Can anybody help me out with getting an expression of the values of $\lambda$ for a matrix $A$ for which $det(A-\lambda I)$ equals the determinant of a matrix with on the main diagonal $-\lambda$, on the diagonal above the main diagonal $\dfrac{1}{2}$ and on the diagonal under the main diagonal $\frac{1}{2} \lambda$.
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2You mean $\det \begin{bmatrix} -\lambda & 1/2 & & & \ 1/2\lambda &- \lambda & 1/2 & & \ & 1/2 \lambda & \ddots & \ddots && \ & & \ddots & &\end{bmatrix}$ ? – flawr Nov 15 '15 at 20:08
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Yes, but -$\lambda$ on the main diagonal instead of $\lambda$ – Roos Jansen Nov 15 '15 at 20:09
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The determinant of such tridiagonal matrices of order $n$ are computed with the linear recurrence of order $2$: $$D_n=-\lambda D_{n-1}-\frac\lambda4 D_{n-2}$$ and the initial conditions $\; D_0=1,\enspace D_1=\lambda$.
How to find a closed formula for $D_n$ :
We look for basic solutions that are geometric progressions $r^n\ (r\ne 0)$; this leads to $$r^n=-\lambda r^{n-1}-\frac\lambda4 r^{n-2}\iff r^2=-\lambda r-\frac\lambda4. $$ So the possible values of $r$ are the roots $r_1,r_2$ of the characteristic equation: $$r^2+\lambda r+\frac\lambda4=0.$$ The general solution is a linear combination of the basic solutions: $\;\alpha r_1^n+\beta r_2^n$, where $\alpha, \beta$ are determined by the initial conditions.
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Thanks for you answer! How can I derivate an explicit expression for $\lambda$ from this? – Roos Jansen Nov 15 '15 at 20:12
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You have to solve for a linear recurrence of order $2$. Do you know about linear differential equations of order? – Bernard Nov 15 '15 at 20:14
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How did you get the linear recurrence? (in case the values on the diagonals change) – Mitch Nov 15 '15 at 22:02
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The general formula is $$\det A_n=a_{nn}\det A_{n-1}-a_{n-1,n}a_{n,n-1}\det A_{n-2}$$ where $\det A_k$ denotes the $k$-th dominant minor. In this formula (easy to prove), the terms are not necessarily constant on the diagonal, nor over nor under the diagonal. – Bernard Nov 15 '15 at 22:16
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The determinant of the tridiagonal matrix can be expressed by the recurrence (link): $$ f_n = -\lambda f_{n-1} -\frac{1}{4}\lambda f_{n-2} \quad (*) $$ and the initial values $f_0=1$, $f_{-1}= 0$.
For $\lambda = 0$ the determinant vanishes. Otherwise:
So $f_1 = -\lambda$, $f_2 = \lambda^2 - \frac{1}{4} \lambda$, $f_3 = -\lambda^3 + \frac{1}{4} \lambda^2+\frac{1}{4}\lambda^2=-\lambda^3+\frac{1}{2}\lambda^2$ and so on.
