• thomas5267
Solve the following linear recurrence relations. $A_n=-x A_{n-1}-\frac{1}{4}A_{n-2}$ This is a recurrence relations of the characteristic polynomial of a particular kind of matrix. EDIT: I need to prove that $$M_n$$ are always diagonalisable. I am only interested in $$n\geq3\text{ and odd}$$. $$M_n$$ is a tridiagonal nxn matrix. $$M_n$$ has 1/2 on all entries on the upper off-diagonal except the last entry, which is equal to 1. It also has 1/2 on all entries on the lower off-diagonal except the first entry, which is also equal to 1. All other entries are zero. For example: $M_5=\begin{pmatrix} 0&\frac{1}{2}&0&0&0\\ 1&0&\frac{1}{2}&0&0\\ 0&\frac{1}{2}&0&\frac{1}{2}&0\\ 0&0&\frac{1}{2}&0&1\\ 0&0&0&\frac{1}{2}&0 \end{pmatrix}\\ M_7=\begin{pmatrix} 0&\frac{1}{2}&0&0&0&0&0\\ 1&0&\frac{1}{2}&0&0&0&0\\ 0&\frac{1}{2}&0&\frac{1}{2}&0&0&0\\ 0&0&\frac{1}{2}&0&\frac{1}{2}&0&0\\ 0&0&0&\frac{1}{2}&0&\frac{1}{2}&0\\ 0&0&0&0&\frac{1}{2}&0&1\\ 0&0&0&0&0&\frac{1}{2}&0 \end{pmatrix}\\$ The characteristic equation of $$M_n$$ is $$-xA_{n-1}-\frac{1}{2}A_{n-2}, A_n=-x A_{n-1}-\frac{1}{4}A_{n-2}$$. $$A_n$$ is the characteristic polynomial of the submatrix of $$M_n$$. The submatrix is generated by dropping the first row and first column of $$M_n$$. For example: M_5=\begin{pmatrix} 0&\frac{1}{2}&0&0&0\\ 1&0&\frac{1}{2}&0&0\\ 0&\frac{1}{2}&0&\frac{1}{2}&0\\ 0&0&\frac{1}{2}&0&1\\ 0&0&0&\frac{1}{2}&0 \end{pmatrix}\\ M_5-xI_5=\begin{pmatrix} -x&\frac{1}{2}&0&0&0\\ 1&-x&\frac{1}{2}&0&0\\ 0&\frac{1}{2}&-x&\frac{1}{2}&0\\ 0&0&\frac{1}{2}&-x&1\\ 0&0&0&\frac{1}{2}&-x \end{pmatrix}\\ \begin{align*} \det(M_5-xI_5)&= -x\begin{vmatrix} -x&\frac{1}{2}&0&0\\ \frac{1}{2}&-x&\frac{1}{2}&0\\ 0&\frac{1}{2}&-x&1\\ 0&0&\frac{1}{2}&-x \end{vmatrix}-\frac{1}{2} \begin{vmatrix} 1&\frac{1}{2}&0&0\\ 0&-x&\frac{1}{2}&0\\ 0&\frac{1}{2}&-x&1\\ 0&0&\frac{1}{2}&-x \end{vmatrix}\\ &=-xA_4-\frac{1}{2}\begin{vmatrix} -x&\frac{1}{2}&0\\ \frac{1}{2}&-x&1\\ 0&\frac{1}{2}&-x\\ \end{vmatrix}\\ &=-xA_4-\frac{1}{2}A_3 \end{align*} $$A_n$$ has a close form solution of $$A_n = \left( \dfrac{-x+\sqrt{x^2-1}}{2}\right)^n + \left(\dfrac{-x-\sqrt{x^2-1}}{2}\right)^n$$ with the initial condition $$A_1=-x,\,A_2=x^2-\frac{1}{2}$$.
Mathematics
• Stacey Warren - Expert brainly.com
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SOLVED
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