Section 1.4

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1. Find a subset that is closed under scalar multiplication but not under addition of vectors.

Solution:
There are many possible answers that work. Here is one of them.
with . Then if and we have

because so that is closed under scalar multiplication. However, and , but so that is not closed under addition of vectors.


2. Find a subset that is closed under vector addition but not under multiplication by complex number.

Solution:
Many possible answers again. Here are a few:

, , , , all are closed under addition, but if you multiply for all of these by then you get .


3. Find a subset that is closed under addition but not scalar multiplication by real scalars.

Solution:
Here them using can be a hint. If you let , the rational numbers, then they will be closed under addition, but not scalar multiplication. That is because and , but .

Another possible answer is . Then this will be closed under addition since the sum of two positive numbers is still positive, but and and .


6. Let be a complex vector space i.e., a vector space where the scalars are . Define as the complex vector space whose additive structure is that of but where complex scalar multiplication is given by . Show that is a complex vector space.

Solution:
To begin we don’t need to show any of the first four axioms are true as they only involve addition of vectors and since has the same additive structure as and is a vector space, the first four axioms will still be true. For the remaining four properties we simply check that they will hold.

5)
6)
7)
8)
Therefore is a complex vector space.

7. Let be the set of polynomials in of degree .
(a) Show that is a vector space.

Solution:
Suppose and . Then and . So that . Also so that is closed under addition and scalar multiplication. Also so addition is commutative. Similarly for associative. The zero vector in is the zero polynomial with all coefficients equal to 0. The additive inverse of is . Also, , , and just by writing each of them out. Therefore is a vector space.


(b) Show that the space of polynomials of degree is and does not form a subspace.

Solution:
First off, the space is equal to because the polynomials that have exactly degree are in but not in and there are no other polynomials in that aren’t also in . Now this set does not form a subspace because it is not closed under addition. and are both polynomials of degree exactly . However, is a polynomial of degree 0 not .


(c) If , show that is a subspace of .

Solution:
To show it is a subspace, we only need to check that it is closed under addition and scalar multiplication. Let and . Then for some polynomial and for some polynomial . Then . But since is just another polynomial in , then is exactly of the form polynomial times . Thus . Also which is again of the form polynomial times so . Therefore is a subspace. 8)
Therefore is a complex vector space.


8. Let . Define addition on by . Define scalar multiplication by .
(a) Show that if we use and , then the first four axioms for a vector space are satisfied.

Solution:
1) so addition is commutative.

2) and addition is associative.
3)
4)


(b) Which of the scalar multiplication properties do not hold?

Solution:
The only property that won’t hold is associativity of scalar multiplication.

which is not the same as .