Solving the recurrence relation:
This is a homogenous linear recurrence relation with characteristic polynomial $$ p(t) = t^2 + \lambda t + \lambda/4 \\ $$ with roots $$ 0 = (t + \lambda/2)^2 + (\lambda -\lambda^2)/4 \iff \\ t_{1,2} = \frac{\pm \sqrt{\lambda(\lambda-1)}-\lambda}{2} $$
Case two roots:
For $\lambda \ne 1$ this leads to solutions $$ f_n = \frac{1}{2^n} \left( c_1 \left(\sqrt{\lambda(\lambda-1)}-\lambda\right)^n + c_2 \left(-\sqrt{\lambda(\lambda-1)}-\lambda\right)^n \right) $$ Inserting $f_0$ and $f_1$ gives $$ 1=c_1+c_2 \\ -\lambda = c_1 t_1 + c_2 t_2 = c_1(t_1-t_2)+ t_2 $$ thus $$ c_1 = \frac{t_2+\lambda}{t_2-t_1} \quad\quad c_2 = \frac{t_1+\lambda}{t_1-t_2} $$ We have \begin{align} t_2 + \lambda &= \frac{-\sqrt{\lambda(\lambda-1)}-\lambda}{2} + \lambda = \frac{-\sqrt{\lambda(\lambda-1)}+\lambda}{2} \\ t_2 - t_1 &= \frac{-\sqrt{\lambda(\lambda-1)}-\lambda}{2} - \frac{\sqrt{\lambda(\lambda-1)}-\lambda}{2} = -\sqrt{\lambda(\lambda-1)} \end{align} so \begin{align} c_1 &= -\frac{-\sqrt{\lambda(\lambda-1)}+\lambda}{2\sqrt{\lambda(\lambda-1)}} = \frac{1}{2} - \frac{1}{2}\sqrt{\frac{\lambda}{\lambda-1}} \\ c_2 &= \frac{1}{2} + \frac{1}{2}\sqrt{\frac{\lambda}{\lambda-1}} \end{align}
Case one root:
For $\lambda = 1$ we have only one root $$ t = -\frac{1}{2} $$ and try $$ f_n = \left(-\frac{1}{2}\right)^n (c_3 + c_4 n) $$ Inserting $f_0 = 1$ gives $c_3 = 1$ and $f_1= -\lambda = -1$ gives $c_4 = 1$.
Result:
This results in the determinant value $$ f_n = \begin{cases} 0 & ; \lambda = 0 \\ \left(-\frac{1}{2}\right)^n \left[ 1 + n \right] & ; \lambda = 1 \\ \frac{1}{2^{n+1}} \left[ \left(1 - \sqrt{\frac{\lambda}{\lambda-1}}\right) \left(\sqrt{\lambda(\lambda-1)}-\lambda\right)^n + \\ \left(1 + \sqrt{\frac{\lambda}{\lambda-1}}\right) \left(-\sqrt{\lambda(\lambda-1)}-\lambda\right)^n \right] & ; \text{else} \\ \end{cases} $$
Test:
The $f_n$ were calculated via the recursive equation $(*)$, $g(n)$ and $h(n)$ are cases of the result forumula above. Both evaluations should match.
Two roots, $n = 3$:
(%i) [f3,expand(radcan(g(3)))];
2 2
L 3 L 3
(%o) [-- - L , -- - L ]
2 2
One root, $n = 3$:
(%i) [ev(f3,L=1),expand(radcan(h(3)))];
1 1
(%o) [- -, - -]
2 2
Two roots, $n=8$:
(%i) [f8,expand(radcan(g(8)))];
7 6 5 4 7 6 5 4
8 7 L 15 L 5 L L 8 7 L 15 L 5 L L
(%o) [L - ---- + ----- - ---- + ---, L - ---- + ----- - ---- + ---]
4 16 32 256 4 16 32 256
One root, $n = 8$:
(%i) [ev(f8,L=1),expand(radcan(h(8)))];
9 9
(%o) [---, ---]
256 256
mvw
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I think your expression for $f_{n}$ in the beginning isn't correct? (see above) – Roos Jansen Nov 15 '15 at 20:58
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Because the product of the off-diagonal elements is equal to $-\dfrac{\lambda}{4}$ instead of $-\dfrac{\lambda^2}{4}$ – Roos Jansen Nov 15 '15 at 21:06
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@mvw: I think the first line of your computation for $a_n$ should be $a_n\lambda^n = -a_{n-1} \lambda^n - \frac{1}{4} a_{n-2} \lambda^{n\color{red}{-1}}. $ – Bernard Nov 15 '15 at 21:22
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I'm curious! I'm also working on it, but don't have the answer yet. – Roos Jansen Nov 15 '15 at 21:51