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MATH 2121 | Linear algebra (Fall 2017) Lecture 7

1 Last time: one-to-one and onto linear transformations

LetT:Rn!Rmbe a function.

The following mean the same thing:

Tislinearis the sense thatT(u+v) +T(u) +T(v) andT(cv) =cT(v) foru;v2Rn,c2R. There is anmnmatrixAsuch thatThas the formulaT(v) =Avforv2Rn. If we are given a linear transformationT, thenT(v) =Avfor the matrix

A=T(e1)T(e2)::: T(en)

whereei2Rnis the vector with a 1 in rowiand 0 in all other rows.

CallAthestandard matrixofT.

The following all mean the same thing for a functionf:X!Y. fisone-to-one.

Ifa;b2Xandf(a) =f(b) thena=b.

Ifa;b2Xanda6=bthenf(a)6=f(b).

fdoes not send dierent inputs to the same output. Similarly, the following all mean the same thing for a functionf:X!Y. fisonto. The range offis equal to the codomain, i.e., range(f) =ff(a) :a2Xg=Y.

For eachy2Ythere is at least onex2Xwithf(x) =y.

Every element of the codomain offis an output for some input. We can detect whether a linear transformation is one-to-one or onto by inspecting the columns of its standard matrix (and row reducing). Theorem.SupposeT:Rn!Rmis the linear transformationT(v) =AvwhereAis anmnmatrix. (1)Tis one-to-one if and only if the columns ofAare linearly independent, which happens precisely whenAhas a pivot position in every column. (2)Tis onto if and only if the span of the columns ofAisRm, which happens precisely whenAhas a pivot position in every row. 1

MATH 2121 | Linear algebra (Fall 2017) Lecture 7

Example.LetT:R2!R2be the linear transformationT(v) =Av. IfAis one of the following matrices, thenTis onto and one-to-one.

Standard matrix ofTPictureDescription ofT

1 0 01 Re ect across thex-axis 1 0 0 1 Re ect acrossy-axis 0 1 1 0 Re ect acrossy=x k0 0 1 (0< k <1) Horizontal contraction 1 0 k1 (k >0) Vertical sheering

2 Operators on linear transformations and matrices

Today's story begins with the observation that linear transformationsRn!Rmare uniquely represented bymnmatrices, and everymnmatrix corresponds to a linear transformationRn!Rm.

There are several simple, natural operations we can use to combine and alter linear transformations to get

other linear transformations. The goal is to translate these function operations into matrix operations.

2

MATH 2121 | Linear algebra (Fall 2017) Lecture 7

Sums and scalar multiples. SupposeT:Rn!RmandU:Rn!Rmare two linear functions with the same domain and codomain. Their sumT+Uis the functionRn!Rmdened by (T+U)(v) =T(v) +U(v) forv2Rn: Ifc2Ris a scalar, thencTis the functionRn!Rmdened by (cT)(v) =cT(v) forv2Rn:

Fact.BothT+UandcTare linear transformations.

Proof.To see thatT+Uis linear, we check that

(T+U)(u+v) =T(u+v) +U(u+v) =T(u) +T(v) +U(u) +U(v) = (T+U)(u) + (T+U)(v) foru;v2Rn, and (T+U)(av) =T(av) +U(av) =aT(v) +aU(v) =a(T+U)(v) fora2Randv2Rn. Since these properties hold,T+Uis linear.

The proof thatcTis linear is similar. (Try this yourself!)Since sums and scalar multiples of linear functions are linear, it follows that dierences and arbitrary

linear combinations of linear functions are linear.

SupposeTandUhave standard matrices

A=2 6 664a

11a12::: a1n

a

21a22a2n:::::::::

a m1am2::: amn3 7

775andB=2

6 664b

11b12::: b1n

b

21b22b2n:::::::::

b m1bm2::: bmn3 7 775
so thatT(v) =AvandU(v) =Bv.

Proposition.The standard matrix ofT+Uis

A+B=2 6 664a

11+b11a12+b12::: a1n+b1n

a

21+b21a22+b22a2n+b2n:::::::::

a m1+bm1am2+bm2::: amn+bmn3 7 775:

The standard matrix ofcTis

cA=2 6 664ca

11ca12::: ca1n

ca

21ca22ca2n:::::::::

ca m1cam2::: camn3 7 775:

This is how wedenesums and scalar multiples of matrices. Note that these operations work in essentially

the same way as for vectors: we can add matrices of the same size, by adding the entries in corresponding

positions together, and we can multiply a matrix by a scalarcby multiplying all entries byc.

