# Eigenvectors as continuous functions of matrix - diagonal perturbations

The general question has been treated here, and the response was negative. My question is about more particular perturbations. The counterexamples given in the previous question have variations not only on the diagonal, and moreover, in the simple example given by A. Quas, the discontinuous change in eigenvectors take place where the eigenvalue is double.

Suppose $$A$$ is a symmetric, positive definite matrix, with simple lowest eigenvalue, and $$D$$ is a diagonal matrix. Does the first eigenvalue and the first eigenvector of $$A+D$$ vary continuously for $$\|D\|_1 <\varepsilon$$ ($$\varepsilon$$ can be as small as possible, and $$\|\cdot \|_1$$ is the $$L^1$$ norm)?

In the case this is true, it is possible to bound the variation of the eigenvector in terms of $$\|D\|_1$$?

• $$A$$ is constant, $$D$$ is a diagonal matrix which is variable with $$N$$ parameters (the matrices are of size $$N \times N$$)

• $$D$$ is just assumed to vary continuously with norm smaller than $$\varepsilon$$.

• What is constant and what is varying there? Is $D$ a matrix-valued function? Of how many parameters? How regular? May 24, 2015 at 10:40
• $A$ is constant, $D$ is diagonal and variable (with $N$ parameters, where $N \times N$ is the size of $A$). Ideally, the perturbation $D$ should only be continuous, but if higher regularity assumptions are needed, I don't mind. May 24, 2015 at 10:53
• The key here, I think, is your assumption that the eigenvalue $\lambda$ is simple. ($A$ needn't be symmetric PD.) Then for $D$ in a small enough neighborhood, there is a continuous (even analytic) map $D\to\lambda(D)$ with $\lambda(0)=\lambda$ and $\lambda(D)$ an eigenvalue of $A+D$. The point is that if you take a simple root of a polynomial, then for sufficiently small perturbations of the coefficients, you get an analytic perturbation of the root. Presumably the same holds for the associated eigenvector, again since the eigenspace is one dimensional. May 24, 2015 at 11:18
• Joe Silverman is correct; this is a job for the implicit function theorem (where the function being defined implicitly is $M \mapsto$its lowest eigenvector). Said theorem requires some derivative to be onto and this eventually turns into the simpleness of the eigenvalue. May 24, 2015 at 14:28
• You don't need $D$ diagonal for this, symmetric is enough. The claim on the eigenvector follows for example by writing the spectral projection as a contour integral of the resolvent. May 24, 2015 at 15:10

As @Joe Silverman said in his comment, the key here is that $\lambda$ is simple.

I copy here a few Theorems on the subject that might be helpful.

From the book Linear Algebra and Its Applications (by Peter D. Lax):

Theorem 7, p.130: Let $A(t)$ be a differentiable square matrix-valued function of the real variable $t$. Suppose that $A(0)$ has an eigenvalue $a_0$ of multiplicity one, in the sense that $a_0$ is a simple root of the characteristic polynomial of $A(0)$. Then for $t$ small enough, $A(t)$ has an eigenvalue $a(t)$ that depends differentiably on $t$, and which equals $a_0$ at zero, that is $a(0)=a_0$.

Theorem 8, p.130: Let $A(t)$ be a differentiable matrix-valued function of $t$, $a(t)$ an eigenvalue of $A(t)$ of multiplicity one. Then we can choose an eigenvector $h(t)$ of $A(t)$ pertaining to the eigenvalue $a(t)$ to depend differentiably on $t$.

From the Book Matrix Analysis (by Roger A. Horn & Charles R. Johnson)

Corollary 6.3.8, p.407: Let $A,E\in \Bbb C^{n\times n}$. Assume that $A$ is Hermitian and $A+E$ is normal, let $\lambda_1,\ldots,\lambda_n$ be the eigenvalues of $A$ arranged in increasing order $\lambda_1\leq \ldots\leq \lambda_n$ and let $\hat \lambda_1,\ldots,\hat \lambda_n$ be the eigenvalues of $A+E$, ordered so that $\Re(\hat\lambda_1)\leq\ldots\leq\Re(\hat\lambda_n)$. Then $$\sum_{i=1}^n |\hat \lambda_i-\lambda_i|^2 \leq \|E\|_F^2,$$ where $\|\cdot\|_F$ is the Frobenius norm.

This is a somehow refined version of Theorem 7 above:

Theorem 6.3.12, p.409: Let $A,E\in \Bbb C^{n\times n}$ and suppose that $\lambda$ is a simple eigenvalue of $A$. Let $x$ and $y$ be, respectively, right and left eigenvectors of $A$ corresponding to $\lambda$. Then
a) for each given $\epsilon>0$ there exists a $\delta>0$ such that, for all $t\in\Bbb C$ such that $|t|<\delta$, there is a unique eigenvalue $\lambda(t)$ of $A+tE$ such that $|\lambda(t)-\lambda-ty^*Ex/y^*x|\leq |t| \epsilon$
b) $\lambda(t)$ is continuous at $t=0$, and $\lim_{t\to 0}\lambda(t)=\lambda$
c) $\lambda(t)$ is differentiable at $t=0$, and $$\left.\frac{\operatorname{d}\lambda(t)}{\operatorname{d}t}\right|_{t=0}=\frac{y^*Ex}{y^*x}$$

• Corollary 6.3.8 looks very nice. Is there a similar result for eigenvectors? I will look in the cited book. May 28, 2015 at 14:44
• @BeniBogosel Unfortunately I did not find any similar results for eigenvectors in the book of Horn (I might have not seen it though). That is why I also reported the results from Lax.
– Surb
May 28, 2015 at 14:51
• I guess this is close of the result I want: math.stackexchange.com/questions/1133071/… There seems to be a Lipschitz bound for the eigenvectors also, but the Lipschitz constant is not too explicit in the references I searched. May 28, 2015 at 22:17