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.

  • $\begingroup$ It is exactly as in the links above - it is the inclusion-exclusion principle, but the last term should be dim$(V\oplus \{0\}\cap \{0\}\oplus V) = 0$ $\endgroup$ – Prahlad Vaidyanathan Dec 10 '13 at 12:08
  • $\begingroup$ Except that inclusion-exclusion does not actually hold in the more general setting where you have more than $2$ vectorspaces (unless they intersect trivially as they do here). So actually applying inclusion-exclusion seems like it is not the best idea (as the important part of that is controlling the double-counting, which does not happen at all here). $\endgroup$ – Tobias Kildetoft Dec 10 '13 at 12:10
  • $\begingroup$ Is it true that if $V=U_1\oplus U_2\oplus...\oplus U_n$ then if $B_i$ is a basis for $U_i$ then $\cup_{i=1}^n B_i$ is a basis for $V$? And thus dim $V= |\cup_{i=1}^n B_i|$? If this is so then in my problem if $B$ is a basis for $V$ then dim $V^m=|\cup B|$ but $\cup B=B$ right? So dim $V^m=dim V$. I'm quite confused. Thanks for the replies. $\endgroup$ – chowching Dec 10 '13 at 12:28

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 $V\times V\times \cdots \times V$.

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  • $\begingroup$ Yes, $V\oplus V\oplus...\oplus V$ is an external direct sum. But how does it follow that $dim V^n=dim V + dim V+...dim V$? $\endgroup$ – chowching Dec 10 '13 at 12:43
  • $\begingroup$ @chowching, see my edited answer. $\endgroup$ – lhf Dec 10 '13 at 12:51
  • $\begingroup$ So $dim V^n=|B\times 0 \times \cdots \times 0 \cup 0 \times B \times 0 \times \cdots \times 0 \cup \cdots \cup 0 \times 0 \times \cdots \times B|$ and applying the principle of inclusion exclusion this is equal to $|B\times 0 \times \cdots \times 0|+|0 \times B \times 0 \times \cdots \times 0|+...+|0 \times 0 \times \cdots \times B|$. This will just be all that will be left because the intersection of any of these sets is just $0\times 0\times\cdots\times 0$ right? And $0\times 0\times\cdots\times 0$ is of dimension $0$? $\endgroup$ – chowching Dec 10 '13 at 13:02

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