# Describe explicitly the linear transformation T from $F^2$ to $F^2$ such that $T\epsilon_1=(a,b),T\epsilon_2=(c,d)$

I have to describe explicitly the linear transformation T from $F^2$ to $F^2$ such that $T\epsilon_1=(a,b),T\epsilon_2=(c,d)$

My try:

We know that $T\epsilon_1=(a,b),T\epsilon_2=(c,d)$ so let $\alpha_1=(x_1,x_2)$ and $\alpha_2=(\lambda_1,\lambda_2)$ and let say that $T\alpha_1=(y_1,y_2)$ and $T\alpha_2=(d_1,d_2)$,so we can write $\epsilon_1$ as $$\epsilon_1=c_1\alpha_1+c_2\alpha_2,$$ for some scalars $c_1,c_2$

Therefore, $$T\epsilon=c_1T\alpha_1+c_2T\alpha_2$$

But we know that $T\alpha_1=(y_1,y_2)$ and $T\alpha_2=(d_1,d_2)$, So this becomes

$$T\epsilon_1=c_1(y_1,y_2)+c_2(d_1,d_2)$$$$=(c_1y_1,c_2y_2)+(c_1d_2,c_2d_2)$$$$=(c_1y_1+c_1d_2,c_2y_2+c_2d_2)$$

Now i can set $a=(c_1y_1+c_1d_2)$ and $b=(c_2y_2+c_2d_2)$,

Therefore $T\epsilon_1=(a,b)$. Similarly we can write for $T\epsilon_2$.

Hence a linear transformation has been described explicitly.

My question: Is what i am doing right?

Assuming that by $\epsilon_1$ and $\epsilon_2$ you mean the standard basis vectors $(1,0)$ and $(0,1)$ for $F^2$, this is way too contorted. You've introduced far too many extraneous letters. Just write down that $T(x_1,x_2)=x_1 T(\epsilon_1)+x_2 T(\epsilon_2) = x_1(a,b)+x_2(c,d)$, which you can expand one step farther.