Let $Bs=\{e_1,e_2,...,e_n\}$ is standard base $\mathbb R^{n}$. If $x_1,x_2,...,x_n$ vectors from space $R^{n}$ such that $e_i\in L(x_1,x_2,...,x_n)$, $i=1:n$ then set $\{x_1,x_2,...x_n\}$ is base of space $\mathbb R^{n}$?.

My answer is yes. Now we know that $\alpha_1e_1+\alpha_2e_2,...,\alpha _ne_n=0_v$ because $e_1,e_2,...,e_n$ linear independent then $\alpha_1=\alpha_2=...=\alpha _n=0$ know we know that $e_i\in L(x_1,x_2,...,x_n)$, $i=1:n$ so then,$\alpha_1^{`}x_1+\alpha_2^{`}x_2,...,\alpha _n^{`}x_n=e_1$, $\beta_1x_1+\beta_2x_2,...,\beta _nx_n=e_2,...,\gamma_1x_1+\gamma_2x_2,...,\gamma _nx_n=e_n$ then $x_1(\alpha_1\alpha_1^{`}+\alpha_2\beta_1...\alpha _n\gamma _1)+x_2(\alpha_2\alpha_2^{`}+\alpha_2\beta_2...\alpha _n\gamma _2)+...+x_n(\alpha_1\alpha _n^{`}+\alpha_2\beta _n...\alpha _n\gamma _n)=0_v$, from here $0x_1+0x_2+..0x_n=0v$ so $\{x_1,x_2,...x_n\} $is base. Is this ok?


It's a bit hard to follow what you're doing.

The set is a basis for $\mathbb{R^n}$ if it spans $\mathbb{R^n}$ and if it is linearly independent.

But because we know that the dimension of $\mathbb{R^n}$ is $n$ and the set $\{x_1,x_2,...x_n\}$ contains exactly $n$ elements, it suffices to show either of these properties since:

  • $n$ linearly independent vectors necessarily span $\mathbb{R^n}$ or, the other way around;
  • a spanning set of $n$ elements for $\mathbb{R^n}$ is necessarily linearly independent.

Now, an arbitrary $x \in \mathbb{R^n}$ can be written as $\alpha_1e_1 + \cdots + \alpha_ne_n$ but all of the $e_i$'s can be written as linear combinations of the $x_i$'s, so substitute and rearrange a bit - it's only a matter of writing it down carefully now.

  • $\begingroup$ Just a small warning. This assumes the OP is already familiar with the dimension theorem for vector spaces, which is probably a safe bet, but for future readers, a small warning might help. en.wikipedia.org/wiki/Dimension_theorem_for_vector_spaces $\endgroup$ – 5xum Aug 23 '18 at 9:32
  • $\begingroup$ I know and I was wondering whether to mention that explicitly; but now this comment does the job - thanks! $\endgroup$ – StackTD Aug 23 '18 at 9:32
  • $\begingroup$ can you write how would you prove? $\endgroup$ – Marko Škorić Aug 23 '18 at 10:03
  • $\begingroup$ @MarkoŠkorić Do you understand the last paragraph; have you tried writing it out? $\endgroup$ – StackTD Aug 23 '18 at 10:48

Your proof is a little confusing. What you need to prove are two things:


You need to prove that every $x\in\mathbb R^n$ can be written as a linear combination of $x_1,\dots, x_n$ (you sort of almost did this part, but you should make it clear).


You need to prove that if $\alpha_1 x_1 + \cdots + \alpha_n x_n = 0$, then $\alpha_i=0$ for all $i$ (you didn't do this part yet).

I suggest you rewrite your proof to make it clear which of the two facts is being proven at which point.


I write that every $ei, i:n$ can write as linear combination, after that I want to see if $x1,x2,...xn$ is linear independent, maybe its confusing why i write $0x1+0x2...+0xn=0$ because first I write $\alpha1=\alpha2=...=\alpha n=0$ so when you multiply with $\alpha1,\alpha2,...\alpha n$ in brackets you have 0, so they are linear independent.


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