# Can a small perturbation of a diagonal matrix increase its smallest eigenvalue to any arbitrarily large value?

Let $$S\in\mathbb{R}^{n\times n}$$ be a diagonal positive semidefinite matrix with exactly $$k$$ positive entries in its diagonal, where $$k. Let $$\epsilon$$ be any arbitrarily small positive real number. Can we find a symmetric matrix $$A=A(\epsilon)\neq 0$$ such that $$\|A\|_{\mathrm{op}} \leq \epsilon\; \lambda_{\textrm{min}}(S+A)$$ ?

Here, $$\|\cdot\|_{\mathrm{op}}$$ denotes the operator norm, and $$\lambda_{\mathrm{min}}$$ is the smallest eigenvalue.

Thank you very much.

• So, $S$ is given and you want to find $A$, such that the inequality holds for any arbitrarily small and $\epsilon$ right?
– KBS
May 6, 2022 at 12:24
• Yes, S is given, and we want to find such A. May 6, 2022 at 12:25
• In those scenarios, the best approach is to try to find a counterexample or a contradiction.
– KBS
May 6, 2022 at 12:25
• I have played with a toy example for n=2 and A is a rank-one perturbation. This leads to some condition on the unique positive diagonal entry of S. However, I'm not sure this would lead to any conclusion as this is just a rank-one perturbation while here one can use any symmetric perturbation. May 6, 2022 at 12:29
• Is $S$ a nonnegative matrix i.e. other diagonal entries are $0$ ? May 6, 2022 at 12:43

As both $$A,S$$ are symmetric, we get $$\lambda_\text{min}(S+A) = \min_{v\in \mathbb{R}^2 : \Vert v \Vert=1} \langle v, (S+A) v \rangle.$$ As $$S$$ has nontrivial kernel, we can pick $$w\in Ker(S)$$ with $$\Vert w \Vert=1$$ and get $$\lambda_\text{min}(S+A) \leq \langle w, (S+A) w\rangle = \langle w,Aw\rangle \leq \Vert A \Vert_\text{op} \Vert w \Vert^2 = \Vert A \Vert_\text{op}.$$ Then your desired inequality would imply that $$\Vert A \Vert_\text{op} \leq \varepsilon \Vert A \Vert_\text{op},$$ which does not work for $$A\neq 0$$ and $$0\leq \varepsilon<1$$.

Note that the same proof works for any symmetric $$S$$ with nontrivial kernel. If $$S$$ admits a negative eigenvalue, we test on the corresponding eigenvector and reach the same contradiction. For your conclusion to hold, we need that the lowest eigenvalue of $$S$$ is strictly positive.

Let us look at the case when $$n=2$$ and $$S=\begin{pmatrix}s & 0\\ 0 & 0\end{pmatrix}$$ for some fixed $$s>0$$.

We are looking for a symmetric $$A$$, therefore we can write $$A=U^T\begin{pmatrix}sx & 0 \\ 0 & sy \end{pmatrix}U$$ for some real numbers $$x,y$$ and some orthogonal matrix $$U$$. Since conjugating a matrix doesn't change its eigenvalues, we need to find $$x,y$$ such that $$s\max(|x|,|y|)\leq \varepsilon \lambda_{min}(B)$$, where $$B:=USU^T+\begin{pmatrix}sx & 0 \\ 0 & sy \end{pmatrix}$$, with $$x,y$$ depending on $$\varepsilon$$.

Now orthogonal matrices in dimension two are either rotations of the form $$\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix}$$ or reflections of the form $$\begin{pmatrix}\cos\theta & \sin\theta \\ \sin\theta & -\cos\theta\end{pmatrix}$$.

In both cases ($$U$$ being either a reflection or a rotation), we get $$\begin{equation*} P_B(\lambda)=\begin{vmatrix} s\cos^2\theta+sx-\lambda & s\sin\theta\cos\theta\\ s\sin\theta\cos\theta & s\sin^2\theta +sy -\lambda\end{vmatrix} = \lambda^2 -s(1+x+y)\lambda+s^2(xy+x\sin^2\theta+y\cos^2\theta), \end{equation*}$$

so $$\lambda_{min}=s\frac{1+x+y-\sqrt{(1+x+y)^2-4(x\sin^2\theta+y\cos^2\theta+xy)}}{2}$$.

Therefore the inequality we want to satisfy reduces to $$\begin{equation*} \frac{2}{\varepsilon}\max(|x|,|y|)\leq 1+x+y-\sqrt{(1+x+y)^2-4(x\sin^2\theta+y\cos^2\theta+xy)}. \end{equation*}$$

By assuming without loss of generality that $$|x|\geq |y|$$ and dividing both sides by $$|x|$$, we see that $$\frac{RHS}{|x|}$$ should be unbounded (as a function of $$x,y$$) for the inequality to be satisfied (for small enough $$\varepsilon$$). But note that $$\frac{RHS}{|x|}\leq 2+\frac{1}{|x|}$$, so we can restrict our search to the case $$|y|\leq|x|<1$$. In fact, setting $$\begin{equation*} f(x,y)=1+x+y-\sqrt{(1+x+y)^2-4(x\sin^2\theta+y\cos^2\theta+xy)}, \end{equation*}$$ the only hope for the inequality to be satisfied for an appropriate choice of $$x,y$$ with $$|y|\leq |x|$$ is that $$\limsup_{x\to 0}\frac{f(x,y)}{|x|}=\infty$$. Clearly, $$\lim_{x\to 0}f(x,y)=0$$, so using L'Hopital's rule, we need that $$\lim_{x\to 0}|\frac{\partial{f}}{\partial{x}}(x,y)|=\infty$$.

However, $$\frac{\partial{f}}{\partial{x}}(x,y)=1-\frac{1+x-y-2\sin^2\theta}{\sqrt{(1+x+y)^2-4(x\sin^2\theta+y\cos^2\theta+xy)}}$$ (assuming my computations are correct :D), and this clearly tends to $$2\sin^2\theta\leq 2$$ as $$x\to 0$$ (recall that $$|y|\leq |x|$$).

Therefore, for $$n=2$$ and for $$\varepsilon$$ sufficiently small, no $$A$$ can be found to satisfy the required inequality.