Does positive definiteness imply $x^{T}Mx \geq \alpha x^{T}x$ for some $\alpha > 0$?

Question

I require an $n \times n$ matrix to satisfy $x^{T}Mx \geq \alpha x^{T}x$ for some $\alpha > 0$ for all $x \in \mathbb{R}^{n}$ where $x \geq 0$ (1).

Is it sufficient to show that M is positive definite (such that $x^{T}Mx > 0$ $\forall x\in \mathbb{R}^{n}$)? If not, what additional properties should matrix M satisfy for this property (1) to hold?

Answer 1

$\alpha$ can be taken to be the lowest eigenvalue of $M$. Observe that $M$ has a full set of orthonormal eigenvectors $v_1,\ldots,v_n$, hence any $x$ can be written as $x=c_1v_1+\ldots+c_nv_n$ and we can compute

$$ x^TMx= c_1^2\lambda_1+\ldots+c_n^2\lambda_n\geq \lambda_{min}||x||^2. $$

Answer 2

Yes, positive definiteness implies that $x^{T}Mx \geq \alpha x^{T}x$ for some $\alpha > 0$.

In particular, the Rayleigh Ritz theroem tells us that if $M$ is symmetric, then we have $x^{T}Mx \geq \alpha x^{T}x$ for $\alpha = \min_{x \neq 0} \frac{x^TMx}{x^Tx} = \lambda_{\min}(M)$.

Answer 3

Let $M$ be positive definite. Let $\{e_i\}_i$ be an orthonormal basis of eigenvectors of $M$ and let $\alpha_i$ be the corresponding eigenvalues, which are positive since $M$ is positive definite. Then any $x$ can be written as $$ x = \sum_i x_i e_i. $$ Thus $$x^T M x = \sum_{ij} x_ix_j \; e_i^T M e_j = \sum_{ij} x_ix_j \alpha_j \; e_i^Te_j = \sum_j \alpha_j x_j^2 \geq \alpha x^T x $$ where $\alpha = \min(\alpha_i)$.