Q:

Linear operator composition and dimensions

Let V be linear space, dim(V) = 50. 
Whether there is such linear operator F on V that dim(Im(F)) = 49 and dim(Im(F∘F) = $47$ ? 
My teacher showed me this: 

$dim(im(F)) = 49 \quad  => \quad  dim(ker(F)) = 1 $
  $dim(im(F∘F)) = 47 \quad  => \quad  dim(ker(F∘F)) = 3$  
  Let $\hat v$ be base of $Ker(F) \quad => \quad$  $F(\hat v) = 0$ 
  $F(F(\hat v)) = F(0) = 0 \quad  =>  \quad \hat v \ \epsilon \   Ker(F∘F) $ 
  So let $\hat v, v_1, v_2$ be base of $F^2$ (by Stenitz exchange lemma). 
  Therefore:  
  $F(F(\hat v)) = 0$ 
  $F(F(v_1)) = 0  \quad  => \quad \exists w_1 \ \ F(v_1) = w_1$ 
  $F(F(v_2)) = 0  \quad  => \quad \exists w_2 \ \ F(v_2) = w_2$ 
  $w_1, \  w_2 \ne 0$ because in such case $F(v_1) = F(v_2) = 0\ $ so $\ v_1, > v_2 \ \epsilon \ ker(F) $ and $ v_1, v_ 2$ are lineary independent with $\hat v  \ =>  \ dim(ker(F)) > 1 \ => \ $  contradition
  So  $F(w_1) = F(w_2) = 0 $ 

And here I'm lost. 
My teacher said that $w_1,w_2$ are lineary independent and that implies that $w_1, w_2 \ \epsilon \   ker(F) \ => \ dim(F) > 1 $  I don't understand why $w_1, w_2$ are lineary independent and how can they be in $ker(F)$ when they belong to space with lower dimension than 50. 
I would really appreciate some explanation.

A:

If $w_1$ and $w_2$ are not linearly independent, we can find $\lambda_1,\lambda_2$ such that $\lambda_1w_1+\lambda_2w_2=0$.  But
$$
\lambda_1w_1+\lambda_2w_2=\lambda_1F(v_1)+\lambda_2F(v_2)=F(\lambda_1v_1+\lambda_2v_2),
$$
so that $\lambda_1v_1+\lambda_2v_2\in\ker F$.  But, we know that $\ker F$ is generated by $v$, so that $\lambda_1v_1+\lambda_2v_2=\lambda_v v$ for some $\lambda_v$.  This shows that $v_1,v_2,v$ is not a linearly independent set, which is a contradiction (as we assumed they are a basis for $\ker F\circ F$).
So, it must be the case that $w_1$ and $w_2$ are linearly independent.  
As for your second question... I'm not sure I follow.  Though they come from a lower-dimensional space, they are still part of your original $V$; they are from a lower-dimensional subspace of $V$. Note that they are contained in $\ker F$ because, for instance,
$$
F(w_1)=F(F(v_1))=(F\circ F)(v_1)=0
$$
by assumption, because $v_1\in\ker F\circ F$.

