# Understanding a linear independence question

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!

• Yes the second statement is false and your counterexample is valid. Commented Nov 10, 2021 at 18:57

Consider the projection $$p:\Bbb R^3\to\Bbb R^2,\ (\alpha_1,\alpha_2,\alpha_3)\mapsto(\alpha_1,\alpha_2)$$.
For any linear map $$f:V\to W$$ if $$f(v_1),\dots,f(v_k)$$ are linearly independent then so are $$v_1,\dots,v_k$$.
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 $$f$$.