Spectral decomposition of a $4\times4$ real nonsymmetric matrix with unknown elements

Question

I'm trying to eigendecompose the following matrix $A$, i.e. to find $Q$ and $\Lambda$ such that $$ A = \begin{bmatrix} -\alpha & \alpha & -\gamma^{-1} & 0\\ \beta & -\beta & 0 & -\gamma^{-1}\\ -1 & 0 & \alpha & -\beta\\ 0 & -1 & -\alpha & \beta \end{bmatrix}=Q\Lambda Q^{-1} $$ where $\alpha>\beta>0$ and $\gamma>0$.

Notice that

• $A$ is a Hamiltonian matrix, i.e. $JA$ is symmetric where $J=\begin{pmatrix}0 & I_2 \\ -I_2 & 0\end{pmatrix}$, $I_2$ is the $2\times2$ identity matrix.
• The characteristic polynomial of a real Hamiltonian matrix is even. Thus, if $λ$ is an eigenvalue of $A$, then $−λ$, $\bar λ$ and $−\bar λ$ are also eigenvalues. It follows that $\text{trace} A=0$.
• $A$ can be written in block notation: $A=\begin{pmatrix}B & -\gamma^{-1}I_2 \\ -I_2 & -B^T \end{pmatrix}$, with $B=\begin{pmatrix}-\alpha & \alpha \\ \beta & -\beta\end{pmatrix}$.
• the elements on the antidiagonal are all $0$.
• sum of row $1$, sum of row $2$, sum of column $3$ and sum of column $4$ are all equal to $-\gamma^{-1}$.
• sum of row $3$ and sum of column $2$ are equal to $\alpha-\beta-1$.
• sum of row $4$ and sum of column $1$ are equal to $\beta-\alpha-1$.

can we use these facts to find $Q$ and $\Lambda$?

Moreover, the characteristic polynomial is $$ p_A(\lambda) = \lambda^4-[(\alpha+\beta)^2+2\gamma^{-1}]\lambda^2+\gamma^{-1}(2\alpha^2+2\beta^2+\gamma^{-1}) $$ hence $$ 2\lambda^2 = (\alpha+\beta)^2+2\gamma^{-1} \pm \sqrt{(\alpha+\beta)^4-4\gamma^{-1}(\alpha-\beta)^2} $$ and the eigenvalues are either real or complex depending on the values of the parameters.

Answer 1

I don't know how explicit or simple you want the expressions for $\Lambda$ and $Q$ to be, but here is a description. Using your notations, we write $A$ as $\begin{bmatrix} B&-\gamma^{-1}{\rm{I}}_2\\ -{\rm{I}}_2&-B^{\rm{T}} \end{bmatrix}$ where $B=\begin{bmatrix} -\alpha&\alpha\\ \beta&-\beta \end{bmatrix}$. Let $\lambda$ be an eigenvalue and $\begin{bmatrix} \mathbf{v}_{2\times 1}\\ \mathbf{w}_{2\times 1} \end{bmatrix}$ an eigenvector for it. We have $$ \begin{bmatrix} B&-\gamma^{-1}{\rm{I}}_2\\ -{\rm{I}}_2&-B^{\rm{T}} \end{bmatrix} \begin{bmatrix} \mathbf{v}\\ \mathbf{w} \end{bmatrix} =\lambda \begin{bmatrix} \mathbf{v}\\ \mathbf{w} \end{bmatrix}\Rightarrow \begin{cases} B\mathbf{v}-\gamma^{-1} \mathbf{w}=\lambda\mathbf{v}\\ -\mathbf{v}-B^{\rm{T}}\mathbf{w}=\lambda\mathbf{w} \end{cases} $$ Solving the second equation for $\mathbf{v}$ and substituting in the first equation yields $$ \mathbf{v}=-\left(B^{\rm{T}}+\lambda{\rm{I}_2}\right)\mathbf{w},\,\quad \left(BB^{\rm{T}}+\lambda(B-B^{\rm{T}})+(\gamma^{-1}-\lambda^2)\,{\rm{I}_2}\right)\mathbf{w}=\mathbf{0}. $$ Thus the eigenvalues are the roots of the quartic $$ {\rm{det}}\left(BB^{\rm{T}}+\lambda(B-B^{\rm{T}})+(\gamma^{-1}-\lambda^2)\,{\rm{I}_2}\right)=0. $$ Notice that changing $\lambda$ to $-\lambda$ changes the matrix $BB^{\rm{T}}+\lambda(B-B^{\rm{T}})+(\gamma^{-1}-\lambda^2)\,{\rm{I}_2}$ to its transpose. Hence (as you mentioned) if $\lambda$ is an eigenvalue, so is $-\lambda$. Writing the roots of the quartic as $\lambda_1,-\lambda_1$ and $\lambda_2,-\lambda_2$, for each $\lambda_i$ the $2\times 2$ matrix $BB^{\rm{T}}+\lambda_i(B-B^{\rm{T}})+(\gamma^{-1}-\lambda_i^2)\,{\rm{I}_2}$ is not full rank and thus there are non-zero column vectors $\mathbf{w}_{ir}$ and $\mathbf{w}_{ic}$ ($i\in\{1,2\}$) with $$ \left(BB^{\rm{T}}+\lambda(B-B^{\rm{T}})+(\gamma^{-1}-\lambda^2)\,{\rm{I}_2}\right)\mathbf{w}_{ic}=\mathbf{0}\quad \mathbf{w}_{ir}^{\rm{T}}\left(BB^{\rm{T}}+\lambda(B-B^{\rm{T}})+(\gamma^{-1}-\lambda^2)\,{\rm{I}_2}\right)=\mathbf{0} $$ Plugging into previous equations, the corresponding eigenvalues could be computed. The result is $$ Q=\begin{bmatrix} -\left(B^{\rm{T}}+\lambda_1{\rm{I}_2}\right)\mathbf{w}_{1c}& -\left(B^{\rm{T}}-\lambda_1{\rm{I}_2}\right)\mathbf{w}_{1r}&-\left(B^{\rm{T}}+\lambda_2{\rm{I}_2}\right)\mathbf{w}_{2c}& -\left(B^{\rm{T}}-\lambda_2{\rm{I}_2}\right)\mathbf{w}_{2r}\\ \mathbf{w}_{1c}&\mathbf{w}_{1r}&\mathbf{w}_{2c}&\mathbf{w}_{2r} \end{bmatrix} $$ which satisfies $$ A=Q \begin{bmatrix} \lambda_1& & &\\ &-\lambda_1& &\\ & & \lambda_2 & \\ & & & -\lambda_2 \end{bmatrix} Q^{-1}. $$