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I just want to know is there any short trick to get all the eigenvalues of the below matrix ? I am expecting some easy way due to its blocks structure, which I don't know much about.

$A=\begin{bmatrix} a & b&e&0\\ c& d&0&e\\ f & 0&0&0\\ 0& g&0&0 \end{bmatrix}$

Thanks in advance. Any help/hint will be appreciated.

user602672
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1 Answers1

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I don't think there are any tricks, but the calculation can be made slightly shorter by using block matrices. Let $X=\pmatrix{a&b\\ c&d}$ and $Y=\pmatrix{f&0\\ 0&g}$. Then $\lambda I-A=\pmatrix{\lambda I-X&-eI\\ -Y&\lambda I}$. Since the two sub-blocks on the bottom row commute, we have \begin{aligned} \det(\lambda I-A) &=\det\left((\lambda I-X)(\lambda I)-(-eI)(-Y)\right)\\ &=\det(\lambda^2 I-\lambda X-eY)\\ &=(\lambda^2-\lambda a-ef)(\lambda^2-\lambda d-eg)-(-\lambda b)(-\lambda c). \end{aligned} This is a quartic polynomial in $\lambda$. There is, in theory, an explicit radical formula for its zeroes.

user1551
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