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I've got two vector spaces $U$ and $V$ over division ring $\mathbb{T}$ .

Space $W$ over division ring $\mathbb{T}$ is defined as $W =\{( u, v ); u \in U, v \in V \}$ with operations $(u_1, v_1) + (u_2, v_2) = (u_1 + u_2, v_1 + v_2)$ and $\alpha (u_1, v_1) = (\alpha u_1, \alpha v_1)$.

How can I find basis and dimension of $W$?

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Let the basis for space $U$ of dimension $d_u$ be $\{a_i\}$ and for vector space $V$ of dimension $d_v$ be $\{b_j\}$. Then basis for $W$ can be $(a_1,0),(a_2,0)...(a_{d_u},0),(0,b_1)....(0,b_{d_v})$. Thus dimension being $d_u+d_v$.
Proof

  1. Let us have a vector $u \in U$ so it can be written as $$\sum_{i=1}^{i=d_u}c_ia_i$$ for some constants $c_i$,similarly let $v \in V$ then it can be written as $$\sum_{j=1}^{j=d_v}l_jb_j$$ for some constants $l_j$. Thus any vector of the form $(u,v)$ can be written as $$\sum_{i=1}^{i=d_u}c_i(a_i,0)+\sum_{j=1}^{j=d_v}l_j(0,b_j)$$
  2. Also all these are linearly independent.( can see easily )
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Hint: let $\{x_1,\dots,x_m\}$ and $\{y_1,\dots,y_n\}$ be bases of $U$ and $V$, respectively. Prove that $$ \{(x_1,0),\dots,(x_m,0),(0,y_1),\dots,(0,y_n)\} $$ is a basis for $W$.

Extend to infinite dimensional spaces, if you have to.

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