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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.

First I write $\alpha_1e_1+\alpha_2e_2+...+\alpha_ne_n=0$. Since $e_i\in L(x_1,x_2,...,x_n)$ we can write every vector in base $Bs$ as linear combination of $\{x_1,x_2,...,x_n\}$. For example:

$\begin{matrix} e_1=a_{11}x_1+a_{21}x_2+...+a_{n1}x_n\\ e_2=a_{12}x_1+a_{22}x_2+...+a_{n2}x_n\\ .........................\\ e_n=a_{1n}x_1+a_{2n}x_2+...+a_{nn}x_n \end{matrix}$

then $0=x1(\alpha_1a_{11}+\alpha_2a_{12}+...+\alpha_na_{1n})+x2(\alpha_1a_{21}+\alpha_2a_{22}+...+\alpha_na_{2n})+...+xn(\alpha_1a_{n1}+\alpha_2a_{n2}+...+\alpha_na_{nn})$

If we suppose opposite that this vectors is linear dependent, then one vector can be write as linear combination of other vectors, for example if I use $x_n$ to write as linear combination, then $(\alpha_1a_{n1}+\alpha_2a_{n2}+...+\alpha_{nn}a_{1n})\not=0$ because we can not divide something with zero, if we put $(\alpha_10+\alpha_20+...+\alpha_na_{nn})\not=0$ then $\alpha_n \not=0$,so now we have $0e_1+0e_2+...+\alpha_ne_n=0$, then $e_n=0$ but it is not true, so $ L(x_1,x_2,...,x_n)$ is base of $\mathbb R^n$, is this ok?

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  • $\begingroup$ What do you mean you "put ($\alpha_1 0 + \alpha_2 0 + \cdots \alpha_n a_{n,n}) \neq 0$"? $a_{i,j}$ are determined once $(x_j)$ is given and cannot be alternated. $\endgroup$ – xbh Sep 2 '18 at 8:14
  • $\begingroup$ I just put that $\alpha1 \alpha2...$ can be zero but one must not be zero $\endgroup$ – Marko Škorić Sep 2 '18 at 8:16
  • $\begingroup$ According to your notation all $\alpha_j = 0$. $\endgroup$ – xbh Sep 2 '18 at 8:17
  • $\begingroup$ i show that it must be because this vector is linear independent $\endgroup$ – Marko Škorić Sep 2 '18 at 8:19
  • $\begingroup$ I think you have messed up the notation. Maybe you want to write $\alpha_1 x_1 + \alpha_2 x_2 + \cdots +\alpha_n x_n = 0$. $\endgroup$ – xbh Sep 2 '18 at 8:21
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I don’t understand how you move from $(\alpha_1a_{n1}+\alpha_2a_{n2}+...+\alpha_{nn}a_{1n})\not=0 $ to $(\alpha_1 0+\alpha_2 0+...+\alpha_{nn}a_{1n})\not=0$?

However I would use the proof below.

Another proof

$(x_1,\dots,x_n)$ spans $(e_1,\dots,e_n)$ and therefore spans $\mathbb R^n$. As $(x_1,\dots,x_n)$ has $n$ elements, $(x_1,\dots,x_n)$ is a minimal family of vectors that spans $\mathbb R^n$. Hence $(x_1,\dots,x_n)$ is linearly independent and is a basis.

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  • $\begingroup$ i just put some number to show that is not zero, yes I know for that, but I just want to show everything because i have very strict proffesor $\endgroup$ – Marko Škorić Sep 2 '18 at 8:14

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