# Prove or disprove a matrix inequality (positive semidefinite)

I want to prove or find a counterexample that there exist constants $$\mu>0, \rho>0$$ such that the following inequality holds: \begin{align} (H + \mu M)^2 \succeq \rho M^2, \end{align} where $$\mu>0, \rho>0$$ are constants to be chosen, $$H\in \mathbb{R}^{n\times n}$$ is a fixed symmetric matrix with bounded eigenvalues $$\lVert H\rVert\le \ell$$, $$M = I_n - Z \in \mathbb{R}^{n\times n}$$ with $$Z = \mathbb{1}_n\mathbb{1}_n^\top/n\in \mathbb{R}^{n\times n}$$, $$\mathbb{1}_n = (1, \dotsc, 1)^\top\in \mathbb{R}^{n}$$, and $$A \succeq B$$ means that $$A-B\succeq 0$$ is a positive semidefinite matrix.

The matrix $$M$$ has some properties that might be helpful: (a) $$M^2 = M$$, (b) the eigenvalues of $$M$$ are $$\lambda =1$$ with multiplicity $$n-1$$ and $$\lambda =0$$ with multiplicity $$1$$, (c) $$M\mathbb{1}_n=0$$.

• Yes, I have tried $n=2$ by hand and we can always find such $\mu$ and $\rho$. May 5, 2021 at 15:44
• @Nicole please replace "Prove of disprove" by "Prove or disprove" in the title of your question. May 5, 2021 at 15:46

This is true. Change coordinates so that $$M$$ is the projection to first $$n-1$$ coordinates: $$Me_n=0$$, $$Me_i=e_i$$ for $$i.

If $$He_n=0$$, then $$H,M$$ both have the same invariant orthogonal decomposition $$e_n\oplus e_n^{\perp}$$, they vanish on the first component and $$M$$ acts as $${\rm Id}$$ on the second component, so large $$\mu$$ and $$\rho=1$$ work.

If $$He_n\ne 0$$, I claim that $$H+\mu M$$ is non-singular for certain $$\mu$$, and for such $$\mu$$ the appropriate $$\rho$$ of course exists. For proving this, we expand $$\det(H+\mu M)$$ as a polynomial in $$\mu$$. The coefficient of $$\mu^{n-1}$$ equals $$H(n,n)$$ (corresponding matrix element) and the coefficient of $$\mu^{n-2}$$ equals $$-\sum_{k=1}^{n-1} H(k,n)^2+H(n,n)\times \text{something}$$. At least one of these two coefficients is non-zero, thus the value of the determinant is non-zero for certain $$\mu$$.

Below goes the (negative) answer fo the uniform-in-$$H$$ version.

Let $$n=2$$. We change the coordinates so that $$M=\begin{pmatrix}1&0\\0&0\end{pmatrix}$$. Choose $$H=\begin{pmatrix}1&1\\1&\frac1{1+\mu}\end{pmatrix}$$ (such matrices are uniformly bounded for $$\mu>0$$, so your $$\ell$$ exists). Then $$H+\mu M$$ has a zero eigenvalue with an eigenvector $$\begin{pmatrix}1\\-1-\mu\end{pmatrix}$$ which is not an eigenvector of $$M$$. So the inequality $$(H + \mu M)^2 \succeq \rho M^2$$ does not hold for no $$\rho$$.

• Hi, thanks for your example. But I am sorry that I didn't make it clear. Here H is a fixed symmetric matrix meaning that it is fixed before we choose $\mu$. While in your example, $H$ changes with $\mu$. May 5, 2021 at 15:43
• Well, but if it is given, why to say that its eigenvalues are bounded? Any finite set is bounded. May 5, 2021 at 15:59
• Yes, you are right. While I was trying to say was that this $H$ matrix is fixed with eigenvalues less than $\ell$, and after we have such an $H$ we then try to find constants $\mu$ and $\rho$. May 5, 2021 at 16:09