Write the linear application Is given the basis of $\mathbb{R^2}$ $v_1=(1,0) v_2=(1,1)$. Write the only linear application $f: \mathbb{R^2} \to \mathbb{R^2}$ with $f(v_1)=(0,1) f(v_2)=(1,0)$.
I tried to solve it like $(x,y)= x(1,0) + y(1,1)$ and I found $f(x,y) = (y,x)$ that is wrong. Where am I wrong?
 A: The linear function ("application") will be of the form $$f(v)=Mv$$ for some matrix $M$.  You want this to satisfy $f(v_1)=(0,1)^T$ and $f(v_2)=(1,0)^T$.  In other words, you want $$M\left(\begin{smallmatrix} 0&1\\1&1\end{smallmatrix}\right)=\left(\begin{smallmatrix} 0&1\\1&0\end{smallmatrix}\right)$$
To find a matrix $M$ to satisfy this property, we calculate $$\left(\begin{smallmatrix} 0&1\\1&1\end{smallmatrix}\right)^{-1}=\left(\begin{smallmatrix} -1&1\\1&0\end{smallmatrix}\right)$$
Multiplying by this on the right, we get 
 $$M=M\left(\begin{smallmatrix} 0&1\\1&1\end{smallmatrix}\right)\left(\begin{smallmatrix} 0&1\\1&1\end{smallmatrix}\right)^{-1}=\left(\begin{smallmatrix} 0&1\\1&0\end{smallmatrix}\right)\left(\begin{smallmatrix} 0&1\\1&1\end{smallmatrix}\right)^{-1}=\left(\begin{smallmatrix} 0&1\\1&0\end{smallmatrix}\right)\left(\begin{smallmatrix} -1&1\\1&0\end{smallmatrix}\right)=\left(\begin{smallmatrix} 1&0\\-1&1\end{smallmatrix}\right)$$
Putting it all together, we get $f(x,y)=(x+0y,-x+y)$ as the desired function.
A: You ended up with the wrong answer because $(x,y)\ne x(1,0)+y(1,1)$. Start with $(x,y)=a(1,0)+b(1,1)$ and figure out what the coefficients $a$ and $b$ must be in terms of $x$ and $y$. You can then apply linearity: $f(x,y)=af(1,0)+bf(1,1)$.
