How to apply linear combination therem of functional analysis in case of $\mathbb R^2$?

Question

Linear combination theorem : Let $\{x_1,....,x_n\}$ be a linearly independent set of vectors in normed space $X$, then there is a number $c>0$ such that, for every choice of scalars $a_1,....,a_n$, we have

$$\|a_1x_1+....+a_nx_n\|\geq c(|a_1|+......+|a_n|).$$

I wanted to to see this how this theorem works. $X=\mathbb R^2$ and $\{(1,0),(0,1)\}$ are linearly independent vectors. By theorem there is a $c$ such that

$$\sqrt{a_1^2+a_2^2}\geq c(|a_1|+|a_2|).$$

How can I find $c$ explicitly ? Please give me hint. Also if any one give me motivation for this theorem that would be very helpful for me.

Answer 1

You have $$\bigl(|a_1|+|a_2|\bigr)^2\leq \bigl(|a_1|+|a_2|\bigr)^2+\bigl(|a_1|-|a_2|\bigr)^2 =2\bigl(|a_1|^2+|a_2|^2\bigr)$$ and therefore $$|a_1|^2+|a_2|^2\geq{1\over2}\bigl(|a_1|+|a_2|\bigr)^2\ .$$ Taking the square root gives your inequality with $c={1\over\sqrt{2}}\,$.

The number $c$ in the general theorem depends on the set of given linearly independent vectors $x_k$. The inequality can be regarded as a quantification of linear independence. A linear combination $\sum_{k=1}^n a_k\,x_k$ is not only $\ne0$ when $a\ne0$, but has a norm which increases linearly with the $|a_k|$.