# Proof for a property of approximating eigenvalue

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

• Just one question, to understand more clearly. Does your matrices infinite dimensional or not? – kolobokish Nov 28 '16 at 16:20
• Hi, its not infinite dimension, in my experiment the dimension is 970*970, and the matrix is a symmetric matrix – Lovingmage Dec 1 '16 at 4:15
• 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. – mathreadler Dec 2 '16 at 18:39

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 $\lambda$.
• @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 $\max_i\{z^i\} = \lambda$. – TZakrevskiy Dec 1 '16 at 12:56
• @Lovingmage the new property you found depends on your choice of a matrix $A$ and a vector $b$. – TZakrevskiy Dec 2 '16 at 19:01