2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.