Dimension of Direct sum of same Vector Spaces 
If $V$ is a finite dimensional vector space and $V^n$ is the vector space $$V\oplus V\oplus ...\oplus V\quad(\text{n summands})$$ then for each $n\geq 1$, $V^n$ is finite dimensional and dim $V^n=n(\text{dim V})$

Hey guys I'm trying to solve this problem.
My idea is to prove it using induction. That is I will first show that if $V^2=V\oplus V$ then dim $V^2$ is dim $V+$ dim $V$. But is this really so? How do we find the dimension if we add two vector spaces? Should it be like the Inclusion-Exclusion Principle in set theory?
$$\text{dim }V^2=\text{dim }V+\text{dim }V-\text{dim }(V\cap V)$$
But then this will be just $\text{dim }V^2=\text{dim }V$. Or should it be like the ones mentioned here:
Dimension of direct sum of vector spaces $(dim (V\oplus W)=dim V+dim W)$
Dimensions of vector subspaces in a direct sum are additive?
The difference is that the vector spaces I'm using now are all the same. Should it still be $dim (V\oplus V)=$ dim $V$ + dim $V$? This looks neater because if I can prove this then I can proceed using induction right?
How do I approach this problem? Thank you.
 A: The notation $U \oplus W$ is somewhat overloaded.
When $U$ and $W$ are subspaces of $V$, $V=U \oplus W$  means that $V=U+W$ and that $U\cap W=0$. We say that $V$ is the internal direct sum of $U$ and $W$.
When $W=U$, you cannot have $V=U \oplus W$ because $U\cap W=U$, unless $U=W=0$.
The other meaning of $U \oplus W$ is a new vector space built from $U$ and $W$. In this context this is $U \times W$ and indeed in this space $U \times W = U' \oplus W'$, where $U'=U \times 0$ and $W' = 0 \times W$. We say that $V$ is the external direct sum of $U$ and $W$.
So $V\oplus V\oplus \cdots \oplus V$ is to be read as an external direct sum, in which case it's is better expressed as the cartesian product $V\times V\times \cdots \times V$.
The dimension of $V\times V\times \cdots \times V$ is clearly $n \dim V$ because if $B$ is a basis for $V$ then $B \times 0 \times \cdots \times 0 \cup 0 \times B \times 0 \times \cdots \times 0 \cup \cdots \cup 0 \times  0 \times \cdots \times B$ is a basis for $V\times V\times \cdots \times V$.
