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

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.

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 $$|a_k|$$.