Given the following linear equation $({\bf A}-\lambda{\bf I}){\bf x}= {\bf b}$, where ${\bf b} = (1, {\sqrt 2}-1,{\sqrt 3}-{\sqrt 2}, ...)^T$. Suppose the largest eigenvalue of matrix ${\bf A}$ is $\lambda_{1}$. Now we pick up a value $\lambda$ that larger than $\lambda_{1}$, and solve the aforementioned linear equation, we can get a vector ${\bf x}$, since $\lambda$ we picked is not the eigenvalue of matrix ${\bf A}$, so this linear equation must have solution. Now we keep decreasing the $\lambda$ we pick with a fixed value to approximate the largest eigenvalue, and in each round we solve the linear equation and get a new ${\bf x}$.
Interesting Finding:
The interesting thing is that with the decreasing of the $\lambda$ we picked, the largest absolute value entry (the value of a vector that ablolute value is the largest) of solved vector ${\bf x}$ will keep increasing until the picked $\lambda$ across the largest eigenvalue.(now $\lambda < \lambda_{1}$). And once the $\lambda$ go across the largest eigenvalue the largest entry of ${\bf x}$ start decreasing. We find this property during our Matlab script, but have no ideal of why it happens. So could any one can help me to make a proof of such property?
Updated Finding: Besides the following findings, I conducted some new experiments, and I found that, when the $\lambda$ we picked is larger than the largest eigenvalue of matrix $\bf A$, the largest entry of solution $\bf x$ of linear equation $({\bf A}-\lambda{\bf I}){\bf x}= {\bf b}$ will always be netive. But after $\lambda$ cross $\lambda_1$(been decreased smaller than the largest eigenvalue). The largest entry of solution $\bf x$ will turned to positive. So anyone can help with the proof of the aforementioned findings?
Thanks
This is quite natural, in fact. We will look at this problem qualitatively.
When you approach the largest eigenvalue, the norm of your solution (and of the equivalent norms is, as you know, the largest absolute value of vector components) in general case must explose (this is what happens here, it seems), because, - by a coarse analogy - you start to divide by zero at this moment.
After you pass to $\lambda < \lambda_1$ you have once again an invertible matrix $A-\lambda I$, hence the solution $x$ exists and is of finite norm, hence its largest component starts to "decrease".
An trivial example that you can study is the solution of an equation $(1-\lambda)x = 1$ for different $\lambda$.