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I have matrices $A\in\mathbb{R}^{n\times n}$, and need to calculate their Eigenvalues. The matrices $A$ are of this form: $A:=\left(\begin{array}{c} c_{1}a\\ c_{2}a\\ \vdots \\ c_{n-m}a \\ b_{1} \\b_{2} \\ \vdots \\ b_{m} \end{array}\right) $with row vectors $a,b_{i}\in\mathbb{R}^{1 \times n}$ and factors $c_{i}\in\mathbb{R}$.

Can I simplify the calculation of their eigenvalues somehow?

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Reasoning on $A^T$ which has the same spectrum as $A$, denoting $V:=a^T$ and $V_k:=b_k$:

$$A^T=[c_1V,c_2V,\cdots c_{n-m}V,V_1,V_2,\cdots,bV_m]$$

Then $A^T$ has $0$ as an eigenvalue with multiplicity $(n-m-1)$, the eigenspace associated with eigenvalue $0$ (i.e., the kernel of $A$) being:

$$\pmatrix{-c_2\\c_1\\0\\\vdots\\0}, \pmatrix{0\\-c_3\\c_2\\\vdots\\0}, \cdots \pmatrix{0\\ \cdots \\0\\-c_n\\c_{n-m-1}}$$

And I think this is all that can be said about the spectrum of $A$.

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    how can I deduce, that $A^{T}$ has eigenvalue 0 with multiplicity $(n-m-1)$? – 2017-02-21
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    Because, if you apply $A^T$ to the vectors I have given, for example the first, you get $[c_1V,c_2V,\cdots c_{n-m}V,V_1,V_2,\cdots,bV_m]\pmatrix{-c_2\\c_1\\0\\\vdots\\0}=-c_2c_1V+c_2c_1V+0+...=0=0\pmatrix{-c_2\\c_1\\0\\\vdots\\0}$ which means that the first vector is an eigenvector associated with eigenvalue $0$. (i.e., belongs to the kernel of $A$).The same for the others. – 2017-02-21