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Thursday, January 10, 2013 CC-BY-NC
Subspaces and direct sums

Maintainer: admin

Lecture notes from lecture #3 of MATH 236. Material covered: some properties of vector spaces, subspaces, sums and direct sums.

1Properties of vector spaces

  1. A vector space V has a unique additive identity element, the zero vector (0V).
  2. Any element in V has a unique additive inverse.
  3. 0v=0 for any vV.
  4. a0=0 for any aF.
  5. (1)v=v for any vV.

Proof of 4:

a0=a(0+0)=a0+a0a0=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:

  1. Contains the zero vector
  2. Closed under vector addition
  3. Closed under scalar multiplication

An example of something that is not a subspace is {(x1,x2,1)R3x,yR}. This is clearly not a subspace of R3 since neither property above is satisfied.

Example: U={(x,3x)xF2}. 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++umuiUi}. 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 vV), then we say that V=U1++um.

Example: U1={(x,y,0)x,yR}, U2={(0,0,z)zR}, U3={(0,y,y)yR}. 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 aR). 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 vV 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=U1Um.

Examples:

  • U1={(x,y,0)}, U2={(0,0,z)}; then U1U2=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:

  1. V is the sum (i.e. any vector in V can be written as a sum of vectors in the Us etc)
  2. 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 vV). We can prove this by contradiction. Assume that there are two distinct ways of writing an element vV, as v=u1++um and v=v1++vm, where uivi 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

  1. 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.