# Dimension $V_1\times…\times V_n$

I have n Vector spaces $V_1,...,V_n$ and would like to show that $\dim(V_1\times...\times V_n)=dimV_1+...+dimV_n$

Is it possible to show this relation using somehow that the dimension of the tensor products $\dim(V_1\bigotimes...\bigotimes V_n)=dim V_1...dimV_n$

If M is set of all function with finite support on $V_1\times...\times V_n$ then $V_1\bigotimes...\bigotimes V_n=M/M_0$ i.e a quotient space so I thought there might be a way to show the dimension formula above.

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There's no need to use the tensor product. The proof that the dimension of a (finite) product of vector spaces is equal to the sum of their dimensions can be done elementarily.

Suppose $V_i$ has basis $\mathcal{B}_i$, and let $$\mathcal{B} = \{ (0, \cdots, 0, v_i, 0, \cdots, 0)\, :\, v_i \in \mathcal{B}_i ,\ 1 \le i \le n\}$$ That is, $\mathcal{B}$ is the set of vectors in $V_1 \times \cdots \times V_n$ whose $i$th components are the basis vectors of $V_i$.

Prove that $\mathcal{B}$ is a basis for $V_1 \times \cdots \times V_n$, and that $\mathcal{B}$ has $|\mathcal{B}_1| + \cdots + |\mathcal{B}_n|$ elements, i.e. there is a bijection between $\mathcal{B}$ and the disjoint union of the $\mathcal{B}_i$s.

This works even when the vector spaces are not finite-dimensional.

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Thanks four your help. B is a basis for $V_1\times...\times V_n$ simply because from $\lambda_1*(v_1,...,0)+...+\lambda_n(0,,,v_n)=0$ it follows that $\lambda_1=...=\lambda_n=0$ correct? I dont see which map is a bijection from B to the disjoint unions of the §B_i§ – Voyage Jan 28 '13 at 0:01
@Voyage: Not quite, but almost. The idea is that if you have any linear combination of vectors from $\mathcal{B}$, it is equal to zero iff each of its components are equal to zero; a given component is zero iff the coefficients for that component are zero; so all the components are zero iff all coefficients are zero. Don't worry too much about the bijection to the disjoint union $-$ that's just my fancy way of saying 'show that there are the right number of elements in the set' and is one of many ways of going about it. – Clive Newstead Jan 28 '13 at 0:03
Thanks but would you mind to write the first part of your comment down mathematically? I also understand that $|B|=|B_1|+..+|B_n|$, it is intuitevly clear to me but I do not know how to write it down formally. – Voyage Jan 28 '13 at 0:09
@Voyage: The elements of $\mathcal{B}$ can be listed as $\sum_{i=1}^n \sum_{j=1}^{m_j} v_{ij}$, where $v_{ij}$ is the vector in the product $V_1 \times \cdots \times V_n$ whose $i$th component is the $j$th basis vector of $V_i$ and all of whose other components are zero. Then $\sum_{i,j} a_{ij}v_{ij} = 0$ if and only if $\sum_j a_{ij}v_{ij}=0$ for each $i$. But this can only occur if $a_{ij}=0$ for all $1 \le j \le m_i$ (check this $-$ use the fact that it's (sort of) a basis for $V_i$), and so it can only hold if $a_{ij}=0$ for all $i,j$. (Now you see why I didn't write it 'mathematically'!) – Clive Newstead Jan 28 '13 at 0:14
(By the way, in the above, $m_j$ is meant to be the dimension of $V_j$.) – Clive Newstead Jan 28 '13 at 0:15