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Math 24

Spring 2012Problems from Monday April 9

First some denitions. If W 1 and W 2 are two subspaces of V , we dene

W 1 + W 2 = {w1 + w2 | w1 ∈W 1 & w2 ∈W 2 }.

In other words, W 1 + W 2 is the collection of all vectors you can get by addingan element of W 1 to an element of W 2 . If W 1 + W 2 = V and W 1 ∩W 2 = {0},then we say V is the direct sum of W 1 and W 2 , and we write V = W 1 ⊕W 2 .

1. Prove that W 1 + W 2 is the smallest subspace containing both W 1 andW 2 . (In other words, W 1 + W 2 is the span of W 1 ∪W 2 .)

To see W 1 + W 2 is a subspace, check closure under addition and undermultiplication by scalars. Let w1 + w2 and w1 + w2 be any elementsof W 1 + W 2 , where w1 , w 1 ∈W 1 and w2 , w 2 ∈W 2 , and let a be anyscalar. Then, since W 1 and W 2 are closed under addition and undermultiplication by scalars,

(w1 + w2 ) + ( w1 + w2 ) = ( w1 + w1 ) + ( w2 + w2 ) ∈W 1 + W 2 ,

a (w1 + w2 ) = aw 1 + aw 2 ∈W 1 + W 2 .

Also, W 1 ⊆W 1 + W 2 , since every w1 ∈W 1 can be written w1 = w1 +0 ∈W 1 + W 2 . For the same reason, W 2 ⊆W 1 + W 2 . We have shown W 1 + W 2is a subspace containing both W 1 and W 2 .Clearly every element w1 + w2 of W 1 + W 2 is in span (W 1 + W 2 ) soW 1 + W 2 ⊆ span (W 1 ∪W 2 ).To show W 1 + W 2 = span (W 1 ∪W 2 ), it remains only to show thatspan (W 1 ∪W 2 ) ⊆ W 1 + W 2 . But this must be true, because we haveshown W 1 + W 2 is a subspace containing W 1 ∪W 2 , and span (W 1 ∪W 2 )is the smallest such subspace.

We could also show span (W 1 ∪W 2 ) ⊆ W 1 + W 2 directly. Let w ∈

span (W 1 + W 2 ). We can write w as a linear combination of elementsof W 1 ∪W 2 ,

w = a 1 u 1 + · · · + a n u n + b1 v1 + · · · bm vm ,

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where u i ∈W 1 and v j ∈W 2 . But then, a 1 u 1 + · · · + a n u n ∈W 1 , and

b1 v1 + · · · + bm vm ∈W 2 , andw = ( a 1 u 1 + · · · + a n u n ) + ( b1 v1 + · · · bm vm ) ∈W 1 + W 2 ,

so span (W 1 ∪W 2 ) ⊆W 1 + W 2 .

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2. Give examples of pairs of subspaces W 1 and W 2 of R 3 , neither of which

is contained in the other, such that:(a) W 1 + W 2 = R 3 . In your example, what is W 1 + W 2 ?

(b) W 1 + W 2 = R 3 , but R 3 is not the direct sum of W 1 and W 2 . Inyour example, what is W 1 ∩ W 2 ?

(c) R 3 is the direct sum of W 1 and W 2 .

This is a homework problem.

3. SupposeW 1

andW 2

are both subspaces of a nite-dimensional vectorspace V . Make a conjecture about the relationship among the dimen-sions of W 1 , W 2 , W 1 ∩ W 2 , and W 1 + W 2 .

dim (W 1 + W 2 ) = dim (W 1 ) + dim (W 2 ) − dim (W 1 ∩ W 2 ).

Intuitively, we add up the number of dimensions in W 1 and W 2 , andthen subtract the number of dimensions in the overlap, because theywere counted twice.

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4. Express M 2 × 2 (C ) as the direct sum of two nonzero subspaces.

There are many possible answers. A straightforward one is:

W 1 = a b0 0 a, b ∈C W 2 = 0 0

c dc, d ∈C

A possibly more interesting solution is to let W 1 be the subspace of matrices with zero trace, and W 2 be the subspace of diagonal matriceswhose two diagonal entries are equal. It’s easy to see their intersectioncontains only the zero matrix. You can see that together they span theentire space by writing down simple bases for W 1 and W 2 ,

1 00 − 1 , 0 1

0 0 , 0 01 0 and 1 0

0 1 ,

and observing that from them you can generate all the standard basiselements.

5. Express P (R ) as the direct sum of two nonzero subspaces in two ways.

(a) One of the subspaces has nite dimension.

(b) Both of the subspaces are innite-dimensional.

This is a homework problem.

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6. Prove the conjecture you made in problem (3). Hint: A basis {x 1 , . . . , x k }

for W 1 ∩ W 2 can be extended to a basis {x 1 , . . . , x k , y1 , . . . y n } for W 1 .It can also be extended to a basis {x 1 , . . . , x k , z 1 , . . . z m } for W 2 . Forhomework, you might want to verify your conjecture by looking atproblem 29(a) of section 1.6 of the textbook. Please make a conjectureyourself rst, though.

This is a challenging homework problem.

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7. Every vector in W 1 + W 2 can be expressed as a sum, w1 + w2 , of vectors

w1∈W 1 and w2

∈W 2 . In what cases is this expression unique ? Proveyour answer is correct.

This answer is unique just in case W 1 ∩ W 2 = {0}; that is, just in casethe sum W 1 + W 2 is a direct sum.

To show this, rst suppose W 1 + W 2 = {0}, and let w be a nonzeroelement of W 1 ∩ W 2 . Then w can be expressed as a sum of a vectorfrom W 1 and a vector from W 2 in more than one way, namely as w + 0and as 0 + w.

Conversely, suppose that W 1 ∩ W 2 = {0}. We must show any vectorw ∈W 1 + W 2 can be expressed as a sum of a vector from W 1 and avector from W 2 in only one way. To do this, suppose we have two suchexpressions w = w1 + w2 and w = w1 + w2 . We must show w1 = w1

and w2 = w2 .

We have w1 + w2 = w1 + w2 , which we can rewrite as w1 − w1 = w2 − w2 .Thus w1 − w1 is in both W 1 and W 2 . The only vector in both W 1 andW 2 is 0, so w1 − w1 = 0, and w1 = w1 . The same argument shows thatw2 = w2 .

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