w16_206_hwk05_solns

8/15/2019 w16_206_hwk05_solns

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Math 206, Spring 2016 Assignment 5 – Solutions Due: February 26, 2016

Part A.

(1) Let

w1 =

11

, w2 =

−1

2

, and w3 =

10

.

Prove that there cannot be T ∈ L(R2,R3) that satisfies

T (w1) =

1

23

, T (w2) =

3

−25

, and T (w3) =

4

17

.

[Hint: Find two distinct ways to express 0 ∈ R2 as a linear combination of w1, w2 and w3. If you

assume that T is linear, what do these two representations for 0 tell you about the value of T (0)?]

Solution. Suppose, for the sake of contradiction, that such a linear transformation T exists. Observethat 0 = 0w1 + 0w2 + 0w3, but also that

0 = 2

11

−

−1

2

− 3

10

= 2w1 − w2 − 3w3.

Now the linearity of T tells us that, on the one hand,

T (0) = T (0w1 + 0w2 + 0w3) = 0

1

23

+ 0

3

−25

+ 0

4

17

=

0

00

.

On the other hand, linearity of T gives

T (0) = T (2w1 − w2 − 3w3) = 2

123

− 1

3−2

5

− 3

417

=

−133

−20

.

Hence our assumption leads to the conclusion that T (0) takes two separate values; but this contradictsthe fact that T is a function. So our initial assumption must be false, and we conclude that there is noT ∈ L(R2,R3) that satisfies the given properties.

(2) Suppose that T ∈ L(Rc,Rr) is given. Prove that im(T ) satisfies the following properties:(a) The vector 0 ∈ R

r is an element of im(T ).

Solution. There are a few ways one could approach this. Here’s one: since T ∈ L(Rc,Rr), thereexists some A ∈ R

r×c so that for all x ∈ Rc we have T (x) = Ax. Hence

T (0

) = A0

= 0Col1(A) + 0Col2(A) + · · · + 0Colc(A) = 0

.Hence 0 ∈ R

r.

Here’s a different proof. We know that for any n ∈ Z+ and any x ∈ R

n, we have 0x = 0. Henceby linearity we have

T (0) = T (0x) = 0T (x) = 0

(where the initial 0 we plugged into T is an element of Rc, and the latter 0 is an element of Rr).

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(b) If v, w ∈ im(T ), then v + w ∈ im(T ).

Solution. Let v, w ∈ im(T ) be given. This means that there exist vectors x1, x2 ∈ Rc so that

T (x1) = v and T (x2) = w. But then observe that

T (x1 + x2) = T (x1) + T (x2) (linearity of T )

= v + w.

Hence v + w ∈ im(T ), since it is the output of the vector x1 + x2Rc.

(c) If v ∈ im(T ) and c ∈ R, then cv ∈ im(T ).

Solution. Let v ∈ im(T ) and c ∈ R be given. By the definition of image, this means that thereexists some x ∈ R

c so that T (x) = v. Note that we therefore have

T (cx) = cT (x) (linearity of T )

= cv

.

Hence cv ∈ im(T ), since it is the output associated to the vector cx ∈ Rc.

(3) For each of the following functions, determine (with proof) if the function is linear. If so, give the matrixto which it corresponds.

(a) The function T : R3 → R2 defined by T

x

yz

=

3x + yx + y − 2z

Solution. We’ll check that T is linear.

First, let

x =

x1

x2

x3

and w =

w1

w2

w3

be given. We then have

T (v + w) = T

x1 + w1

x2 + w2

x3 + w3

(definition of vector addition)

=

3(x1 + w1) + (x2 + w2)

(x1 + w1) + (x2 + w2) − 2(x3 + w3)

(definition of T )

=

3x1 + x2

x1 + x2 − 2x3

+

3w1 + w2

w1 + w2 − 2w3

(definition of vector addition)

= T

x1

x2

x3

+ T

w1

w2

w3

(definition of T )

= T (x) + T (w).

This verifies the first property of linearity.






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For the second, let

x =

x1

x2

x3

and c ∈ R

be given. Then we have

T (cx) = T

cx1

cx2

cx3

(definition of scaling)

=

3(cx1) + (cx2)

(cx1) + (cx2) − 2(cx3)

(definition of T )

= c

3x1 + x2

x1 + x2 − 2x3

(definition of scaling)

= cT (x).

This verifies the second property of linearity.

Now that we know T is linear, we can compute the matrix to which it corresponds. By a theoremfrom class, we know this matrix is

T (e1) · · · T (ec)

=

3 1 01 1 −2

.

(b) The function T : R2 → R3 defined by T

xy

=

x

1y

Solution. This is not linear. Note that

T

00

=

0

10

.

