Vector space/Product/Dimension/Exercise/Solution

The product space has dimension ${}n+m$. To proof this let ${}v_{1},\ldots ,v_{n}$ be a basis of ${}V$ and let ${}w_{1},\ldots ,w_{m}$ be a basis of ${}W$. We claim that the elements

$(v_{j},0),\,j\in \{1,\ldots ,n\},{\text{ and }}(0,w_{i}),\,i\in \{1,\ldots ,m\},$

form a basis of ${}V\times W$.

Let ${}(v,w)\in V\times W$. Then there are representations

$v=\sum _{j=1}^{n}a_{j}v_{j}{\text{ and }}w=\sum _{i=1}^{m}b_{i}w_{i}.$

Therefore

${}{\begin{aligned}(v,w)&=(\sum _{j=1}^{n}a_{j}v_{j},\sum _{i=1}^{m}b_{i}w_{i})\\&=(\sum _{j=1}^{n}a_{j}v_{j},0)+(0,\sum _{i=1}^{m}b_{i}w_{i})\\&=\sum _{j=1}^{n}a_{j}(v_{j},0)+\sum _{i=1}^{m}b_{i}(0,w_{i}),\end{aligned}}$

which means that we have a generating system.

To show linear independence, suppose that

${}\sum _{j=1}^{n}a_{j}(v_{j},0)+\sum _{i=1}^{m}b_{i}(0,w_{i})=0=(0,0)\,.$

The same computation shows

${}(\sum _{j=1}^{n}a_{j}v_{j},\sum _{i=1}^{m}b_{i}w_{i})=(0,0)\,$

and this means

$\sum _{j=1}^{n}a_{j}v_{j}=0{\text{ and }}\sum _{i=1}^{m}b_{i}w_{i}=0.$

Since each family forms a basis, we conclude ${}a_{j}=0$ for all ${}j$ and ${}b_{i}=0$ for all ${}i$.

To the exercise