Bilinear forms on normed vector spaces

Question

For a positive definite bilinear form a: $\mathbb{R}^n\times\mathbb{R}^n\to \mathbb{R}$ defined by: $$a(x,y)=\sum^n_{i,j=1}a_{ij}x_{i}y_{j}$$ $a_{ij}\in \mathbb{R}\ x,y,$ two vectors of length $n$, prove that: $$a(x,x)\ge \alpha\lVert x \rVert_2^2 \quad \forall x\in \mathbb{R}^n\quad\alpha>0$$ After rewriting a bit we can see, that it is equivalent to saying that $$a\left(\frac{x}{\lVert x\rVert_2},\frac{x}{\lVert x\rVert_2}\right) \ge \alpha$$ Doesn't this already show that due to the bilinearity, such an $\alpha$ exists, since we can always pick it such a way, that:$$\alpha = \frac 1{\lVert x\rVert_2^2}a(x,x)$$or if need be: $$\alpha = \frac 1 {\lVert x\rVert_2^2+1}a(x,x)$$ to ensure that it is less than the bilinear form? Since all factors are $>0$ it seems to me like it would work but it is so simple that I'm uncertain about it. Is there anything I overlooked/missed?

Answer 1

Reducing the problem to vectors of norm $1$ is the first step. The second step is to notice that your bilinear form has a positive minimum on the unit sphere $\mathbb{S}^{n-1}=\{x\in\mathbb{R}^n:\|x\|=1\}$, because it is continuous and the unit sphere is compact.