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How to get the inverse of this matrix:

$\left(\begin{array}{ccccccc} 2&-1\\-1&2&-1\\&-1&2&-1\\&&&\ddots\\&&&&\ddots\\&&&&-1&2&-1\\&&&&&-1&2 \end{array}\right)$

where the blank elements are all zeros. Thank you.

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    $\begingroup$ en.wikipedia.org/wiki/Tridiagonal_matrix#Inversion $\endgroup$ – Amzoti May 7 '13 at 4:35
  • $\begingroup$ Oh my God, thank you very much! $\endgroup$ – user76300 May 7 '13 at 4:38
  • $\begingroup$ @Amzoti: a smaller question: when $i=j$, what does $b_i\cdots b_{j-1}$ mean? $\endgroup$ – user76300 May 7 '13 at 4:58
  • $\begingroup$ I mean, how to calculate $b_i\cdots b_{j-1}$? For example, if $i=j=1$, what does $b_0$ mean? $\endgroup$ – user76300 May 7 '13 at 5:11
  • $\begingroup$ But if we try to find the inverse of $\left(\begin{array}{cc} 2&3\\3&2 \end{array}\right)$, and consider $i=j=1$, $b_1b_{-1}$ would be $3\times3$ which is 9, but the correct value suggests that the product should take the value 1. What's the matter? $\endgroup$ – user76300 May 7 '13 at 5:23
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Lets do an example using your matrix with dimension $4x4$, symmetric, tridiagonal.

We have the matrix:

$$\displaystyle A = \begin{bmatrix}2&-1&0&0\\-1&2&-1&0\\0&-1&2&-1\\0&0&-1&2\end{bmatrix}.$$

The inverse of this matrix:

$$\displaystyle A^{-1} = \frac{1}{5}\begin{bmatrix}4 & 3 & 2 & 1\\3 & 6 & 4 & 2\\2 & 4 & 6 & 3\\1 & 2 & 3 &4\\\end{bmatrix}.$$

Look at the structure of the "first and fourth" and "second and third" rows.

If you are looking for the eigenvalues and eigenvectors, we have:

$$\displaystyle \lambda_1 = \frac{1}{2} (5+\sqrt{5}), ~v_1 = (-1, \frac{1}{2} (1+\sqrt{5}), \frac{1}{2} (-1-\sqrt{5}), 1)$$

$$\displaystyle \lambda_2 = \frac{1}{2} (3+\sqrt{5}), ~v_2 = (1, \frac{1}{2} (1-\sqrt{5}), \frac{1}{2} (1-\sqrt{5}), 1)$$

$$\displaystyle \lambda_3 = \frac{1}{2} (5-\sqrt{5}), ~v_3 = (-1, \frac{1}{2} (1-\sqrt{5}), \frac{1}{2} (-1+\sqrt{5}), 1)$$

$$\displaystyle \lambda_4 = \frac{1}{2} (3-\sqrt{5}), ~v_4 = (1, \frac{1}{2} (1+\sqrt{5}) \frac{1}{2} (1+\sqrt{5}), 1)$$

That should help you follow the Wiki algorithm and @joriki provided the algorithm for using the normalized eigenvector and eigenvalue as another approach.

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  • $\begingroup$ Too much effort (and quality) to go un-noticed! +++ $\endgroup$ – amWhy May 8 '13 at 0:36
  • $\begingroup$ You're getting me interested in numerical analysis/methods and I had hitherto excluded those topics from consideration (even "ignoring" some relevant tags)...that's is now changing; indeed my interest in and knack for comp sci (discovered late in the whole context of things) has contributed to that as well! $\endgroup$ – amWhy May 8 '13 at 0:43
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Let $n$ be the dimension of the matrix. It has normalized eigenvectors $v_j$ with

$$v_{jk}=\sqrt\frac2{n+1}\sin\frac{jk\pi}{n+1}$$

corresponding to eigenvalues

$$ \lambda_j=4\sin^2\frac{j\pi}{2(n+1)}, $$

so it can be represented as $V\Lambda V$, and its inverse as $V\Lambda^{-1}V$.

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  • $\begingroup$ Thank you. I'm dull in math so please give me a couple of days to fully understand it. $\endgroup$ – user76300 May 7 '13 at 22:22
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this matrix inverse is $A=A_{ij}$ and $$A_{ij}=\min{\{i,j\}}$$

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