$V = \sum_{i=1}^{n} \oplus U_i \Rightarrow {\rm dim}\ (V) = \sum_{i=1}^{n}{\rm dim}\ (U_i) $

Question

Claim : For the finite vector space V and its finite subspaces $U_i, i =1,2,...,n $

$V = \sum_{i=1}^{n} \oplus U_i \Rightarrow {\rm dim}\ (V) = \sum_{i=1}^{n} {\rm dim}\ (U_i) $

Proof

Let $\beta_i$ be the basis of subspace $U_i$ and denote each element of $\beta_i$ $\beta_{i0}, \beta_{i1},....,\beta_{in}$

Since $V = \sum_{i=1}^{n} \oplus U_i$, it follows that $\beta:= \cup_{i}^{n}\beta_i$ spans $V$ and $\cap_{i}^{n}\beta_i = \emptyset$.

Then $\mid\beta\mid =\dim(V)$ and $\mid \cup_{i}^{n}\beta_i\mid = \sum_i^n dim(U_i) $ whcih concludes that

${\rm dim}\ (V) = \sum_{i=1}^{n} {\rm dim}\ (U_i) $ Q.E.D.

Answer 1

You need to prove that for any $i$, $V_i\cap(\sum_{j\ne i} V_j)=\{0\}$.

For this, observe that $\dim (\sum_{j\ne i} V_j) = \sum_{j\ne i} \dim(V_j)$ (if not, \sum_{i} \dim(V_i) can't be $\dim(V)$), and $\dim(\sum_{j} V_j)=\dim(V_i)+\sum_{j\ne i} \dim(V_j) - \dim(V_i\cap(\sum_{j\ne i} V_j))$.

This means $\dim(V_i\cap(\sum_{j\ne i} V_j))=0$, and you're done.

Answer 2

One characterisation of direct sums is that the union of bases of the subspaces (is always a disjoint union, and) gives a basis of the sum. From this it follows that the dimension of any direct sum of subspaces is the sum of the dimensions of those subspaces. Apply this here to the sum being equal to the whole space.

Here is why the union of bases forms a basis of the (direct) sum. By definition of a sum, the union of the bases spans the sum. So it suffices to show the union of basis is an independent set. Let some linear combination of the vectors in the union be given whose value is the zero vector; we need to show that all its coefficients are $0$. Grouping the terms of the linear combination according to the subspace their vector came from, we obtain a list of vectors $v_i\in U_i$ such that $v_1+v_2+\cdots+v_n=0$; from the definition of a direct sum, this can only happen if $v_i=0$ individually for $i=1,2,\ldots,n$. But $v_i$ was obtained as a linear combination of vectors from a basis of$~U_i$, and this can only be zero of all its coefficients are zero. QED

Answer 3

Proof

Let $\beta_i$ be the basis of subspace $U_i$ and denote each element of $\beta_i$ $\beta_{i0}, \beta_{i1},....,\beta_{in}$

Since $V = \sum_{i=1}^{n} \oplus U_i$, it follows that $\beta:= \cup_{i}^{n}\beta_i$ spans $V$ and $\cap_{i}^{n}\beta_i = \emptyset$.

Then $\mid\beta\mid =\dim(V)$ and $\mid \cup_{i}^{n}\beta_i\mid = \sum_i^n dim(U_i) $ whcih concludes that

${\rm dim}\ (V) = \sum_{i=1}^{n} {\rm dim}\ (U_i) $ Q.E.D.