2
$\begingroup$

$U$ and $V$ are vector spaces, their Cartesian product is defined by

$$U\times V = [(u,v): u \in U, v \in V]$$

addition and multiplication defined by

$$(u_1,v_1) + (u_2,v_2) = (u_1+u_2, v_1+v_2)$$ $$r(u,v) = (ru,rv)$$

Let the $T_1: V \rightarrow W$ and $T_2:V\rightarrow W$ be linear transformations, and define a transformation $S: V \times V \rightarrow W \times W$ by

$$S[(v_1,v_2)] = (T_1(v_1),T_2(v_2)$$

how do I go about proving the linear transformation here? I know the idea behind proving linear transformations for composition when they are $T_1(T_2(v_1))$ for example but I am not sure how to prove it in this case.

To prove addition I have gotten this far but I don't know where to go from here:

$$S[(v_1,v_2)+(v_3,v_4)] = S[(v_1,v_2)] + S[(v_3,v_4)]$$ $$=[T_1(v_1),T_2(v_2)] + [T_1(v_3),T_2(v_4)]$$

$\endgroup$
1
  • $\begingroup$ The first line of what you wrote is exactly what you want to prove. It's better to begin the the left-hand side of your first line, apply definitions and properties, to arrive at the right-hand side. $\endgroup$ Jan 29, 2018 at 2:50

1 Answer 1

3
$\begingroup$

\begin{align*} S[(v_{1},v_{2})+(v_{3},v_{4})]&=S(v_{1}+v_{3},v_{2}+v_{4})\\ &=(T_{1}(v_{1}+v_{3}),T_{2}(v_{2}+v_{4}))\\ &=(T_{1}(v_{1})+T_{1}(v_{3}),T_{2}(v_{2})+T_{2}(v_{4}))\\ &=(T_{1}(v_{1}),T_{2}(v_{2}))+(T_{1}(v_{3}),T_{2}(v_{4}))\\ &=S(v_{1},v_{2})+S(v_{3},v_{4}). \end{align*}

$\endgroup$
1
  • $\begingroup$ Scalar case is dealt with similar technique. $\endgroup$
    – user284331
    Jan 29, 2018 at 2:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .