# Lower bound on quadratic form with positive definite matrix

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?

• Symmetric positive definite matrix is orthogonally diagonalizable so $M = QDQ^T$ with $Q^TQ = QQ^T = I$ and so your question is equivalent to showing: $$x'(QDQ^T - D)x\geq 0$$ or equivalently $M-D$ is positive semidefinite. I am not sure if this is true in general Commented Jun 14, 2023 at 19:54
• $$M = \left( \begin{array}{rr} 5 & -4 \\ -4 & 5 \\ \end{array} \right)$$ and $$x = \left( \begin{array}{r} 0 \\ 1 \\ \end{array} \right)$$ Commented Jun 14, 2023 at 20:03
• Your definition of $D$ is ambiguous because you don't give the order in which you rank the eigenvalues. Commented Jun 14, 2023 at 21:51

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 $$M-D=0 \ \iff \ M=D$$.

• Thanks for your answer. We have that $M$ is symmetric and positive definite, and $-D$ is diagonal and negative definite. Can you explain how to infer that the sum $M - D$ is semi-positive definite? Commented Jun 15, 2023 at 7:55
• Just because a matrix $A$ is semi-positive definite iff $\forall x, x^T\!\!Ax \ge 0$... Commented Jun 15, 2023 at 9:10
• ok, this one was clear :-) Commented Jun 15, 2023 at 10:25

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.