# Characteristic polynomial of matrix with complex numbers

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.

• 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... Commented Aug 24, 2020 at 7:44
• @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? Commented Aug 24, 2020 at 8:06
• @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 $$\begin{bmatrix} -\frac{R_S+R_R}{L_L} & \frac{R_R}{L_M\cdot L_L}-i\frac{1}{L_L}\omega_m \\ R_R & -\frac{R_R}{L_M}+i\omega_m \end{bmatrix}$$ Commented Aug 24, 2020 at 8:06
• @Steve Your computation is also correct. It is justified in this case because the $2 \times 2$ block-entries commute Commented Aug 24, 2020 at 8:08
• @BenGrossman thank you for your reaction. Does it mean that the substitution mentioned above is in your opinion correct? Commented Aug 24, 2020 at 8:36