is linear combination unique for each vector?

Question

each vector $\vec x$∈$R^n$ can be expressed as a sum $$\vec x = \vec x_1\vec e_1+···+\vec x_n\vec e_n$$ Show that this expression is unique, that is, there does not exist other, different linear combination of vectors $$\vec e_1,\vec e_2,...,\vec e_n$$ which is equal to $\vec x$. I've test vector $$\begin{bmatrix}16 \\32 \\\end{bmatrix} = 1\begin{bmatrix}1 \\1 \\\end{bmatrix} + 2\begin{bmatrix}8 \\16 \\\end{bmatrix}$$ $$\begin{bmatrix}16 \\32 \\\end{bmatrix} = 2\begin{bmatrix}2 \\8 \\\end{bmatrix} + 4\begin{bmatrix}3 \\4 \\\end{bmatrix}$$ and wrote it in such different ways. So I don't get what's happening here, is not this against uniqueness? Or if I am wrong then give me some hints how to prove it.

Answer 1

This statement means that for a given basis of linearly independent vectors $\vec e_1, \dots, \vec e_n$ there is a unique decomposition of any vector $\vec x$ into a linear combination $\vec x = x_1\vec e_1 + \dots + x_n \vec e_n$. Note that $x_i$ are not vectors, but real numbers: coordinates of the vector $\vec x$ in the basis $\vec e_1, \dots, \vec e_n$.

In your example you considered two different bases, so of course the coordinates are different.