I need help: Consider the matrix B(x) = (x,1;1,x) and let n ≥ 2 be an integer. Calculate the product of matrices B(2)B(3) · · · B(n).
- Thread starter vicsan21
- Start date
- Joined
- Feb 4, 2004
- Messages
- 16,582
What are your thoughts? What have you tried? How far have you gotten?
Please be complete. Thank you!
blamocur
Elite Member
- Joined
- Oct 30, 2021
- Messages
- 3,101
What do you get when you diagonalize B(x) ?
- Joined
- Feb 4, 2004
- Messages
- 16,582
The original poster (OP) will almost certainly not be responding here, as OP was already provided the complete solution on MathForums.com.
blamocur
Elite Member
- Joined
- Oct 30, 2021
- Messages
- 3,101
I worked out the "closed form" answer at the time, and since it's been one week after OP I am posting it here for "completeness sake":
[math]B(x) = \left(\begin{array}{cc}x & 1\\1 & x\end{array}\right) = \frac{1}{2} \left(\begin{array}{cc}1 & -1\\1 & 1\end{array}\right) \left(\begin{array}{cc}x+1 & 0\\0 & x-1\end{array}\right) \left(\begin{array}{cc}1 & 1\\-1 & 1\end{array}\right)[/math][math]\prod_{k=2}^n B(k) = \frac{1}{2} \left(\begin{array}{cc}1 & -1\\1 & 1\end{array}\right) \left(\begin{array}{cc}a_n & 0\\0 & b_n\end{array}\right) \left(\begin{array}{cc}1 & 1\\-1 & 1\end{array}\right)[/math][math]= \frac{1}{2} \left(\begin{array}{cc}a_n & -b_n\\a_n & b_n\end{array}\right) \left(\begin{array}{cc}1 & 1\\-1 & 1\end{array}\right) = \frac{1}{2} \left(\begin{array}{cc}a_n+b_n & a_n-b_n\\a_n-b_n & a_n+b_n\end{array}\right)[/math][math]a_n = \prod_{k=2}^n (k+1) = \frac{1}{2}(n+1)![/math][math]b_n = \prod_{k=2}^n(k-1) = (n-1)![/math][math]a_n + b_n =\frac{1}{2}\left(n^2+n+2 \right) (n-1)![/math][math]a_n - b_n =\frac{1}{2}\left(n^2+n-2 \right) (n-1)![/math][math]\prod_{k=2}^n B(k) = \frac{(n-1)!}{4}\left(\begin{array}{cc}n^2+n+2 & n^2+n-2\\n^2+n-2 & n^2+n+2\end{array}\right)[/math]
[math]B(x) = \left(\begin{array}{cc}x & 1\\1 & x\end{array}\right) = \frac{1}{2} \left(\begin{array}{cc}1 & -1\\1 & 1\end{array}\right) \left(\begin{array}{cc}x+1 & 0\\0 & x-1\end{array}\right) \left(\begin{array}{cc}1 & 1\\-1 & 1\end{array}\right)[/math][math]\prod_{k=2}^n B(k) = \frac{1}{2} \left(\begin{array}{cc}1 & -1\\1 & 1\end{array}\right) \left(\begin{array}{cc}a_n & 0\\0 & b_n\end{array}\right) \left(\begin{array}{cc}1 & 1\\-1 & 1\end{array}\right)[/math][math]= \frac{1}{2} \left(\begin{array}{cc}a_n & -b_n\\a_n & b_n\end{array}\right) \left(\begin{array}{cc}1 & 1\\-1 & 1\end{array}\right) = \frac{1}{2} \left(\begin{array}{cc}a_n+b_n & a_n-b_n\\a_n-b_n & a_n+b_n\end{array}\right)[/math][math]a_n = \prod_{k=2}^n (k+1) = \frac{1}{2}(n+1)![/math][math]b_n = \prod_{k=2}^n(k-1) = (n-1)![/math][math]a_n + b_n =\frac{1}{2}\left(n^2+n+2 \right) (n-1)![/math][math]a_n - b_n =\frac{1}{2}\left(n^2+n-2 \right) (n-1)![/math][math]\prod_{k=2}^n B(k) = \frac{(n-1)!}{4}\left(\begin{array}{cc}n^2+n+2 & n^2+n-2\\n^2+n-2 & n^2+n+2\end{array}\right)[/math]