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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

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  • $\begingroup$ Just one question, to understand more clearly. Does your matrices infinite dimensional or not? $\endgroup$ – kolobokish Nov 28 '16 at 16:20
  • $\begingroup$ Hi, its not infinite dimension, in my experiment the dimension is 970*970, and the matrix is a symmetric matrix $\endgroup$ – Lovingmage Dec 1 '16 at 4:15
  • $\begingroup$ A symmetric matrix has all eigenvalues real and non-negative. Maybe you could check up the power method to find largest eigenvalue of a matrix. I think it is faster than what you propose. $\endgroup$ – mathreadler Dec 2 '16 at 18:39
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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 $\lambda$.

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  • $\begingroup$ Hi thanks for your answer. I just want to know, is that possible to make a mathematical proof for this property. Since if this property can be proofed, we may use this algorithm as a efficient method to approximate eigenvalues. $\endgroup$ – Lovingmage Dec 1 '16 at 4:22
  • $\begingroup$ @Lovingmage there's a well-known algoritm working with a similar approach. Let $\lambda$ be a positive eigenvalue of a symmetric matrix $A$ and its absolute value is strictly maximal among other eigenvalues. Let also vectors $x^i$ form a basis of your vector space. For each vector $x^i$ consider a sequence $x^i_0 = x^i$, $x^i_{n+1} = \frac{Ax^i_n}{\|x^i_n\|_2}$. You have now a limit $z^i = \lim_{n\to\infty}\|Ax^i_n\|$. The final step is $\max_i\{z^i\} = \lambda$. $\endgroup$ – TZakrevskiy Dec 1 '16 at 12:56
  • $\begingroup$ Hi I have some wrong description on my question, and I found a new property, could you please check it out and help me with the proof? Thanks $\endgroup$ – Lovingmage Dec 2 '16 at 18:31
  • $\begingroup$ @Lovingmage the new property you found depends on your choice of a matrix $A$ and a vector $b$. $\endgroup$ – TZakrevskiy Dec 2 '16 at 19:01

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