# inequality involving eigendecomposition

Let $A \in \mathbb{R}^{n \times n}$ be an Hermitian matrix. This means the eigenvalues $\lambda_{i}, i = 1, ..., n$ are real and the eigenvectors $v_i, i = 1 ... n$ are orthogonal on each other. We assume that alle eigenvalues are simple and we assume the eigenvalues are in increasing order (so $\lambda_1 < \lambda_2< ...< \lambda_n$), we define $V := [v_1 ... v_n]$ and $\Lambda = diag(\lambda_1, ..., \lambda_n)$ and we note $V_m = [v_1 ... v_m]$ and $\Lambda_m = diag(\lambda_1, ..., \lambda$. This means we can write $A = V \Lambda V^*$ In the paper I'm reading they state the following: $$\min_{u \in span{V_m}^\bot, \left \| u \right \| = 1} u^* A u \geq \min_{u \in span{V_m}^\bot, \left \| u \right \| = 1 }u^* \left( V_m \Lambda_m V_m^* + \lambda_{m+1} (I - V_m V_m^*)\right) u$$ I don't know how to use the fact that $u \in span{V_m}^\bot$. Can anyone help me?

• I'm probably not going to be able to solve it anyway, but what does $spanV^{\bot}$ mean? Seems to me like $spanV$ is the whole space. – Callus Aug 28 '17 at 12:29
• You are right. It needed to be $span V_m ^\bot$ so I adapt my question. Thanks for the comment! – Koen Aug 28 '17 at 12:48
• If $u\in\text{span}V_{m}^{\perp}$, then shouldn't the first term on the RHS be zero and the second term be just $\lambda_{m+1}$? – nemo Aug 28 '17 at 13:01
• Yeah, the right hand side should just be $\lambda_{m+1}$ as you say, and the LHS is also equal to $\lambda_{m+1}$ in this case, I believe. – Callus Aug 28 '17 at 13:07
• I think the LHS is bigger than $\lambda_{m+1}$, see my answer. Does that make sense? – nemo Aug 28 '17 at 14:52

I hope you agree that the RHS is $\lambda_{m+1}$, since the first term is zero and the second term is $\lambda_{m+1}$ (since $I-V_{m}V_{m}^{\top}$ is projection onto $V_{m}^{\perp}$ and $u\in\text{span}V_{m}^{\perp}$). Also assume that $u=\Sigma_{i=m+1}^{n}\alpha_{i}v_{i}$. For the LHS: