# Dimension of the direct product of two vector spaces

The direct product of the vector space $$\mathcal V$$ of order n and the vector space W of order m is the vector space of order n+m on the set of vectors $$\{(v_1,\dots,v_n,w_1,\dots,w_m:\textbf v\in V,\textbf w \in W\}$$. Note that $$\dim(V \text{direct product} W)=\dim (V)+\dim (W)$$.

Why is this? It looks like if we have n=m=3, we have 3x3=9 dimension. (I know this is not possible with a vector of length 6.) Consider a set of three lin indep vectors $$\textbf v_1,\textbf v_2,\textbf v_3$$ in V and $$\textbf w_1,\textbf w_2,\textbf w_3$$ in W. Appending $$\textbf v_1,\textbf v_2,\textbf v_3$$ to $$\textbf w_1$$, we get three lin indep vectors compared to appending them to $$\textbf w_2,\textbf w_3$$. Why doesn't this argument work?

• The construction you have in mind is probably the one of tensor product. This instead is the cartesian product.
– Bob
Commented Jun 18, 2022 at 5:02
• @Bob that helps, so the basis should be (vi,0,0,0),(0,0,0,wi),i=1,2,3
– Vons
Commented Jun 18, 2022 at 5:13

$$\begin{pmatrix}\textbf v_1\\\textbf w_1\end{pmatrix}= \begin{pmatrix}\textbf v_1\\ \textbf w_2\end{pmatrix}+ \begin{pmatrix} \textbf v_2\\ \textbf w_1\end{pmatrix}-\begin{pmatrix} \textbf v_2\\ \textbf w_2\end{pmatrix}$$