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I have the following matrix: $$ A = \left[ \begin{matrix} -(a_1 + a_2) & a_2 \\ a_2 & -(a_2 + a_3) & a_3\\ & & \ddots & \\ &&a_{n-1}& -(a_{n - 1} + a_{n}) & a_n\\ &&&a_n & -(a_n + a_{n + 1}) \end{matrix} \right] $$ with $a_i > 0$. I want to prove that $A$ is invertible. Unfortunately, it is not strictly diagonally dominant, so Gerschgorin theorem only says that $\lambda \leq 0$ (since we know that $\lambda\in \mathbb{R}$).

I also tried to write $A$ as the sum of two triangular matrices (with eigevalues $-a_1, \dots, -a_n$ and $-a_2$, $\dots$, $-a_{n + 1}$), but that does not seem lead anywhere. I was also unable to directly compute $\det A$ or the null space of $A$: the expressions quickly become too ugly.

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  • $\begingroup$ Hey, I have a matrix that is minus your matrix! What are the odds of that? Mine comes from a linear regression problem where the data consist of differences between observations at consecutive time points (math.stackexchange.com/questions/3228824/…, though I don't explain the problem there). How did you get yours? I'm wondering if these kind of matrices can be inverted analytically. $\endgroup$ – becko May 17 '19 at 22:14
  • $\begingroup$ refer to the answer to this post $\endgroup$ – G Cab May 18 '19 at 22:16
  • $\begingroup$ @becko 1st Newton's law and some objects, connected by springs :) $\endgroup$ – Antoine Jan 17 '20 at 16:45
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$A$ is the sum of $n+1$ negative semidefinite matrices $A_1+A_2+\ldots+A_{n+1}$, where \begin{cases} A_1=-a_1e_1e_1^T,\\ A_k=-a_k(e_{k-1}-e_k)(e_{k-1}-e_k)^T,&k=2,3,\ldots,n,\\ A_{n+1}= - a_{n+1}e_ne_n^T. \end{cases} So, if $Ax=0$, we must have $x^TAx=\sum_{j=1}^{n+1} x^TA_jx=0$. Since each $A_j$ is negative semidefinite, each summand $x^TA_jx$ is non-positive. So, for the sum to be zero, we must have $x^TA_jx=0$ for each $j$. Solving this system of quadratic equations, we get $x=0$ and hence $A$ is invertible.

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  • $\begingroup$ Only to be sure: the first and the last equation respectively give $x_1 = 0$ and $x_{n} = 0$. The other give us $x_{j - 1} = x_j$? $\endgroup$ – Antoine Sep 1 '17 at 19:08
  • $\begingroup$ @Antoine Yes, exactly. In short, $x^TAx=-a_1x_1^2-\sum_{k=2}^na_k(x_{k-1}-x_k)^2-a_{n+1}x_n^2$. It is zero iff $x=0$. $\endgroup$ – user1551 Sep 1 '17 at 19:27

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