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I have been attempting to determine characteristic polynomial of below given matrix

$ \mathbf{A} = \begin{bmatrix} -\frac{R_S+R_R}{L_L}\cdot\mathbf{E}_2 & \frac{R_R}{L_M\cdot L_L}\cdot\mathbf{E}_2-\frac{1}{L_L}\omega_m\cdot\mathbf{J} \\ R_R\cdot\mathbf{E}_2 & -\frac{R_R}{L_M}\cdot\mathbf{E}_2+\omega_m\cdot\mathbf{J} \end{bmatrix} $

where

$ \mathbf{E}_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $

$ \mathbf{J} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} $

I have found following derivation

$$ \begin{align} \mathbf{A}-\lambda\cdot\mathbf{E}_4 = \begin{bmatrix} -\frac{R_S+R_R}{L_L}\cdot\mathbf{E}_2 & \frac{R_R}{L_M\cdot L_L}\cdot\mathbf{E}_2-\frac{1}{L_L}\omega_m\cdot\mathbf{J} \\ R_R\cdot\mathbf{E}_2 & -\frac{R_R}{L_M}\cdot\mathbf{E}_2+\omega_m\cdot\mathbf{J} \end{bmatrix} - \begin{bmatrix} \lambda\cdot\mathbf{E}_2 & \mathbf{Z} \\ \mathbf{Z} & \lambda\cdot\mathbf{E}_2 \end{bmatrix} = \begin{bmatrix} -\frac{R_S+R_R}{L_L}\cdot\mathbf{E}_2-\lambda\cdot\mathbf{E}_2 & -\frac{1}{L_L}\left(-\frac{R_R}{L_M}\cdot\mathbf{E}_2+\omega_m\cdot\mathbf{J}\right) \\ R_R\cdot\mathbf{E}_2 & \left(-\frac{R_R}{L_M}\cdot\mathbf{E}_2+\omega_m\cdot\mathbf{J}-\lambda\cdot\mathbf{E}_2\right) \end{bmatrix} \end{align} $$

where

$ \mathbf{Z} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} $

$ \mathbf{E}_4 = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $

Then below given substitution has been used

$$ \delta=\frac{1}{L_L} $$

$$ a=-\frac{R_S+R_R}{L_L}\cdot\mathbf{E}_2 $$

$$ b=-\frac{R_R}{L_M}\cdot\mathbf{E}_2+\omega_m\cdot\mathbf{J} $$

$$ c=R_R\cdot\mathbf{E}_2 $$

The substitution resulted in following matrix

$$ \begin{align} \mathbf{A}-\lambda\cdot\mathbf{E}_4 = \begin{bmatrix} a-\lambda & -\delta\cdot b \\ c & b-\lambda \end{bmatrix} \end{align} $$

Based on last matrix following characteristic polynomial has been written

$$ \lambda^2-(a+b)\cdot\lambda+(a+\delta\cdot c)\cdot b $$

I have doubts regarding the mathematical correctness of the above mentioned derivation. The problem which I have is related to the substitution in the elements $-\frac{R_S+R_R}{L_L}\cdot\mathbf{E}_2-\lambda\cdot\mathbf{E}_2 = a-\lambda$ and $\left(-\frac{R_R}{L_M}\cdot\mathbf{E}_2+\omega_m\cdot\mathbf{J}-\lambda\cdot\mathbf{E}_2\right) = b-\lambda$. Does anybody have any idea how to find the characteristic polynomial in mathematical correct way? Thanks in advance for any ideas.

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  • $\begingroup$ The matrix $A$ is of dimension $4$ if I understand well. Hence its characteristic polynomial is of degree $4$. Computing it by hand won’t be a special fun... $\endgroup$
    – mathcounterexamples.net
    Commented Aug 24, 2020 at 7:44
  • $\begingroup$ @mathcounterexamples.net thank you for your reaction. Yes, you are correct. The matrix $\mathbf{A}$ is $4\times 4$ matrix. What do you think about the mathematical correctness of the above mentioned derivation of the characteristic polynomial? $\endgroup$
    – Steve
    Commented Aug 24, 2020 at 8:06
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    $\begingroup$ @Steve You can use the exact same trick that I describe in my answer to your previous question. That is, it suffices to calculate the eigenvalues of the complex matrix $$ \begin{bmatrix} -\frac{R_S+R_R}{L_L} & \frac{R_R}{L_M\cdot L_L}-i\frac{1}{L_L}\omega_m \\ R_R & -\frac{R_R}{L_M}+i\omega_m \end{bmatrix} $$ $\endgroup$
    – Ben Grossmann
    Commented Aug 24, 2020 at 8:06
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    $\begingroup$ @Steve Your computation is also correct. It is justified in this case because the $2 \times 2$ block-entries commute $\endgroup$
    – Ben Grossmann
    Commented Aug 24, 2020 at 8:08
  • $\begingroup$ @BenGrossman thank you for your reaction. Does it mean that the substitution mentioned above is in your opinion correct? $\endgroup$
    – Steve
    Commented Aug 24, 2020 at 8:36

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