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I have a matrix $A$ ($n \times n$) defined as follows:

$$A = \{ 0 \text{ if } i

This is an upper triangular matrix, and I want to solve a system $Ax =b$ -- thus in a sense invert $A$.

I was wondering if a general inverse exists for this problem.

Any help appreciated, thanks in advance..

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    So what you mean is that the $(i,j)$ entry is $\dbinom i j$?2012-11-12
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    @MichaelHardy \choose(j,i) prob^{i} *(1-prob)^{j-i}2012-11-12
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    You're using computer code instead of mathematical notation. You can write $\dbinom j i p^i (1-p)^{j-i}$.2012-11-12

2 Answers 2

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Let $q=1-p$ and $r = \frac pq$. Then $$ A_{ij} = \begin{cases}{j\choose i} r^i q^j, & i\le j,\\0 & i>j.\end{cases} $$ Therefore $A = \mathrm{diag}(r,r^2,\ldots,r^n)\ B\,\ \mathrm{diag}(q,q^2,\ldots,q^n)$ where $$ B_{ij} = \begin{cases}{j\choose i}, & i\le j,\\0 & i>j.\end{cases} $$ The matrix $B$ is intimately related to the definition of Pascal matrix. The entries of $C=B^{-1}$ are known to take the following form: $$ C_{ij} = \begin{cases}{j\choose i} (-1)^{i+j}, & i\le j,\\0 & i>j.\end{cases} $$ Hence the entries of the $M=A^{-1}$ are given by $$ M_{ij} = \begin{cases}{j\choose i} (-1)^{i+j} q^{-i}r^{-j}, & i\le j,\\0 & i>j.\end{cases} $$

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    Thank you very much, I just figured that out myself as well. Still, I really appreciate your answer.2012-11-13
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I suspect it is a lower triangular matrix. Anyway, append it with an identity matrix and start eliminating from bottom to top (top to bottom). See Gauss-Jordan elimination.

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    it is actually upper-triangular the way I defined it. I know how to solve the problem with elimination and backsubstition. The issue is when I do so, some of the x variables exceed double max-min limits thus can not be computed. For that reason, I am investigating whether a nice form of inv(A) exists2012-11-12
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    @Roark In your definition, both of inequalities are of $j \ge i$ form. Is it a typo?2012-11-12
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    No, it's not. The diagonal entries are then prob^{i}2012-11-12