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4.4: Sums and direct sum

( \newcommand{\kernel}{\mathrm{null}\,}\)

Throughout this section, V is a vector space over F, and U1,U2V denote subspaces.

Definition 4.4.1: (subspace) sum

Let U1,U2V be subspaces of V . Define the (subspace) sum of U1

figure 4.4.1 - Copy.jpg

Figure 4.4.1: The union UU of two subspaces is not necessarily a subspace.

and U2 to be the set

U1+U2={u1+u2|u1U1,u2U2}.

Check as an exercise that U1+U2 is a subspace of V . In fact, U1+U2 is the smallest subspace of V that contains both U1 and U2 .

Example 4.4.2. Let

U1={(x,0,0)F3|xF},U2={(0,y,0)F3|yF}.

Then

U1+U2={(x,y,0)F3|x,yF}.

If, alternatively, U2={(y,y,0)F3|yF}, then Equation (4.4.2) still holds.

If U=U1+U2 , then, for any uU, there exist u1U1 and u2U2 such that u=u1+u2.

If it so happens that u can be uniquely written as u1+u2 , then U is called the direct sum of U1 and U2.

Definition 4.4.3: Direct Sum

Suppose every uU can be uniquely written as u=u1+u2 for u1U1 and u2U2 . Then we use

U=U1U2

to denote the direct sum of U1 and U2.

Example 4.4.4. Let
U1={(x,y,0)R3|x,yR},U2={(0,0,z)R3|zR}.
Then R3=U1U2. However, if instead
U2={(0,w,z)|w,zR},
then R3=U1+U2 but is not the direct sum of U1 and U2 .
Example 4.4.5. Let
U1={pF[z]|p(z)=a0+a2z2++a2mz2m},U2={pF[z]|p(z)=a1z+a3z3++a2m+1z2m+1}.
Then F[z]=U1U2.

Proposition 4.4.6. Let U1,U2V be subspaces. Then V=U1U2 if and only if the following two conditions hold:

  1. V=U1+U2;
  2. If 0=u1+u2 with u1U1 and u2U2 , then u1=u2=0.

Proof.
() Suppose V=U1U2. Then Condition 1 holds by definition. Certainly 0=0+0, and, since by uniqueness this is the only way to write 0V , we have u1=u2=0.

() Suppose Conditions 1 and 2 hold. By Condition 1, we have that, for all vV , there exist u1U1 and u2U2 such that v=u1+u2 . Suppose v=w1+w2 with w1U1 and w2U2 . Subtracting the two equations, we obtain

0=(u1w1)+(u2w2),

where u1w1U1 and u2w2U2. By Condition 2, this implies u1w1=0 and u2w2=0, or equivalently u1=w1 and u2=w2 , as desired.

Proposition 4.4.7. Let U1,U2V be subspaces. Then V=U1U2 if and only if the following two conditions hold:

  1. V=U1+U2;
  2. U1U2={0}.

Proof.
() Suppose V=U1U2. Then Condition 1 holds by definition. If uU1U2 , then 0=u+(u) with uU1 and uU2 (why?). By Proposition 4.4.6, we have u=0 and u=0 so that U1U2={0}.

() Suppose Conditions 1 and 2 hold. To prove that V=U1U2 holds, suppose that

0=u1+u2, where u1U1 and u2U2.

By Proposition 4.4.6, it suffices to show that u1=u2=0. Equation (4.3) implies that u1=u2U2. Hence u1U1U2, which in turn implies that u1=0. It then follows that u2=0 as well.

Everything in this section can be generalized to m subspaces U1,U2,Um, with the notable exception of Proposition 4.4.7. To see, this consider the following example.

Example 4.4.8. Let

U1={(x,y,0)F3|x,yF},U2={(0,0,z)F3|zF},U3={(0,y,y)F3|yF}.

Then certainly F3=U1+U2+U3 , but F3U1U2U3 since, for example,

(0,0,0)=(0,1,0)+(0,0,1)+(0,1,1).

But U1U2=U1U3=U2U3={0} so that the analog of Proposition 4.4.7 does not hold.


This page titled 4.4: Sums and direct sum is shared under a not declared license and was authored, remixed, and/or curated by Isaiah Lankham, Bruno Nachtergaele, & Anne Schilling.

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