Finding root of a cubic equation.

Question

I was solving a physics statistical mechanics problem of an interacting system.

In that question, I have to find the eigenvalues of a matrix P whose elements are given by $$P= \begin{bmatrix} e^{x} & 1 & e^{-x}\\ 1 & 1 & 1\\ e^{-x} & 1 & e^{x} \end{bmatrix}$$ I obtained the eigenvalue equation as $$\lambda^3-\lambda^2(e^{2x} - e^{-x}+1) - \lambda(e^{-2x} - e^{2x} - 2e^{x}+1) - (e^{2x} - e^{-2x} - 2e^{x} + 2e^{-x})=0$$

I cannot go further; the solution in the textbook is stated without showing how to calculate the roots.
The eigenvalues are $$\lambda_1 = 2\sinh (x)$$ $$\lambda_{2,3}:=\frac{1}{2}\bigg[1+2\cosh(x)\pm\sqrt{4\cosh^{2}(x)-4\cosh(x)+9}\bigg]$$

I tried finding info on how to get eigenvalues of a symmetric, definite matrix. I found that for such matrices, the eigenvalue is bounded by the lowest sum of a row and the highest sum of the row. But since the sum of rows is distinct, I can't use that to deduce the exact value of the eigenvalue.
Can anyone help me how to solve this?

Answer 1

Given the form of the answer, it may be easier to work with $\sinh$ and $\cosh$ from the start. To save space, I'm going to write $c$ and $s$ for $\cosh x$ and $\sinh x$. A bit of determinant manipulation gives \begin{eqnarray} \left|\begin{matrix} c + s - \lambda & 1 & c - s \\ 1 & 1-\lambda & 1 \\ c - s & 1 & c + s -\lambda\end{matrix}\right| &=& \left|\begin{matrix} c + s - \lambda & 1 & c - s \\ 1 & 1-\lambda & 1 \\ \lambda - 2s & 0 & 2s -\lambda \end{matrix}\right| \\&=& (\lambda - 2s)\left(\left|\begin{matrix}1 & c-s\\1-\lambda & 1\end{matrix}\right|-\left|\begin{matrix}c+s-\lambda & 1 \\1 & 1-\lambda\end{matrix}\right|\right) \\ &=&(\lambda-2s)[1 - (1-\lambda)(c-s)-(c+s-\lambda)(1-\lambda) + 1] \\ &=& (\lambda - 2s)[2(1 - c) + (2c+1)\lambda-\lambda^2] \end{eqnarray} The first equality comes from the fact that adding a constant multiple of one row to another doesn't change the determinant, and the second comes from the Laplace expansion along the bottom row. So we need to solve the polynomial equation $$ (\lambda - 2\sinh x)[\lambda^2 - (2\cosh x + 1)\lambda - 2(1-\cosh x)] = 0 $$ The linear factor's solution is obvious, and the quadratic factor's solutions can be gotten through the quadratic formula.

Answer 2

The $x$ only appears in your matrix in the context of $e^x$, so call that $a$ and your matrix is now $$ A = \begin{pmatrix} a&1&1/a\\1&1&1\\1/a&1&a \end{pmatrix}. $$ This is so nice that fiddling around a bit leads us to the "obvious" eigenvector $(1,0,-1)$: $$ \begin{pmatrix} a&1&1/a\\1&1&1\\1/a&1&a \end{pmatrix}\begin{pmatrix}1\\0\\-1\end{pmatrix} = \begin{pmatrix}a-1/a\\0\\1/a-a\end{pmatrix} = (a-1/a)\begin{pmatrix}1\\0\\-1\end{pmatrix}. $$

This is why one of your eigenvalues is $2\sinh(x) = 2(e^x-e^{-x})/2 = e^x - e^{-x} = a - 1/a$.

The characteristic polynomial of the above $3 \times 3$ matrix is
$$ \det(A - \lambda I_3) = -\lambda^3 + (2a + 1)\lambda^2 + (a^2 + 2a - 2 - 1/a^2)\lambda + (a^2 - 2a + 2/a - 1/a^2). $$ This is not the cubic polynomial you wrote down: my coefficient of $\lambda^2$ is $2a+1 = 2e^x + 1$ but yours has an $e^{2x}$ term. I checked that my characteristic polynomial has root $a - 1/a$.

In any case, once you have one root $a-1/a$ of a cubic, you can factor out a linear factor $x - (a-1/a)$ and you're left with a quadratic factor, whose roots (the remaining eigenvalues) can be found with the quadratic formula.