# Showing $\min\limits_{j=1,\dots,n}|\lambda-\lambda_j|\le ||C||_p||C^{-1}||_p||B||_p$

Let $$A$$ be a diagonalizable $$n\times n$$ matrix with eigenvalues $$\lambda_1,\dots, \lambda_n$$, $$B$$ an $$n\times n$$ matrix, and $$\lambda$$ an eigenvalue of $$A+B$$. Show that $$\min\limits_{j=1,\dots,n}|\lambda-\lambda_j|\le ||C||_p||C^{-1}||_p||B||_p$$ where $$C$$ is a nonsingular matrix such that $$C^{-1}AC$$ is diagonal and $$p=1,2,\infty$$.

I'm having difficulty figuring out where to start. If given some guidance I'm sure I can easily get the rest. I know that under the assumption $$A$$ is diagonalizable gives $$C^{-1}AC=diag(\lambda_1,\dots,\lambda_n)$$, but I am failing to see how I will use the other assumptions. Any input would be greatly appreciated!

• What is the context of this question? Is exercise contained any book? Or is it a problem whose solution contributes to some of your research? How do you know it's not a false statement? Please clarify these questions. – MathOverview Feb 2 at 14:35
• It is a problem in the book Numerical Analysis by Rainer Kress – user906357 Feb 2 at 14:39

Here's an argument for the $$p=\infty$$ case.
Suppose that $$(A+B)v=\lambda v$$. Since $$C^{-1}AC$$ is diagonal, we have $$C^{-1}ACe_j=\lambda_je_j$$, where $$e_1,\ldots,e_n$$ are the canonical basis. So $$Av_j=\lambda_jv_j$$, where $$v_j=Ce_j$$.
Assume $$A$$ is invertible (when $$A$$ is not invertible, we may tweak the eigenvalues slightly so that it is, and the we approximate the estimates). Then $$v_1,\ldots,v_n$$ are a basis. Write $$v=\sum_jr_jv_j$$ for some coefficients $$r_j$$. Now \begin{align} Bv&=\lambda v-Av=\sum_j r_j \lambda v_j-\sum_jr_j Av_j=\sum_jr_j(\lambda-\lambda_j)v_j\\ \ \\ &=\sum_jr_j(\lambda-\lambda_j)Ce_j. \end{align} Thus $$\tag1 \sum_j r_j(\lambda-\lambda_j)e_j=C^{-1}Bv.$$ Since $$v\ne0$$, $$r_k=\max\{|r_j|:\ j\}\ne0$$. Then $$\tag2 |r_k|\,|\lambda-\lambda_k|\leq\left\|\sum_jr_j(\lambda-\lambda_j)e_j\right\|_\infty=\|C^{-1}Bv\|$$ Also, since $$v=\sum_jr_jv_j=\sum_jr_jCe_j$$, we have that $$\tag3 \|v\|=\left\| C\,\sum_jr_je_j\right\|\leq\|C\|\,\left\| \sum_jr_je_j\right\| _\infty=\|C\|\,|r_k|$$ From $$(2)$$ and $$(3)$$, $$|\lambda-\lambda_k|\leq\frac1{|r_k|}\,\|C^{-1}Bv\|\leq\frac1{|r_k|}\,\|C^{-1}\|\,\|B\|\,\|v\|\leq\|C^{-1}\|\,\|B\|\,\|C\|.$$
• Thank you for the answer. How is it we know $|r_k|\ge1$? – user906357 Feb 3 at 2:57
• We don't. Look at $(3)$. – Martin Argerami Feb 3 at 2:58