{m+1}$

Where, $A$ is any square matrix of order $n$,

$O$ is zero matrix of order $n$

$I$ is identity matrix of order $n$,

$k \in \mathbb{N}$

As well as one eigenvalue of above matrix is zero.

I think if we can convert above matrix in terms of either Kronecker product or Kronecker sum of two matrices then we can find eigenvalues of above matrix by taking multiplication or addition of two matrices respectively.

The other way is might be if we can convert above matrix in block diagonal matrix then we can find eigenvalues easily.

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    $\begingroup$ If you subtract $kI$ from any matrix, it will subtract $k$ from all of the eigenvalues, so your problem reduces to $\left[\matrix{(m-1)kI-A & -A & -A &\cdots&-A\cr -A&-A&0&\cdots&0\cr -A&0&-A&\cdots&0\cr \vdots&\vdots&\vdots&\ddots&\vdots\cr -A&0&0&\cdots&-A\cr}\right]$. $\endgroup$ – Christopher Carl Heckman Apr 25 '16 at 4:00
  • $\begingroup$ ya ok then what I have to do solve it further $\endgroup$ – kalpeshmpopat Apr 25 '16 at 5:31
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At first, to summarize this problem, let replace diagonal factor matrices as:

$$ \begin{bmatrix} X \\ Y \end{bmatrix} = \begin{bmatrix} \lambda-mk & 1 \\ \lambda-k & 1 \end{bmatrix} \begin{bmatrix} I \\ A \end{bmatrix} $$

where $\lambda$ is eigenvalue. Therefore, we can suppose the following determinant:

$$ \begin{aligned} \det \begin{bmatrix} X & A & A & \cdots & A & A \\ A & Y & O & \cdots & O & O\\ A & O & Y & \cdots & O & O \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ A & O & O & \cdots & Y & O \\ A & O & O & \cdots & O & Y \end{bmatrix} \end{aligned} $$

Calculation determinant has an important property that each rows can be added or reduced from another row. Thus, top row can be subtracted from bottom row to extract the diagonal factor as $\det(Y)$. Then the block matrix becomes smaller from $(m+1)n$ square to $mn$ square:

$$ \begin{aligned} \det \begin{bmatrix} X-AY^{-1}A & A & A & \cdots & A & O \\ A & Y & O & \cdots & O & O\\ A & O & Y & \cdots & O & O \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ A & O & O & \cdots & Y & O \\ A & O & O & \cdots & O & Y \end{bmatrix} =& \det \begin{bmatrix} X-AY^{-1}A & A & A & \cdots & A \\ A & Y & O & \cdots & O \\ A & O & Y & \cdots & O \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ A & O & O & \cdots & Y \end{bmatrix} \det(Y) \end{aligned} $$

Likewise, we repeat this operation until the block matrix becomes $n$ orders matrix. Eventually, the below determinant is obtained:

$$ \det(Y)^m\det(X-mAY^{-1}A) $$

Now, we attempt to factorize the second determinant $\det(X-mAY^{-1}A)$ using this consideration. Then the whole result is shown below:

$$ \det(\lambda I-(kI+A))^{m-1} \det\biggl(\lambda I-\cfrac{B+\sqrt{C}}{2}\biggr) \det\biggl(\lambda I-\cfrac{B-\sqrt{C}}{2}\biggr) $$

where:

$$ \begin{cases} B=-2A+k(m+1)I \\ \\ C=4mA^2-5k(m+1)A+k^2(1+m)^3I \end{cases} $$

In conclusion, this $n(m+1)$ square matrix has $3n$ kinds eigenvalues. However, $C=O$ case is $2n$ kinds exceptionally because of multiple solution.

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