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My issue is that when calculating the eigenvalues of a matrix, there are cases where the eigenvalue is $0$, and so I was trying to confirm this by calculating the determinant and checking when it is equal to $0$.

The setup is: $a^2 - 4bc \neq 0$, $bc \neq 0$, $(a,b,c) \in \mathbb R^3$, and $n \geq 1$ is an integer and I have this expression: \begin{equation} \dfrac1{\sqrt{a^2-4bc}} \left( \left( \dfrac{a + \sqrt{a^2-4bc}}2\right)^{n+1} - \left( \dfrac{a - \sqrt{a^2-4bc}}2\right)^{n+1}\right) = 0 \end{equation}

Which I simplified to \begin{equation} \left( \dfrac{a + \sqrt{a^2-4bc}}2\right)^{n+1} = \left( \dfrac{a - \sqrt{a^2-4bc}}2\right)^{n+1} \implies \sqrt{a^2-4bc} = -\sqrt{a^2-4bc} \end{equation}

From here, I am unsure how to get a solution, since $a^2 - 4bc \neq 0$, and for real numbers, $x$, I think the only case for $+x = -x$ is when $x = 0$, which is not true here. So since the square root can be complex, I am unsure if there are cases where this can be true.

The eigenvalue expression is \begin{equation} \lambda_k = a + 2b\sqrt{\frac{c}{b}}\text{cos}(\frac{k\pi}{n+1}), \qquad k \in \{1, \dots, n\} \end{equation}

I am not experienced with complex numbers, but using the eigenvalue equations, it seems like the case of $b = 1 \text{, } c = 1, a= -2\cdot\text{cos}(\frac{1\pi}{n+1})$ (for any $n$, but I used $n = 5$), should lead to a $0$ determinant/singular matrix, but this results (for $n=5$) in $a^2 - 4bc = -1$, and so it'd require $\sqrt{-1} = -\sqrt{-1}$, but I can't see how that'd be true.

Thanks in advance for the help! (Also if you want to see the matrix I am working with, this question has the determinant expression: How to compute the determinant of a tridiagonal matrix with constant diagonals?)

Slade
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1 Answers1

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From $$\left(\frac{a+\sqrt{a^2-4bc}}2\right)^{n+1}=\left(\frac{a-\sqrt{a^2-4bc}}2\right)^{n+1} $$ we get, by dividing by the non-zero(!) rught hand side $$ \left(\frac{a+\sqrt{a^2-4bc}}{a-\sqrt{a^2-4bc}}\right)^{n+1}=1,$$ or after making the denominator rational, $$ \left(\frac{(a+\sqrt{a^2-4bc})^2}{4bc}\right)^{n+1}=1,$$ so the condition is rather $$ \tag1\frac{(a+\sqrt{a^2-4bc})^2}{4bc}=e^{\frac{2k\pi i}{n+1}},\quad 0\le k\le n.$$ Now distinguish cases depending on the value of $bc$.

If $bc<0$, the left hand side in $(1)$ is real and negative, and we arrive at $$ \left|a+\sqrt{a^2-4bc}\right|=2\sqrt{-bc},\quad n\text{ odd}, k=\frac{n+1}2.$$

If $0<bc<\frac14a^2$, the left hand side in $(1)$ is real and positive, and we arrive at $$ \left|a+\sqrt{a^2-4bc}\right|=2\sqrt{bc},\quad n\text{ arbitrary}, k=0.$$ Finally, if $bc>\frac14a^2$, we really encounter complex solutions, namely for $$ a=\sqrt{4bc}\cos \phi, \quad(n+1)\phi\equiv 0\pmod{2\pi}.$$

Hagen von Eitzen
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  • I'm trying to see how this can correspond with $\lambda_k = 0 = a + 2b\sqrt{\frac{c}{b}}\text{cos}(\frac{k\pi}{n+1}), \quad k \in {1, \dots, n}$, which is the eigenvalue equation. But it doesn't seem to match up. – Slade Apr 29 '19 at 17:11
  • I got it actually! Thanks so much! – Slade Apr 29 '19 at 18:48