Is it possible to determine the eigenvalues of this rather complicated matrix? How many eigenvalues are stable/ unstable

eigenvalues-eigenvectorsmatrices

Without giving you the lengthy and really nasty computations, let $n\in\mathbb{N}$, $\beta(n):=\frac{n(n+1)(2n+1)}{6}$ and $\alpha(n):=\sqrt{1/\beta(n)}$. The $(n\times n)$-matrix $A(n)$ is then given by the following entries:

The first row of $A(n)$ is given by

\begin{align*}
a_{1,i}=\frac{a(n)}{\beta(n)}\cdot\begin{cases}-\beta(n)-1, & i=1\\
-1, & 2\leqslant i\leqslant n-1\\
n^2+2n, & i=n
\end{cases}
\end{align*}
The last row is given by
\begin{align*}
a_{n,i}=\frac{a(n)}{\beta(n)}\cdot\begin{cases}n(n^2+2n-\beta(n)), & i=n\\
n(\beta(n)-1), & i=n-1\\
-n, & \textrm{else}
\end{cases}
\end{align*}
And, for $2\leqslant j\leqslant n-1$ and $1\leqslant i\leqslant n$, the entries are
\begin{align*}
a_{j,i}=\frac{a(n)}{\beta(n)}\cdot\begin{cases}-j(1+\beta(n)), & i=j\\
j(\beta(n)-1), & i=j-1\\
-j, & i=j+1<n\\
n^3+n^2-2n, & i=j+1=n\\
-j, & 1\leqslant i\leqslant j-2\\
jn(n+2), & j+2\leqslant i=n\\
-j, & j+2\leqslant i<n
\end{cases}
\end{align*}

Question: Is there any chance to determine the eigenvalues of this matrix? Or, if not determining explicitly, is it possible to say how many of the eigenvalues are stable/ unstable?

I observe some similarities between the entries.

Let me illustrate this for $n=4$. Without computing the concrete values, the matrix is

$$
A(n)=\frac{\alpha(n)}{\beta(n)}\underbrace{\begin{pmatrix}
\color{green}{-(1+\beta(n))} & \color{red}{-1} & \color{red}{-1} & \color{magenta}{n(n+2)}\\
\color{blue}{j(\beta(n)-1)} & \color{green}{-j(1+\beta(n))} & \color{red}{-j} & \color{magenta}{jn(n+2)}\\
\color{red}{-j} & \color{blue}{j(\beta(n)-1)} & \color{green}{-j(1+\beta(n))} & \color{magenta}{(n-1)n(n+2)}\\
\color{red}{-n} & \color{red}{-n} & \color{blue}{n(\beta(n)-1)} & n(n^2+2n-\beta(n))
\end{pmatrix}}_{=C(n)}
$$

  • The red entries are all of the form $-j$, where $j$ is the row.

  • The lower off-diagonal elements (blue) are all of the form $j(\beta(n)-1)$.

  • The diagonal entries in green are all of the form $-j(1+\beta(n))$.

    • The entries in the last column in magenta are all of the form $jn(n+2)$.

So the only entry which "marches to a different drummer" seems to be $c_{nn}$.

Maybe these observations are helpful to determine the eigenvalues of $C(n)$.

Maybe Laplace expansion? Maybe Gauss elimination?

Update (due to discussion in the comments)

  1. In the meantime, it turned out that $\vec{c}(n):=(1,2,\ldots,n)$ is an eigenvector of $C(n)$ to the corresponding positive eigenvalue $\lambda=2\beta(n)$. So there exists at least one unstable eigenvalue.

  2. Note also that $C(n)$ can be written as the product
    $$
    C(n)=\textrm{diag}(1,2,\ldots,n)\cdot C'(n),
    $$
    where $C'(n)$ is much easier than $C(n)$ because the row-dependent factor $j$ disappears. Maybe this is helpful.

For example, due to this product structure one might estimate the Gershgorin circles, as mentioned here.

Best Answer

EDIT : you can factorize the whole matrix by $6$, replacing $b(n)$ by $b(n)/6$ (which is an integer).

Well, I've tried writing off the matrix for $n=4$ (calculating the values after factorization), and get (if I'm correct)

$$\begin{pmatrix}-31&-1&-1&24\\58& -62 & -2& 48 \\ -3 & 87 & -93 & 72 \\ -4 & -4 & 116 & -24 \end{pmatrix}$$

for which WolframAlpha gave me the following eigenvalues : $(-120, -90, -60, 60)$. As you can see, only one is positive, so your intuition might happen to be true.

Actually, the eigenvector for the positive eigenvalue is $(1,2,3,4)$ so I wonder to which extent the vector $(1,\dots, n)$ can be an eigenvector for all $n$. Maybe it is only a happy coincidence, or maybe there is something deeper behind it.

I will try to edit this answer as soon as I get time to explore further. Please let met know if you advance somehow.

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