# If $A + \Delta A$ has eigenvalue $\lambda$, show that $A$ has eigenvalue $\lambda + \Delta \lambda$.

$$A$$ is symmetric and can be diagonalised as $$A = V\Lambda V^{-1}$$, where $$\Lambda$$ is a diagonal matrix with eigenvalues of $$A$$ and $$V$$ is an orthogonal matrix with column vectors equal to eigenvectors of $$A$$.

If $$A + \Delta A$$ has eigenvalue $$\lambda$$, show that $$A$$ has eigenvalue $$\lambda + \Delta \lambda$$ with

$$\Delta \lambda = 0 \qquad\text{or}\qquad |\Delta \lambda| = ||(\lambda I - \Lambda)^{-1}||_2^{-1}.$$

My attempt:

$$A u = (\lambda + \Delta \lambda)u$$ $$A u = \lambda u + \Delta \lambda u$$ $$- \Delta \lambda u = (\lambda I - A)u$$ $$-(\lambda I - A)^{-1} \Delta \lambda u = u$$ $$||u||_2 = ||(\lambda I - A)^{-1} \Delta \lambda u||_2$$

Where do I go from here?

• What is $\Delta A$? Is $\Delta A$ necessarily symmetric? Dec 7, 2020 at 1:32
• If $\Delta A$ is an arbitrary symmetric matrix, then the most we can say is that Weyl's inequalities hold. There is essential information missing from the question. Dec 7, 2020 at 1:34
• @BenGrossmann $\Delta A$ is not necessarily symmetric. What other information would you need? Dec 7, 2020 at 16:09
• I thought that there was a counterexample to the question as stated but I realize now I had confused $\lambda$ with $\lambda + \Delta \lambda$ Dec 7, 2020 at 16:16
• I think we should instead have $|\Delta \lambda |= ||(\lambda I - \Lambda)^{-1}||_2^{-1}$. Could you check the question for a missing absolute value sign (and possibly an inequality)? Dec 7, 2020 at 16:33

The key to this problem, I think, is to realize that $$A + \Delta A$$ doesn't tell us anything about $$A$$.
By replacing $$A$$ with $$(A - \lambda I)$$, suppose without loss of generality that $$\lambda = 0$$.
We are given that $$A + \Delta A$$ has eigenvalue $$0$$, and we want to show that it must either hold that $$A$$ has eigenvalue $$0$$ or $$A$$ has eigenvalue $$\Delta \lambda$$ with $$|\Delta \lambda| = \|\Lambda^{-1}\|^{-1}$$. Equivalently, if $$A$$ does not have $$0$$ as an eigenvalue (i.e. $$A$$ is invertible), then $$A$$ has an eigenvalue $$\Delta \lambda$$ with $$|\Delta \lambda| = \|\Lambda^{-1}\|^{-1}$$.
This is easy to see: suppose that $$A$$ is invertible. We write $$\Lambda = \pmatrix{\lambda_1 \\ & \ddots \\ && \lambda_n},$$ where each $$\lambda_i$$ is an eigenvalue of $$A$$. We note that $$\Lambda^{-1} = \pmatrix{\lambda_1^{-1}\\ & \ddots \\ && \lambda_n^{-1}},$$ from which it follows that $$\|\Lambda^{-1}\|_2 = \max_{j=1,\dots,n} |\lambda_j|^{-1} = (\min_{j=1,\dots,n} |\lambda_j|)^{-1}.$$ Thus, we have $$\|\Lambda^{-1}\|_2^{-1} = \min_{j=1,\dots,n}|\lambda_j|$$. So, it is indeed the case that $$A$$ has an eigenvalue $$\Delta \lambda = \lambda_j$$ for which $$|\lambda_j| = \|\Lambda^{-1}\|^{-1}$$, which is what we wanted to show.
A less "matrix-dependent" approach: again, suppose WLOG that $$\lambda = 0$$, and suppose that $$A$$ is invertible. We note that for $$z \in \Bbb C$$, $$A - zI \text{ is invertible }\iff I - z A^{-1} \text{ is invertible }.$$ For $$z$$ with $$|z|$$ sufficiently small, we see that $$I - zA^{-1}$$ must be invertible because the inverse can be expressed via the convergent Neumann series $$(I - zA^{-1}) = \sum_{k=0}^\infty A^{-k} z^k.$$ The radius of convergence of this series is equal to $$\rho(A^{-1})^{-1}$$, which is equal to $$\|\Lambda^{-1}\|_2^{-1}$$ (since $$A$$ is real and symmetric, this is in turn equal to $$\|A^{-1}\|_2^{-1}$$). By the analyticity of holomorphic functions, there exists a $$z \in \Bbb C$$ with $$|z| = \|\Lambda^{-1}\|^{-1}$$ at which the function $$(I - zA^{-1})^{-1} = A^{-1}(A - zI)^{-1}$$ has a singularity, which is to say that $$A - zI$$ fails to be invertible. In other words, such a $$z$$ must be an eigenvalue of $$A$$.