Evaluate the Basis and Dimension of $V \times W$ I was coming across a question where I had to find the basis and dimension of $V \times W$, where $V$ and $W$ both are finite dimensional vector spaces.
Consider a particular case where $V = \mathbb{R}^2$ and $W = M_2(\mathbb{R})$ over the field $\mathbb{R}$.
I am aware of the dimesions of the vector spaces that I mentioned in the above particular case. I came across with What is the difference between Cartesian and Tensor product of two vector spaces as well. But I couldn't find what exactly the basis will look like in the above case. Also there was a second thought that where exactly the problem occurs if I conclude above's dimension as 8 and not 6, since basis is the smallest spanning and largest linearly independent then there must be a violation for the same as concluding dimension to be 8 will also include 2 linearly dependent vectors. Therefore its tempting to look for the structure of basis. If someone can help?
Thanks in advance :)
 A: In case you are dealing with a finite product of vector spaces $V_{1}\times V_{2}\times...\times V_{m}$ then often this is denoted by $\displaystyle\bigoplus_{i=1}^{m}V_{i}$ (direct sum) . where $V_{i}$ have dimension $n_{i}$ (say) . Let $\{v_{k}^{i}\}_{k=1}^{n_{i}}=\{v_{1}^{i},...,v_{n_{i}}^{i}\}$ be a basis for $V_{i}$ for each $i=1,2,...m$
Let $(z_{i,j}^{k})_{j=1}^{m}=\begin{cases} v_{k}^{i}\,,i=j\\0\,,i\neq j\end{cases}=(0,0,...,v_{k}^{i},0,...,0)$ for $k=1,2,...,n_{i}$ . That is for a fixed $i$ $z_{i,j}^{k}$ represents the element in the product whose $i$-th component has the $k$-th basis vector  and has $0$ in all other components.
Then $\displaystyle\bigcup_{i,k}\{(z_{i,j}^{k})_{j=1}^{m}\}$ i.e the collection of all $(z_{i,j}^{k})_{j=1}^{m}$ as $i$ varies over $1,2,..m$ and $k$ varies over $1,2,...,n_{i}$ , is a basis for $\displaystyle\bigoplus_{i=1}^{m}V_{i}$  and it's cardinality is $\displaystyle\sum_{i=1}^{m}n_{i}=\sum_{i=1}^{m}\dim(V_{i})$.
Now it will take you a little work to prove that the set as I described above is a basis. Namely you will need to work componentwise and use linear independence of each basis . Showing it is a spanning set is relatively easier. You just take a vector $(x_{1},...,x_{m})$ and write each $\displaystyle x_{i}=\sum_{k=1}^{n_{i}}c^{i}_{k}v_{k}^{i}$ and  write it as $\displaystyle(\sum_{k=1}^{n_{1}}c^{1}_{k}v_{k}^{1},\sum_{k=1}^{n_{2}}c^{2}_{k}v_{k}^{2},...,\sum_{k=1}^{n_{m}}c^{m}_{k}v_{k}^{m})$ and then use linearity.
From these it directly follows that your space will have dimension $4+2$ .
Note that this was a general construction. You will someday learn about Direct Sums or Coproducts and see that a basis is given by $\bigsqcup_{i\in I} \mathcal{B}_i $ for $\bigoplus_{i\in I} V_{i} $ where $I$ can be any indexing set (countable or uncountable) and $\mathcal{B_{i}}$ is a basis for $V_{i}$ for each $i\in I$. Ignore this part if it confuses you right now. Even my notation is not entirely correct but as a set the basis is simply $\bigsqcup_{i\in I} \mathcal{B}_i $.
