0
$\begingroup$

My problem: Let $M$ be the set of vectors in $\Bbb F^n$. We can create the set of vectors $N$ in $\Bbb F^{n-m}$ by cutting last $m$ components of all vectors in $M$. Proof, that if $N$ is linearly independent thus $M$ is linearly independent.

  • Well, my English terminology might not be right, but I will try to explain my problem further by an example, if I understood it right.

{*The solution is in the comments. I will edit this question as soon as possible. My previous attempt was a total nonsense so I deleted.}

  • 0
    in both of your examples dimension cannot be defined because they are not vector subspaces. And your second example is not a counterexample of the statement – 2017-02-08
  • 0
    Yes I see it actually doesn't work, but how to fix/prove it? – 2017-02-08
  • 0
    Say you have a set of linearly independent vectors $v_1,\dots,v_n$ where $v_i$ are in some $\mathbb{F}^m$ (so $m\ge n$, otherwise set of $n$ vectors cannot be linearly independent). You somehow put $k$ numbers at the end of $v_i$'s so that you make them into a vector in $\mathbb{F}^{m+k}$, whose first $m$ entries we know. Call them $u_i$'s. Now if $\lambda_1 u_1 + \cdots +\lambda_n u_n =0$, in particular we know that $\lambda_1 v_1+\cdots \lambda_n v_n=0$, so you require all scalars to be $0$. – 2017-02-08
  • 0
    Is it all? It looks so easy and great. I wanted to fix it by the rank theorem but it is a nonsense, thank you. – 2017-02-08

0 Answers 0