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1.if $(\alpha_1 , \alpha_2)$ and $(\beta_1 , \beta_2)$ they are linearly independent vectors in $R^2$ , then for every $\alpha_3 ,\beta_3$ $\in \Bbb R$ the vectors $(\alpha_1 , \alpha_2 ,\alpha_3 )$ and $(\beta_1 , \beta_2 , \beta_3)$ they are also linearly independent

  1. if $(\alpha_1 , \alpha_2 ,\alpha_3 )$ and $(\beta_1 , \beta_2 , \beta_3)$ are linearly independent then $(\alpha_1 , \alpha_2)$ and $(\beta_1 , \beta_2)$ are linearly independent as well.

What I did -

let $\lambda_i \in \Bbb R$ $i \in \Bbb N$ for the first question it is given that the vectors so $ \lambda_1\cdot (\alpha_1 , \alpha_2) + \lambda_2 (\beta_1 , \beta_2)=(0,0)$ then we get \begin{cases} \lambda_1\cdot \alpha_1 + \lambda_2 \beta_1 = 0\\ \lambda_1\cdot \alpha_2 + \lambda_2 \beta_2 = 0 \end{cases} given the information that it is independent then $\lambda_1 , \lambda_2$ has to be $\lambda_1 = \lambda_2 =0$ which means $(\alpha_1 , \alpha_2 ,\alpha_3 )$ and $(\beta_1 , \beta_2 , \beta_3)$ are also independent because $\lambda_1 = \lambda_2 =0$ \begin{cases} \lambda_1\cdot \alpha_1 + \lambda_2 \beta_1 = 0\\ \lambda_1\cdot \alpha_2 + \lambda_2 \beta_2 = 0\\ \lambda_1\cdot \alpha_3 + \lambda_2 \beta_3 = 0\\ \end{cases} So the statement is true.

for the second question we also have $ \lambda_1\cdot (\alpha_1 , \alpha_2 , \alpha_3) + \lambda_2 (\beta_1 , \beta_2 , \beta_3)=(0,0,0)$but if $(\alpha_1 , \alpha_2 , \alpha_3) , (\beta_1 , \beta_2 , \beta_3)$ are linearly independent is doesn't necessarily mean that $(\alpha_1 , \alpha_2)$ and $(\beta_1 , \beta_2)$ are also becaumse assume $\alpha=(0,1,0)$ and $\beta = (0,1,1)$ then the vectors in $\Bbb R^3$ are linearly independent but in $\Bbb R^2$ are dependent because $1\cdot(0,1) -1\cdot (0,1)=(0,0)$

Is my way correct? is there a general way or an explanation other than the counter example(if correct) on why any of this happens (also for the first part) ? thank you!

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    $\begingroup$ Yes the second statement is false and your counterexample is valid. $\endgroup$
    – podiki
    Commented Nov 10, 2021 at 18:57

1 Answer 1

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Consider the projection $p:\Bbb R^3\to\Bbb R^2,\ (\alpha_1,\alpha_2,\alpha_3)\mapsto(\alpha_1,\alpha_2)$.
It's a linear map with nontrivial kernel.

The first statement generalizes as:

For any linear map $f:V\to W$ if $f(v_1),\dots,f(v_k)$ are linearly independent then so are $v_1,\dots,v_k$.

The (negation of the) second statement:

If a linear map $f:V\to W$ has nontrivial kernel (i.e. isn't injective) then there are linearly independent vectors in $V$ which are mapped to linearly dependent vectors in $W$ by $f$.

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