Lecture notes from lecture #3 of MATH 236. Material covered: some properties of vector spaces, subspaces, sums and direct sums.
1Properties of vector spaces¶
- A vector space V has a unique additive identity element, the zero vector (→0∈V).
- Any element in V has a unique additive inverse.
- 0→v=0 for any →v∈V.
- a→0=→0 for any a∈F.
- (−1)→v=−→v for any →v∈V.
Proof of 4:
a⋅→0=a⋅(→0+→0)=a⋅→0+a⋅→0∴a⋅→0=→0
The proofs for the others are left as exercises.
2Subspaces¶
A subspace of a vector space V is a subset U of V that is in itself a vector space.
Properties of a subspace U:
- Contains the zero vector
- Closed under vector addition
- Closed under scalar multiplication
An example of something that is not a subspace is {(x1,x2,1)∈R3∣x,y∈R}. This is clearly not a subspace of R3 since neither property above is satisfied.
Example: U={(x,3x)∣x∈F2}. Verifying that this is a subspace is trivial and is left as an exercise for the reader.
Note that any non-trivial vector space V (i.e., a vector space that is not {0}) has at least two (distinct) subspaces: the set consisting of only the zero vector ({0}), and V itself. The trivial vector space {0} has only one distinct subspace. The empty set is of course not a subspace.
3Sums¶
Let U1,…,Um be subspaces of V. Their sum, U1+…Um, is the set of all possible linear combinations (since subspaces are closed under scalar multiplication): {u1+…+um∣ui∈Ui}. The sum of subspaces is a subspace of V in itself.
If any element in V can be written as a combination of vectors in Ui (so v=u1+…+vm for any v∈V), then we say that V=U1+…+um.
Example: U1={(x,y,0)∣x,y∈R}, U2={(0,0,z)∣z∈R}, U3={(0,y,y)∣y∈R}. Clearly, U1+U2+U3=V. We can write the zero vector in more than one way: either as the sum of zero vectors, or as (0,a,0)+(0,0,a)+(0,−a,−a) (for any a∈R). From this, we deduce that one of the subspaces is superfluous. We'll look into this more in the next section.
3.1Direct sums¶
If any element v∈V can be written uniquely as a sum of vectors in U1,…,Um, then we say that V is the direct sum of U1,…,Um, which we write as V=U1⊕…⊕Um.
Examples:
- U1={(x,y,0)}, U2={(0,0,z)}; then U1⊕U2=F3.
- Ue(F)=a0+a2x2+…+a2m+x2m (even polynomials), Uo(F)=a1+a3x3+…+a2m+1x2m+1 (odd polynomials); then Ue(F)⊕Uo(F)P(F) (the polynomial vector space).
3.1.1Verifying direct sums¶
Proposition 1.8: To check that a vector space V is the direct sum of subspaces U1,…,Um, we need to check that the following conditions are satisfied:
- V is the sum (i.e. any vector in V can be written as a sum of vectors in the Us etc)
- The zero element can be written uniquely as a sum of vectors in the Us (0+…+0)
This proposition will be useful for homework questions.
Proof: (→) Assume V is the direct sum. Then 1) comes for free. 2) is also kind of free.
(←) Assume that 1) and 2) hold. Then we need to prove that V is the direct sum (such that there is a unique decomposition for any element v∈V). We can prove this by contradiction. Assume that there are two distinct ways of writing an element v∈V, as v=u1+…+um and v=v1+…+vm, where ui≠vi for at least one i. Consider −v (which we know is an element of V, by property 5 of vector spaces). −v+v=0, since the zero vector is the additive identity. So then we have two distinct decompositions for the zero vector: 0=−v+u1+…+um, and 0=−v+v1+…+vm. This contradicts condition 2) above. ◼1
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This differs somewhat from the proof shown in class, which I believe was not a proof by contradiction. I don't think he finished the proof though so who knows. ↩