w16_206_hwk07_solns

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Math 206, Spring 2016 Assignment 7 – Solutions Due: March 11, 2016

Part A.

(1) Suppose that T ∈ L(Rc,Rr) and S ∈ L(Rr,Rq). Prove that ker(T ) ⊆ ker(S ◦ T ), and that im(S ◦ T ) ⊆

im(S ).

Solution. We’ll first prove that ker(T ) ⊆ ker(S ◦ T ). For this, let x ∈ ker(T ) be given. This meansthat T (x) = 0. Note further that since S is linear, we have that S (0) = S (00) = 0S (0) = 0, where thefinal equality is because any vector scaled by 0 is 0. Hence we must have

(S ◦ T )(x) = S (T (x)) (definition of composition)

= S (0) (since x ∈ ker(T ))

= 0 (since S (0) = 0).

Hence we have that x ∈ ker(S ◦ T ).

Now we prove that I’m(S ◦ T ) ⊆ I’m(S ). For this, let b ∈ im(S ◦ T ) be given. This means thatthere exists some x ∈ Rc so that (S ◦ T )(x) = b. But the definition of composition then tells us thatS (T (x)) = b, and hence b is the output (under the function S ) of the element T (x) ∈ Rr. Henceb ∈ im(S ), as desired.

(2) Suppose that T ∈ L(Rc,Rr) is injective, and suppose that {v1, · · · , vk} ⊆ Rc is a linearly independentcollection. Prove that {T (v1), · · · , T (vk)} is linearly independent.

Solution. To show that {T (v1), · · · , T (vk)} is independent, let c ∈ Rk be some linear relation. Inother words, suppose that

c1T (v1) + · · · + ckT (vk) = 0.

If we can show that c = 0, then we will have shown the set is independent.

Notice that the linearity of T turns our given relation into the following equation

T (c1v1 + · · · + ckvk) = 0.

Of course, we also know that T (0) = 0 [see, for instance, the proof in the previous problem, or theargument we gave in class]. But since T is injective, the equality

T (c1v1 + · · · + ckvk) = T (0)

forces the equality

c1v1 + · · · + ckvk = 0.

In other words, we have shown that c is a linear relation of {v1, · · · ,vk} as well. Since we’re told that

{v1, · · · , vk} is independent, we conclude that c = 0, as desired.

(3) () Your friend has just heard that a subspace W ⊆ Rn is a set which satisfies: (1) 0 ∈ W ; (2) if w1, w2 ∈ W , then w1 + w2 ∈ W ; and (3) if w ∈ W and k ∈ R, then kw ∈ W . [This is the definition wegave in class.] She tries to convince you that the first condition is unnecessary and should be omittedso that the definition of a subspace is simpler and more elegant. Here’s her argument:

“We know that W is closed under scaling. So if you pick up any w ∈ W and scale it by 0,then the result has to be in W . But we already said in class that if you scale anything by 0,the result is 0. Hence condition (1) is just a consequence of condition (3).”

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Explain to your friend why her argument isn’t quite right.

Solution. Her argument breaks down precisely in the case where W is the empty set. In this case

the procedure “pick up any w

∈ W ” can’t be completed, so there’s no vector to scale which will allowyou to produce 0. Other than this her argument works perfectly well. [And, in fact, one could replacecondition (1) from class with the seemingly more tame assertion that W = ∅.]

Part B.

(1) Suppose that {v, z, w} is a basis for some subspace W in Rn. Prove that {v + w,w + z, v + z} is alsoa basis for W .

Solution. We need to verify that {v + w,w + z, v + z} is linearly independent and that they spanW . We’ll resolve independence first. Suppose that c ∈ R3 is some relation amongst these vectors:

c1(v

+w

) + c2(w

+z

) + c3(v

+z

) = 0.

We will prove the set is independent by showing that c must, in fact, be 0. To do this, note that ourequation is equivalent (after distributing and rearranging terms) to

(c1 + c3)v + (c1 + c2)w + (c2 + c3)z = 0.

Since we know that {v, w, z} is a basis, it must be independent, and hence we must have c1 + c3 =c1 + c2 = c2 + c3 = 0. But this is system of equations which we can solve via row reduction:

1 0 1 01 1 0 00 1 1 0

−ρ1+ρ2−→

1 0 1 00 1 −1 0

0 1 1 0

−ρ2+ρ3−→

1 0 1 00 1 −1 0

0 0 2 0

(1/2)ρ3−→

1 0 1 00 1 −1 00 0 1 0

ρ3+ρ2−→ 1 0 1 00 1 0 00 0 1 0

−ρ3+ρ1−→ 1 0 0 00 1 0 00 0 1 0

.Hence we get c1 = c2 = c3 = 0, as desired.

Now we’ll show that {v + w, w + z, v + z} spans W . Let x ∈ W be given; our goal is to findd1, d2, d3 ∈ R so that

x = d1(v + w) + d2(w + z) + d3(v + z).

Recall that we’re told {v, w, z} is a basis for W , and hence must be a spanning set. Therefore we doknow that there are constants b1, b2, b3 ∈ R so that

x = b1v + b2w + b3z.

