Prove that the dimension of $U\times V$ is equal to $m+n$

Question

Prove that the dimension of $U\times V$ is equal to $m+n$.

Suppose the $dim(U) = m$ and $dim(V) = n$. since $U \times V$ = {$(u,v)|u \in U$ and $v \in V$}, we can represent any element of $U \times V$ by using the corresponding basis. That is $(u,v) = (t_1u_1 + t_2u2+...+t_mu_m, k_1v_1+k_2v_2+...+k_nv_n)$.

It is defined by the exercise conditions that $(u_1,v_1)+(u_2,v_2) = (u_1+u_2,v_1+v_2) \in U$ and $\alpha(u,v) = (\alpha u,\alpha v)$.

$(u,v) = (t_1u_1 + t_2u2+...+t_mu_m, k_1v_1+k_2v_2+...+k_nv_n) = t_1(u_1,0)+...+t_m(u_m,0)+k_1(0,v_1)+...+k_n(0,v_n)$.

So $(u_1,0),...,(u_m,0),(0,v_1),...,(0,v_n)$ spans $U\times V$

Please, can you show me why it is independent too?

Answer 1

Given the set $\{(u_1,0),\ldots,(u_m,0),(0,v_1),\ldots,(0,v_n)\}$ where $\{u_1,\ldots,u_m\}$ is a basis of $U$ and $\{v_1,\ldots,v_n\}$ is a basis of $V$, both over the field $K$.

Consider the equation $$k_1(u_1,0)+\ldots+k_m(u_m,0)+l_1(0,v_1)+\ldots+l_n(0,v_n)=0,$$ where $k_1,\ldots,k_m,l_1,\ldots,l_n\in K$. Then $$(k_1u_1,0)+\ldots+(k_mu_m,0)+(0,l_1v_1)+\ldots+(0,l_nv_n)=0$$ and so $$(k_1u_1+\ldots+k_mu_m,l_1v_1+\ldots+l_nv_n)=(0,0).$$ Since $\{u_1,\ldots,u_m\}$ is a basis of $U$, $k_1=\ldots=k_m=0$, and since $\{v_1,\ldots,v_n\}$ is a basis of $V$, $l_1=\ldots=l_n=0$. Done.

Answer 2

Suppose $(t_1u_1+...+t_mu_m,k_1v_1+...+k_nv_n) = (0,0)$, then specifically, $t_1u_1+...+t_mu_m = 0$ and $k_1v_1+...+k_nv_n$. Now, everything follows from the independency of $U$ and $V$.