The product space has dimension
. To proof this let
be a
basis
of
and let
be a basis of
. We claim that the elements
-
form a basis of
.
Let
.
Then there are representations
-
Therefore
![{\displaystyle {}{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f320ddefe30e769f2f02df0bc099554f9f737790)
which means that we have a generating system.
To show linear independence, suppose that
-
![{\displaystyle {}\sum _{j=1}^{n}a_{j}(v_{j},0)+\sum _{i=1}^{m}b_{i}(0,w_{i})=0=(0,0)\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d2711009cd520f16322602e727f4084142bc56e)
The same computation shows
-
![{\displaystyle {}(\sum _{j=1}^{n}a_{j}v_{j},\sum _{i=1}^{m}b_{i}w_{i})=(0,0)\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1908099f35647df3bed16075d59147e2fc028e6e)
and this means
-
Since each family forms a basis, we conclude
for all
and
for all
.