Of course we know that for 0 ∈ R2 we have 0 + 0 = 0, but observe that

T (0 + 0) = T (0) =

0

10

whereas

T (0) + T (0) = 01

0

+ 01

0

= 02

0

.

Since T (0 + 0) = T (0) + T (0), we see that T fails to be linear.

Part B.

(a) Let v =

1

−22

, and for a given w ∈ R

3 define w⊥ = w − proj

vw.






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(i) Prove that w⊥ is orthogonal to v.

Solution. We will show that w⊥ · v = 0. To help in this calculation, recall that proj

vw =

w·v

v·vv

. Hence we getw⊥ · v =

w −

w · v

v · v v

· v

= w · v − w · v

v · v v · v (linearity of dot product)

= w · v − w · v (cancellation of v · v)

= 0.

(ii) Prove that the function T : R3 → R3 given by T (w) = w

⊥ is a linear transformation.

Solution.

We will check the relevant axioms. First, suppose that w

1 and w

2 are given inR3. Then we have

T (w1 + w2) = (w1 + w2) − projv

(w1 + w2) (definition of T )

= (w1 + w2) − projv

w1 − projv

w2 (linearity of proj from class)

= (w1 − projv

w1) + (w2 − projv

w2 commutativity of vector addition)

= T (w1) + T (w2).

This verifies the first condition of linearity.

For the second, let w ∈ R3 and c ∈ R be given. Then we have

T (cw) = (cw) − projv

(cw)(definition of T )

= cw − cprojv

w (linearity of proj from class)

= c(w − projv

w) (scaling distributes across addition)

= cT (w) (definition of T ).

This verifies the second condition of linearity.

(iii) Compute the matrix that corresponds to T .

Solution. By a theorem from class, we know this matrix is T (e1) · · · T (ec)

=

e1 − e1·v

v·vv e2 − e2·v

v·vv e3 − e3·v

v·vv

=

8/9 2/9 −2/9

2/9 5/9 4/9−2/9 4/9 5/9

.

(iv) Is T injective?

Solution. We’ll compute the reduced row echelon form of the matrix associated to T : 1 0 −1/2

0 1 10 0 0

.

(You can reach this form by performing the following sequence of operations: 9

8ρ1, −2

9ρ1 +

ρ2, 29

ρ1 + ρ3, 2ρ2, −1

2ρ2 + ρ3, −1

4ρ2 + ρ1.) Note that this matrix has rank 2. By a theorem






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from class, since the rank is less than the number of columns, we have that T is not injective.

(b) Suppose that v ∈ span {w1, w2, · · · , wk}. Prove that

span {w1, w2, · · · , wk} = span {w1, w2, · · · , wk, v} .

Solution. We’ll prove this result using a dual containment argument. First, we argue that{w1, w2, · · · , wk} ⊆ span {w1, w2, · · · , wk, v}. So let z ∈ {w1, w2, · · · , wk} be given; our goal isto show that z ∈ span {w1, w2, · · · , wk, v}. By definition of span, our hypothesis tells us that z isa linear combination of the vectors w1, · · · , wk. This means that there exist d1, · · · , dk ∈ R so that

z = d1w1 + · · · + dkwk.

But observe that then we also have

z = d1w1 + · · · + dkwk + 0 = d1w1 + · · · + dkwk + 0v.

Hence we have exhibited z as a linear combination of the vectors w1, · · · , wk, v. Therefore z ∈span {w1, · · · , wk, v}. This verifies the first containment.

For the containment span {w1, w2, · · · , wk, v} ⊆ {w1, w2, · · · , wk}, suppose that z ∈ span {w1, · · · , wk, v

This means that z is a linear combination of w1, · · · , wk, v, and hence (by definition of linear com-bination) there exist d1, · · · , dk, dk+1 ∈ R so that

z = d1w1 + · · · + dkwk + dk+1v.

Furthermore, we’re told that v ∈ span {w1, · · · , wk}. Hence v is a linear combination of w1, · · · , wk,and so there are real numbers c1, · · · , ck satisfying

v = c1w1 + · · · + ckwk.

We plug this expression for v into our expression for z above to find

z = d1w1 + · · · + dkwk + dk+1v

= d1w

1 + · · · + dkw

k + dk+1(c1w

1 + · · · + ckw

k)= (d1 + dk+1c1)w1 + · · · + (dk + dk+1ck)wk.

Now since each of the di and ci are real numbers, each coefficient di + dk+1ci above is a realnumber. So we have exhibited z as a linear combination of w1, · · · , wk. Therefore we have z ∈span {w1, · · · , wk}. This verifies the second containment.






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