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

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.

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 $$A=Q \begin{bmatrix} \lambda_1& & &\\ &-\lambda_1& &\\ & & \lambda_2 & \\ & & & -\lambda_2 \end{bmatrix} Q^{-1}.$$
• Thank you very much! Could you explain why the matrix $M = BB^{\rm{T}}+\lambda_i(B-B^{\rm{T}})+(\gamma^{-1}-\lambda_i^2)\,{\rm{I}_2}$ is of rank one? And why do you consider the two equations $M w_{ic} = 0$ and $w_{ir}^T M = 0$? Mar 28 '20 at 10:54
• @soundwave I edited my answer; I want $BB^{\rm{T}}+\lambda(B-B^{\rm{T}})+(\gamma^{-1}-\lambda^2)\,{\rm{I}_2}$ to be singular, so the rank is $0$ or $1$. The $\lambda_i$'s are defined to be the roots of the quartic equation given by the vanishing of the determinant. Once you have such a $\lambda_i$, you use my computations of eigenvectors in the beginning of the solution to express the upper part $\mathbf{v}_i$ in terms of the lower part $\mathbf{w}_i$ that should be a non-zero (column) vector in the null space of $BB^{\rm{T}}+\lambda_i(B-B^{\rm{T}})+(\gamma^{-1}-\lambda_i^2)\,{\rm{I}_2}$. Mar 28 '20 at 22:16
• Finally, notice that $-\lambda_i$ is also among the eigenvalues, and substituting $\lambda_i$ with its opposite in $BB^{\rm{T}}+\lambda_i(B-B^{\rm{T}})+(\gamma^{-1}-\lambda_i^2)\,{\rm{I}_2}$ results in the transpose matrix. A vector in the column null space of the transpose is the transpose of a vector in the row null space of the original matrix. That's where $\mathbf{w}_{ic}$ and $\mathbf{w}_{ir}^{\rm{T}}$ come from. Mar 28 '20 at 22:24
• Thank you for the explanation! With "column null space" and "row null space" do you mean, respectively, the cokernel and the kernel? I'm reading that the kernel of a matrix $A$ coincides with the cokernel of $A^\rm T$, and vice versa the cokernel of $A$ coincides with the kernel of $A^\rm T$. But wouldn't this mean that $w_{ic} = w_{ir}$? I'm a bit confused since I don't understand very well. Mar 29 '20 at 15:13
• @soundwave By the "column null space" I mean the kernel as if you think of the matrix as a linear map; i.e. the subspace of vectors $\mathbf{w}$ with $A\mathbf{w}=\mathbf{0}$. The "row null space" is defined by $\{\mathbf{w}|\mathbf{w}^{\rm{T}}A=\mathbf{0}\}$. What I do is very concrete, I'm picking vectors with these properties. As for the linear algebra fact you mentioned, you must be careful about what you mean by "coincide". Do you mean an isomorphism? In that case you don't get the same spaces, but spaces that may be identified. And again, there is no reason to appeal to such a thing. Apr 2 '20 at 0:48