## Halmos Chapter one Sections eight and nine: Dimension and Isomorphism

1.

**Problem:**

**a) ** What is the dimension of considered as a real vector space?

**b) **Every complex vector space is intimately associated with a real vector space ; the space obtained from by refsing to multiply vectors of by anything but real scalars. If what is

**Proof:**

**a) **Clearly and are l.i. in and given any we have that and thus is a basis for . So,

*Remark: *The subscript is used to indicate that is to be taken over . Note that if we consider as a vector space over then is a basis and so . In fact, this clearly generalized that if is a field then since is a basis.

**b) **Using the above notation the question is asking given what is , we intend to prove that

To see this it suffices to produce a basis of cardinality for . So, by assumption there exists a basis for , consider then (remembering that if then ) the set . We claim this is a basis for . To see this first suppose that

for . Now, if we think again of we see that says

so then by the l.i. of in we see that this implies that

which is true iff from where the l.i. of in follows. Now, to see that

To see this note that for any there exists such that

which implies that

and so is a linear combination of with coefficients in . The conclusion follows.

*Remark:* If one is reading this, there may be a feeling that some kind of mathematical legerdemain is going on here. I assure you there is not. Let me take time to clarify this. Remember that being a vector space over means there is a scalar multiplication such that

Or, if we instead of writing the function by concatenation think of it as the function

(the subscript of will make sense later) so that . Then, says that

Which in words says that “If we multiply and (using the FIELD multiplication of ) and then multiply by the result (using SCALAR multiplication) we get the same element of that we would get if we first multiply by (using SCALAR multiplication) and then multiply the result (which lives in ) by (using SCALAR multiplication)”.

Now, when we consider all we’ve really done is considered the same set of vectors except now we’ve created a new function

such that

except that may ONLY be real. In other words (the scalar multiplication of ) is just the restriction of the scalar multiplication on to . Put symbolically

So now, we can apply this less deceptive way of writing things to resolve the l.i. part of the above proof (you can apply the same logic to the spanning part). We first rewrite the claimed basis of as

where now it makes more sense since each and thus a legitimate vector choice. Now, to prove l.i. we suppose there are such that

But, for each instance of we note that the first slot is taken up by a real number (either or ) and the second slot by a vector (either or ) and so by the above discussion we see that gives the same value as and so may be rewritten as

but remembering that since the domain of is we may apply our previous discussion to conclude

and so (using more axioms for the scalar multiplication)

and so may be rewritten as

but the l.i. of in says that this is true if and only if

but by the definition of the complex numbers this is true if and only if

And thus, remembering this is precisely what we wanted to prove in we’re done.

As you can see, the actual proof is easy but subtleties are abound in this problem. I hope I’ve clarified any misunderstandings in the above proof.

2.

**Problem:** Is the set of all real numbers a finite dimensional vector space over ?

**Proof:** No it’s not. Suppose there existed which is a basis for over . Then, define

where are the unique rational numbers such that

Now, to see that is injective suppose that

then,

from where injectivity immediately follows. Surjectivity is an immediate consequence of

Thus, it follows that . This is a contradiction though since and

*Remark: *In fact, it is fairly easy to show that given a vector space with that

(same mapping as above) is a linear isomorphism. In other words, it’s bijective and

3.

**Problem: **How many vectors are there in an-dimensional vector space over the field .

**Proof: **Clearly by what was said in the last problem it’s

4.

**Problem:** Discuss the following assertion: if two rational vector spaces have the same cardinal number then they are isomorphic.

**Proof: **This is completely ridiculous. Note that

yet

since

and since dimension is invariant under linear isomorphism the conclusion follows.

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