Eigenvalues of a complicated matrix

Question

For all $n\in \mathbb{N}$, and any $q=(q_1,q_2)\in\mathbb{R}^2$ we consider the matrix

$$\begin{pmatrix} 2|q|^{2n}+14nq_1^2|q|^{2(n-1)}+8n(n-1)q_1^4|q|^{2(n-2)}&4nq_1q_2|q|^{2(n-1)}+8n(n-1)q_1^3q_2|q|^{2(n-2)}\\4nq_1q_2|q|^{2(n-1)}+8n(n-1)q_1^3q_2|q|^{2(n-2)}&2nq_1^2|q|^{2(n-1)}+8n(n-1)q_1^2q_2^2|q|^{2(n-2)} \end{pmatrix}$$ where $|q|=\sqrt{q_1^2+q_2^2}$

I want to find the real eighenvalues of this matrix.

Please help me to do so. Thanks in advance.

Answer 1

You can simply compute $\det(A-\lambda I)$ and find the roots, it's a quadratic equation, should be simple.

However, your matrix is very ugly. You can avoid a lot of pointless computation by being smart:

$$A=2|q|^{2(n-2)} \begin{pmatrix} |q|^{4}+7nq_1^2|q|^{2}+4n(n-1)q_1^4&2nq_1q_2|q|^2+4n(n-1)q_1^3q_2\\ 2nq_1q_2|q|^{2}+4n(n-1)q_1^3q_2&nq_1^2|q|^{2}+4n(n-1)q_1^2q_2^2 \end{pmatrix}$$ It's still ugly, but not nearly as much as before. Just compute the determinant of the leftover matrix, find its eigenvalues and add the factor back at the end.

EDIT: as the matrix is symmetric, eigenvalues are guaranteed to be real and also the expressions are a little easier to calculate. Te final expression will further simplify when you use $|q|^2=q_1^2+q_2^2$:

For a $2\times 2$ matrix, the characteristic polynomial $\det (B-\lambda I)$ can be more simply written as

$$\lambda^2-\lambda \operatorname{Tr} B + \det B=0$$

I used $B$ for just your simplified matrix without the factor in front. You need:

$$\operatorname{Tr} B=|q|^4+8n q_1^2|q|^2+4n(n-1)q_1^2(q_1^2+q_2^2)=|q|^4+4n(n+1)q_1^2|q|^2$$

$$\det B=\{|q|^{4}+7nq_1^2|q|^{2}+4n(n-1)q_1^4\}\{nq_1^2|q|^{2}+4n(n-1)q_1^2q_2^2\}-\{2nq_1q_2|q|^2+4n(n-1)q_1^3q_2\}^2=$$ $$=|q|^4(nq_1^2|q|^2+4n(n-1)q_1^2q_2^2)+7n^2 q_1^4|q|^4+28n^2(n-1)q_1^4q_2^2|q|^2+4n^2(n-1)q_1^6|q|^2+16n^2(n-1)^2q_1^6q_2^2-4n^2q_1^2q_2^2|q|^4-16n^2(n-1)^2q_1^6q_2^2-16n^2(n-1)q_1^4q_2^2|q|^2=$$ $$=|q|^4(nq_1^2|q|^2-4nq_1^2q_2^2+7n^2 q_1^4)+4n^2(n-1)q_1^4|q|^2(3q_2^2+q_1^2)=$$ $$=n|q|^2q_1^2(q_1^4(1+3n+4n^2)+q_1^2q_2^2(-2-5n+12n^2)-3q_2^4)$$

I hoped for something nicer but that number $7$ makes things hard to simplify.

The eigenvalues don't simplify either.

Answer 2

Let $$B=\begin{pmatrix} |q|^{4}+7nq_1^2|q|^{2}+4n(n-1)q_1^4&2nq_1q_2|q|^4+4n(n-1)q_1^3q_2\\ 2nq_1q_2|q|^{2}+4n(n-1)q_1^3q_2&nq_1^2|q|^{2}+4n(n-1)q_1^2q_2^2 \end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$

The eigenvalues of $A$ are given by $\lambda_1(q)=2|q|^{2(n-2)}\frac{(a+b)+\sqrt{(a+b)^2-4(ab-cd)}}{2}$ $\lambda_2(q)=2|q|^{2(n-2)}\frac{(a+b)-\sqrt{(a+b)^2-4(ab-cd)}}{2}$

We have $$\lambda_1(q)\ge 0,\;\text{for all},\;q\in\mathbb{R}^2$$

I want to know if $\lambda_2(q)\le 0$ for all $q\in\mathbb{R}^2$.