Published on 31 Jan 2019

School

Department

Course

Professor

Math 46

Solutions for Final Exam Review

1. In the interests of brevity, I’m not going to write out either the question

or the answer to the various parts of problem one. The answers to the parts

(a), (b), and (c) should all be in your notes, and as for part (d), well ... I

don’t think I should answer that just yet. Maybe after the exam.

2. Let A∈Mm×n. Assume that for all x∈Rnthat Ax=0. Prove that A

is the zero matrix.

Solution: Well, we know that Ax=0for every single vector x∈Rn. That

means we can pick any xwe want, and plug it in. Me, I want to pick x=ei.

Because then we see that Ax=Aeiis the ith column of A. But that’s equal

to 0by hypothesis! So every column of Ais zero, so the whole matrix has

to be zero. ♣

3. Solve the following linear system:

2x+y−2z= 10

3x+ 2y+ 2z= 1

5x+ 4y+ 3z= 4

(a) by row reduction.

(b) by Cramer’s Rule.

Solution: I’m going to omit the details of this, since you can see multitudi-

nous examples of row reduction and Cramer’s Rule in the textbook and in

your notes. However, the answer is that there is only one solution to this

system, and it’s given by x= 1, y= 2, and z=−3. ♣

4. Let Abe an m×nmatrix and let TA:Rn→Rmbe deﬁned by TA(x) = Ax.

Are the following statements true or false? If true, give a proof. If false,

explain why.

(a) dim NulA≤n.

Solution: True. The null space of Ais a subspace of Rn, so its dimension

has to be at most n.♣

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(b) rankA≤m.

Solution: True. The rank of Ais equal to the number of pivots in the

reduced row-echelon form of A. Since each pivot is in a diﬀerent row, the

number of pivots is no more than the number of rows of A, which is m.♣

(c) If n > m then the linear transformation TAcannot be one-to-one.

Solution: True. Recall that TAis one-to-one if and only if Ahas a pivot

in every column (after row-reduction). This is only possible if Ahas at least

as many rows as columns ... which means that if TAis one-to-one, then we

can’t have n > m.♣

(d) If n < m then TAcannot be onto.

Solution: True. Recall that TAis onto if and only if Ahas a pivot in every

row (after row-reduction). This is only possible if Ahas at least as many

columns as rows ... which means that if TAis onto, then we can’t possibly

have n < m.♣

(e) If TAis one-to-one and m=n, then TAmust be onto.

Solution: True. (Yes, all the answers to question four were “true”.) If

m=n, then Ais a square matrix. If Ais square, then the IMT applies, and

that means that TAis onto if and only if it’s one-to-one. So we’re done. ♣

5. Let Abe an n×nmatrix satisfying A3=In, where Inis the n×nidentity

matrix. Show that det A= 1.

Solution: Since A3=I, we know that det A3= det I= 1. This means that

(det A)3= 1, since det(AB) = det(A) det(B) for any two matrices Aand B.

But if (det A)3= 1, then assuming that the entries in Aare real numbers

(which, in this course, they always will be), we must have det A= 1, as

desired. ♣

6. For the following problems, prove the statement or give a speciﬁc coun-

terexample:

(a) W={p∈P2|(p(3))2+p(3) = 0}is a subspace of P3.

Solution: This is obviously not true, since Wis deﬁned to be a set of

polynomials in P2, so it can’t be a subspace of P3.

If you ignore that typo, and ask if Wis a subspace of P2instead, then

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the answer is still no. To check if Wis a subspace, you need to check three

things: the existence of 0∈W, closure under addition, and closure under

scalar multiplication. As it turns out, the zero vector is in W, but Wfails

the second and third tests.

For example, let p(t) = −1, the constant polynomial. Then certainly

(p(3))2+p(3) = (−1)2+ (−1) = 1 −1 = 0, so p(t)∈W. But if we multiply

p(t) by the scalar 3, we see that (3p(3))2+ 3p(3) = 9 + 3(−1) = 6 6= 0. Thus,

3p(t) is not in W, so Wis not closed under scalar multiplication. ♣

(b) W={(x, y)∈R2|x+y= 0}is a subspace of R2.

Solution: This is true. Recall that to check if Wis a subspace of R2, we

need to check three things: the existence of 0∈W, closure under addition,

and closure under scalar multiplication. Let’s check them, one by one.

First, it is clear that 0= (0,0) is in W. This follows immediately from

the fact that 0 + 0 = 0.

Second, we need to check that Wis closed under addition. To do this,

pick any two vectors (x1, y1) and (x2, y2) in W. We want to show that

(x1, y1) + (x2, y2) = (x1+x2, y1+y2) is also in W. This means showing that

x1+x2+y1+y2= 0, because that’s the deﬁnition of W.

But we already know that x1+y1= 0, because (x1, y1)∈W. And

x2+y2= 0, because (x2, y2)∈W. So that means that x1+x2+y1+y2= 0,

so (x1+x2, y1+y2)∈W, as desired.

Lastly, we need to check that Wis cloesd under scalar multiplication. To

do this, pick any vector (x, y)∈W, and pick any scalar α∈R. We want to

show that α(x, y) = (αx, αy) is in W.

This amounts to showing that αx+αy = 0. But that follows immediately

from the fact that x+y= 0 (which is the deﬁnition of what it means for

(x, y) to be in W), so we’re done.

Since we’ve shown that Wsatisﬁes all three criteria for being a subspace

of R2, we’re done. ♣

(c) Let Vbe a vector space and let f1:V→Rand f2:V→Rbe linear

transformations. Deﬁne T:V→R2by T(v) = (f1(v), f2(v)). Then Tis

linear.

Solution: This is true.

In order to prove that Tis linear, we need to check two things: that T

respects addition, and that it respects scalar multiplication.

First, we need to show that Trespects addition. This means showing

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## Document Summary

Solutions for final exam review: in the interests of brevity, i"m not going to write out either the question or the answer to the various parts of problem one. The answers to the parts (a), (b), and (c) should all be in your notes, and as for part (d), well I don"t think i should answer that just yet. Maybe after the exam: let a mm n. Assume that for all x rn that ax = 0. Solution: well, we know that ax = 0 for every single vector x rn. That means we can pick any x we want, and plug it in. Me, i want to pick x = ei. Because then we see that ax = aei is the ith column of a. So every column of a is zero, so the whole matrix has to be zero: solve the following linear system: