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I'm trying to solve a linear system using Cholesky decomposition, but I get stuck when solving the system.

Here is the matrix : $$M = \begin{pmatrix} 2 & -1 & 0 & \cdots & 0 \\ -1 & 2 & -1 & \cdots & 0 \\ \vdots & -1 & 2 & \ddots & \vdots \\ 0 & \cdots &\ddots & \ddots & -1 \\ 0 & \cdots & \cdots & -1 & 2 \end{pmatrix}$$

(Sorry, it is quite ugly but I don't manage $\LaTeX$ and matrices well).

So for the Cholesky matrix ($T$ such that $TT*=M$) I find a matrix with a lower double diagonal with $-\sqrt{\frac{k-1}{k}}$ and $\sqrt{\frac{k +1}{k}}$.

So I'm stuck there, I can't manage to find a solution to $Tx=B$ with $B= \begin{pmatrix}1 & -1 & \cdots & 1 & -1 \end{pmatrix}*$.

I find $x_1 = \frac{\sqrt{2}}{2}$ and $x_{k+1}=(-1)^k \sqrt{\frac{k+1}{k+2}} + \sqrt{\frac{k}{k+2}}x_k$ but I need a not-reccurent forme to apply the step with $T*$.

How can I do ?

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  • $\begingroup$ That's the Cartan matrix of type $A_n$, which has determinant $n+1$ and its inverse is pretty easy to write down. Why do you need Cholesky decomposition? $\endgroup$ – Marc van Leeuwen Feb 28 '15 at 17:05
  • $\begingroup$ @Marc : I'm a begginer in matrices, don't know about Cartan matrix and the goal of the exercise is to use Cholesky decomposition. $\endgroup$ – servabat Feb 28 '15 at 17:06

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