Lower bound on quadratic form with positive definite matrix

Question

I came accross the following lower bound and I am looking for a proof that uses only basic properties or well-established results of linear algebra (or other domains of math).

Let $M \in \mathbb{R}^{d\times d}$ be a symmetric positive definite matrix and $D = \text{diag}\{\lambda_1, \dots, \lambda_d\}$ the diagonal matrix of the $d$ eigenvalues of $M$. For any vector $x \in \mathbb{R}^d$ it should hold that $$x'Mx \geq x'Dx$$

Can somebody help with proving this?

Answer 1

Writing the condition under the form :

$$\forall x, \ x^T(M-D)x \ge 0,$$

a consequence is that $M-D$ is a (symmetric) semi-definite positive matrix, which is traceless ($\operatorname{trace}(M-D)=\operatorname{trace}(M)-\operatorname{trace}(D) = 0$).

This is impossible for a non-zero matrix, as established here.

The only possible case is therefore when $M-D=0 \ \iff \ M=D$.

Answer 2

The inequality is true iff $M=D.$ The proof will be performed by induction on $d.$ WLOG we may assume that $\lambda_1$ is the greatest eigenvalue. Then $M\le \lambda_1I.$ By assumption $e_1'Me_1\ge \lambda_1.$ Thus $e_1'(M-\lambda_1I)e_1\ge 0.$ As $\lambda_1I-M$ is positive definite, we get $(M-\lambda I) e_1=0,$ i.e. $Me_1=\lambda_1e_1.$ As $M$ is positive definite the subspace $e_1^\perp$ is invariant for $M,$ and clearly for $D.$ Now we may restrict to $e_1^\perp$ and apply the induction hypothesis to show that $M$ restricted to $e_1^\perp$ is diagonal. Thus $M$ is diagonal.