Find equation for an isomorphism such that In $\mathbb{R}^4$  plane V is given, $V=span(\alpha_1,\alpha_2)$ where $\alpha_1=[1,3,4,1]$, $\alpha_2=[1,2,2,3] $
a) Find the formula for isomorphism $\varphi:\mathbb{R}^4\rightarrow\mathbb{R}^4$, such that $\varphi(V)=span(\epsilon_1,\epsilon_2)$
b) Find the system of equations which will desribe V
In b) probably i will have to use some properties of isomorphism, but i am not certain. How to deal with question of that type?
 A: Do you know how to extend $\alpha_1, \alpha_2$ to a basis consisting of 4 vectors? 
If so, you can build a matrix $M$ with those four vectors as its columns, and the transformation $x \mapsto Mx$ will send
$e_i$ to $\alpha_i$ for $i = 1, \ldots, 4$. 
Therefore the transformation defined by $x \mapsto M^{-1} x$ will send $\alpha_i$ to $e_i$, and you'll be done. 
For part b, if you take the vectors you found in part a and apply the Gram-Schmidt process to them, you'll get four orthonormal vectors $\beta_i$ ($i = 1, \ldots, 4$) such that the span of 
$$
\beta_1, \ldots, \beta_k
$$
is the same as the span of 
$$
\alpha_1, \ldots, \alpha_k
$$
for each $k = 1, 2, 3, 4$. (You might even want to use the $\beta_i$ as the columns of $M$, for then $M^{-1} = M^t$, which is easy to compute!)
Because the vectors $\beta_3$ and $\beta_4$ are independent and orthogonal to $\alpha_1$ and $\alpha_2$, the equations
$$
\beta_3 \cdot x = 0 \\
\beta_4 \cdot x = 0 
$$
determine a plane, and that plane contains $\alpha_1$ and $\alpha_2$, adn therefore must be the plane spanned by those two. 
