Prove that the sum of the squares of the coefficients of the bases are independent of the bases.

Question

Let {$x_1, ..., x_n$} & {$y_1, ..., y_n$} be bases of a vector space $X \subseteq \mathbb{R}^n$. Let $x \in X$ so that $x=a_1x_1+...+a_nx_n=b_1y_1+...b_ny_n$.

Was wondering if $[a_1^2+...+a_n^2]^{1/2}=[b_1^2+...+b_n^2]^{1/2}$

If so, could someone provide a proof that uses basic linear algebra concepts since I'm fairly new to the topic.

Answer 1

Not true. If $(x_i)$ is a basis so is $(2x_i)$. See what happens in this case. However if the bases are orthonormal bases then the conclusion is true and both sums are equal to $\|x\|$.