With this in mind, observe that

b1 + b2 − b3

2 (v + w) +

−b1 + b2 + b3

2 (w + z) +

b1 − b2 + b3

2 (v + z)

=

b1 + b2 − b3

2 +

b1 − b2 + b32

v +

b1 + b2 − b3

2 +

−b1 + b2 + b32

w +

−b1 + b2 + b3

2 +

b1 − b2 + b32

z

= b1v + b2w + b3z = x.

Hence x is a linear combination of {v + w,w + z,v + z}, and this set spans W as desired.

[Note: in the proof of that the given set spans above, it may seem that we pulled some prettysophisticated constants out of thin air which magically produced our desired vector. However, thisis only because arguing that a certain vector is in the span of a collection amounts to verifying some



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existential statement. As we’ve seen before, the work in proving existential statements often comes fromdoing some hard work on a scratch piece of paper to find the quantities you claim exist. In this case,we knew that there were constants b1, b2, b3 ∈ R so that

x = b1

v + b2

w + b3

z,

and we wanted constants d1, d2, d3 ∈ R so that

x = d1(v + w) + d2(w + z) + d3(v + z).

If we take that last statement and manipulate it (something we can’t do in a proof, since that would bemanipulating an equation we want to be true!), it would yield

x = (d1 + d3)v + (d1 + d2)w + (d2 + d3)z.

Hence to find d1, d2 and d3 which satisfy the equation we want to be true, we really need to solve thesystem

d1 + d3 = b1d1 + d2 = b2d2 + d3 = b3

.

We produced the constants in our proof by simply solving this linear system.](2) Suppose that V and W are subspaces in Rn, and define

V + W = {v + w : v ∈ V , w ∈ W }.

Prove that V + W is a subspace of Rn.

Solution. We need to verify the three axioms of a subspace.

First, we must show that 0 ∈ V + W . For this, recall that since V and W are subspaces, we have0 ∈ V and 0 ∈ W . Hence we have that 0 + 0 ∈ V + W . But of course 0 + 0 = 0, so we have shown0 ∈ V + W as desired.

Now we will show that V + W is closed under addition. Let z1, z2 ∈ V + W be given. To say thatz1 ∈ V + W means there exist v1 ∈ V and w1 ∈ W so that z1 = v1 + w1. Likewise there exist v2 ∈ V and w2 ∈ W with z2 = v2 + w2. We then get

z1 + z2 = (v1 + w1) + (v2 + w2) = (v1 + v2) + (w1 + w2).

But observe that since V is closed under addition, we have v1 + v2 ∈ V ; likewise since W is closedunder addition, we get w1 + w2 ∈ W . Hence we have exhibited z1 + z2 as the sum of a vector from V and a vector from W , and therefore is in V + W .

Finally, we show that V + W is closed under scaling. let z ∈ V + W and k ∈ R be given. By thedefinition of V + W , there exist some v ∈ V and w ∈ W with z = v + w. Note that

kz = k(v + w) = kv + kw.

Since V is closed under scaling, we have kv ∈ V ; likewise since W is closed under scaling, we have

kw ∈ W . So we have exhibited kz as a sum of a vector from V and a vector from W , and hencekz ∈ V + W .

(3) Suppose that W is a subspace of Rr, and that T ∈ L(Rc,Rr). Prove that

T −1(W ) = {w ∈ Rc : T (w) ∈ W }

is a subspace of Rc.¡br¿[Note: One reads ”T −1(W )” as ”the preimage of W under T . This notationis defined even when T is not a bijection, because we’re not attempting to define some function T −1 :Rr → Rc. Instead, T −1(W ) is simply meant to represent the set of al inputs of T whose corresponding

output is in W . In particular, in the special case where W = {0}, note that T −1({0}) is just ker(T ).]



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Solution. We’ll verify the three axioms of a subspace.

[For this first axiom, I’ll use the notation 0c to denote the zero vector in Rc, and 0r to denote the zero

vector in Rr. This will just help alleviate any confusion about where particular elements “live.”] First,

we must argue that 0c ∈ T −1(W ). For this, note that — since T is linear — we know that T (0c) = 0r.Furthermore we know that 0r ∈ W since W is a subspace of R

r. Hence we have T (0c) ∈ W , which iswhat it means to say that 0c ∈ T

−1(W ).

Now we’ll argue that T −1(W ) is closed under addition. Let v1, v2 ∈ T −1(W ) be given. This means

that T (v1), T (v2) ∈ W . Since W is closed under addition, this means that T (v1) + T (v2) ∈ W . Thelinearity of T tells us that T (v1 + v2) = T (v1) + T (v2), and hence we have T (v1 + v2) ∈ W . But thisis what it means to say that v1 + v2 ∈ T −

1(W ), so T −1(W ) is closed under addition.

Now we’ll argue that T −1(W ) is closed under scaling. Let v ∈ T −1(W ) and k ∈ R be given. Thismeans that T (v) ∈ W . Since W is closed under scaling, this means that kT (v) ∈ W . The linearity of T tells us that kT (v) = T (kv), and hence we have T (kv) ∈ W . But this is what it means to say thatkv

∈ T −1(W ), so T −1(W ) is closed under scaling.


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