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Math 217:x2.4 Invertible linear maps and matrices

Professor Karen Smith

Denitions:LetXandYbe any sets, andX!Yany mapping.

The mappingX!Yisinjectiveif for ally2Y, there is at most onex2Xsuch that(x) =y. Put dierently, no twoxin the source map to the sameyin the target under. The mappingX!Yissurjective (or onto)if for ally2Y, there is somex2Xsuch that(x) =y. The mappingX!Yisinvertible (or bijective)if for eachy2Y, there is auniquex2Xsuch that(x) =y. Whenis invertible, we can dene theinverse mappingY !Xto be the map sending eachyto that uniquex with(x) =y.

A. For each linear mapping below, consider whether it is injective, surjective, and/or invertible. If

it is invertible, give the inverse map.

1. The linear mappingR3!R3which scales every vector by 2.Solution note:This is surjective, injective, and invertble. The inverse scales by12

The matrix of the inverse is2

412
0 0 0 12 0 0 0 12 3 5

.2. The linear mappingR3!R3which rotates every vector byaround thex-axis.Solution note:Invertible (hence surjective and injective). The inverse rotates by.3. The mappingR2!R2dened by projection onto a lineL.Solution note:Not surjective, since the image is the lineL. Not injective, since all

points on a given line perpendicular toLhave the same image. Not invertible.4. The mappingR2!R2dened by re

ection over the lineL.Solution note:This is invertible (so injective and surjective). It is its own inverse!5. The shearR2!R2dened by multiplication by the matrix1 5

0 1 :What is the matrix of the inverse map?Solution note:Invertible. The inverse is15 0 1 :B.The point is:For alinear transformation:Rn!Rmwe can say is surjective if and only if for all~yin the target, the equation(~x) =~yhas at least one solution, and is injective if and only if for all~yin the target, the equation(~x) =~yhas at most one solution, and is bijective if and only if for all~yin the target, the equation(~x) =~yhas exactly one solution.

1. Rephrase the three bullet points above in terms of solving a system of linear equations in-

volving the matrix of.Solution note:LetAbe the matrix ofT. ThenTis surjective if and only if for all~y2Rn, the system of linear equationsA~x=~yhas at least one solution (ie. is consistent). ThenTis injective if and only if for all~y2Rn, the system of linear equationsA~x=~yhas at most one solution. ThenTis invertle if and only if for all

~y2Rn, the system of linear equationsA~x=~yhas exactly one solution.2. LetT:R3!R3be the linear transformation given by left multiplication by2

41 4 1

0 1 1

0 1 13

5 :Use row-reduction to determine whether or not there is an vector~xsuch thatT(~x) =2 40
2 13 5 :Solution note:We want to know whether or not there is an~x=2 4x 1 x 2 x 33
5 such that T(2 4x 1 x 2 x 33
5 ) =2 40
2 13 5 :That is, we want to know if the system A 2 4x 1 x 2 x 33
5 =2 40
2 13 5 has a solution or not. The augmented matrix is 2

41 4 1 0

0 1 1 2

0 1 1 13

5 ;which can be row- reduced to 2

41 4 1 0

0 1 1 2

0 0 013

5 by replacing row 3 by "row 3 minus row 2". This is not yet in row reduce echelon form, but already we see that we have an inconsistent system, since the last row stands for the equation 0x1+0x2+0x3=1, which obviously has no solutions. So there is no solution to this system, and this means there is no~xsuch thatT(~x) =2 40
2 13 5 :This means that2 40
2 13 5 is not in the image ofTand thatTis not surjective.C. ConsiderT:R3!R3given byT(~x) =2 4x

1+x2+x3

x 2 x 1+x33 5

1. IsTlinear? If so, what is its matrixA? Why do you think some people call the matrix the

\coecient matrix" ofT?Solution note:Yes,Tis linear. Its matrix isA=2

41 1 1

0 1 0

1 0 13

5 ;which is the matrix of

coecients of linear forms dening the map.2. What are the source and target ofT?Solution note:The source and target space are all ofR3.3. Explain why

~b2R3is in the image ofTif and only if the system of linear equationsA~x=~b is consistent.Solution note:This is really just repeating what we did in problem B. Again: Suppose that~b2R3is in the image ofT. This means that there is some~xin the sourceR3 such thatT(~x) =~b. For that~x, we have2 4x

1+x2+x3

x 2 x 1+x33 5 =2 4b 1 b 2 b 33
5 =~bshowing that ~xis a solution toA~x=~b. Conversely, ifA~x=~bis consistent, then there is vector~x which satises2 4x

1+x2+x3

x 2 x 1+x33 5 =2 4b 1 b 2 b 33
5 showing thatA~x=T(~x) =~b, so~bis in the

image ofT.4. The image ofTis a plane ofR3. Find the equation of this plane. Also, write the image in set

builder notation. Hint: Use what you know about solving linear equations and row reduction.Solution note:The image is any vectorT(~x) inR3; this will be any vector of the

form~y=x1+x2+x3x2x1+x3T. Thus the image will be any~yfor which we can nd anx1,x2andx3solving this equality. Row reducing the augmented matrix, we have 2

41 1 1y

1

0 1 0y

2

1 0 1y

33
5 !2

41 0 1y

1y2

0 1 0y

2

0 0 0y1+y2+y33

5 so the image isf~y2R3y

3=y1y2g. Geometrically, this is a plane through the

origin inR3with normal vector [111]tr.5. IsTis invertible? Is its matrix invertible? Explain. Also, without computing, why is the

rank less than 3? Solution note:No! We just veried that the image of the mapTis a plane, which is NOT the whole targetR3. Any~yin the target which does not satisfyy36=y1y2 will not be hit byT, that is, we have no solution toT(~x) =~y. SoTis not surjective, hence it can't be invertible (since invertible means SURJECTIVE AND INJECTIVE. Alternatively, we can also say thatTis not injective: note thatT(2 40
0 03 5 ) =T(2 41
0 13 5 2 40
0 03 5 :So there are two dierent vectors taken to the same vector byT, and it is not injective. HenceTis not invertible. So also the matrix ofTis not invertible. Its rank can not be three, because if the rank of a 33 matrix is 3, then it is invertible.D.

1. True or False: A linear transformation is invertible if and only its matrix is invertible. Explain.

2. Explain how to nd the inverse ofAusing row reduction (or to tell that no such inverse

exists).

3. Demonstrate your technique by nding the inverse of

2

41 2 4

0 1 2 0 03 5 :Solution note:1. TRUE. Suppose:Rn!Rnis invertible, with matrixA. Let :Rn!Rnbe the inverse, with matrixB. We know that the composition and are both the identity maps. On the other hand, the compositions have matricesABandBArespectively, soAis invertible, with inverseB. Conversely, if Ais the matrix ofand we know thatAis invertible. Then the map given by left multiplication by the inverse matrixBis the inverse of. Indeed, the compositions will have matrixABandBAwhich are both the identity matrix, hence dene the identity map.

2). To nd the inverse of annnmatrixAwe form then2nmatrix [A In] and

row reduce until we have either a row of zeros in the rstnnspot (in which case Ais not invertible) OR the identity matrix, in which case the secondnnmatrix will beA1:See the book: Theorem 2.4.5.

3) Row reduce:2

41 2 4 1 0 0

0 1 2 0 1 0

0 00 0 13

5

We get

2

41 0 0 12 0

0 1 0 0 12=

0 0 1 0 0 1=3

5

So the inverse is

2 412 0
0 12=

0 0 1=3

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