Example.We have4 0 5

1 3 2 +1 1 1 3 5 7 =5 1 6 2 8 9 and 4 0 5 1 3 2 + 21 1 1 3 5 7 =4 05 132
+2 2 2

6 10 14

=2 23

7 7 12

3

MATH 2121 | Linear algebra (Fall 2017) Lecture 7

SupposeT;U;Vare linear transformationsRn!Rmwith standard matricesA;B;C. Leta;b2R.

The following properties then hold:

FunctionsMatrices

1.T+U=U+V A+B=B+A.

2. (T+U) +V=T+ (U+V) (A+B) +C=A+ (B+C).

3.T+ 0 =Twhere 0 :Rn!Rmis the map 0(v) = 02Rm.A+ 0 =A.

4.a(T+U) =aT+aU a(A+B) =aA+aB.

5. (a+b)T=aT+bT(a+b)A=aA+bA.

6.a(bT) = (ab)T.a(bA) = (ab)A.

Composition.SupposeU:Rn!RmandT:Rm!Rkare linear.

Note that we assume the codomain ofUis equal to the domain ofT.

ThecompositionTUis the functionRn!Rkgiven by

(TU)(v) =T(U(v)) forv2Rn:

Fact.SinceTandUare linear,TUis linear.

Proof.To see thatTUis linear, we check that

(TU)(u+v) =T(U(u+v)) =T(U(u) +U(v)) =T(U(u)) +T(U(v)) = (TU)(u) + (TU)(v) foru;v2Rn, and (TU)(cv) =T(U(cv)) =T(cU(v)) =cT(U(v)) =c(TU)(v) forc2Randv2Rn.Important note:UTis not dened unlessk=n. Even ifk=nso that bothTUandUTare dened, there is no reason to expect thatTU=UT.

Example.Ifn=m=k= 1 andT(x) = 2xandU(x) =x2, then

(TU)(x) =T(x2) = 2x2but (UT)(x) =U(2x) = 4x2: SinceTUis a linear transformationRn!Rk, there is a uniqueknmatrixCsuch that (TU)(v) =Cvforv2Rn: IfAis the standard matricx ofTandBis the standard matrix ofU, then we dene thematrix product AB=C: Note how this denition works: ifAiskmandBismnthen wedeneABto be the uniquekn matrixCsuch thatCv=A(Bv) for allv2Rn. How do we actually compute the rectangular array which isAB, fromAandB? Theorem.SupposeBhas columnsb1;b2;:::;bn2Rmso thatB=b1b2::: bn.

ThenAB=Ab1Ab2::: Abn.

4

MATH 2121 | Linear algebra (Fall 2017) Lecture 7

Note that this makes sense asAiskm.

Proof.ABis the standard matrix of the linear functionTU, so

AB=(TU)(e1) (TU)(e2)(TU)(en)=A(Be1)A(Be2)A(Ben)

Ab1Ab2Abn:Example.IfA=2 3

15 andB=4 3 6 12 3 , thenb1=4 1 ,b2=3 2 ,b3=6 3 , so

AB=Ab1Ab2Ab3=11 0 21

1 139 The quick rule for computingAB: if theith row ofAandjth column ofBare a1a2::: amand2 6 664b
1 b 2::: b m3 7 775
then the entry in theith row andjth column ofABis a1a2::: am2 6 664b
1 b 2::: b m3 7

775=a1b1+a2b2++ambm:

Example.SupposeA=1 2 3 4

5 6 7 8

andB=2 6

641 2 3

4 5 6 7 8 9

9 9 93

7 75.

The entry in the 2nd row and 2nd column ofABis

52 + 65 + 78 + 89 = 10 + 30 + 56 + 72 = 168:

WriteInfor thennmatrix

I n=2 6 6641
1 13 7 775
which has 1 in each diagonal position, and zeros in all other positions. The matrixInis the standard matrix of theidentity mapRn!Rn. This is the linear functionTwithT(v) =vfor allv2Rn.

Proposition.LetA;B;Cbe matrices.

AssumeAismn,Bisnl, andCislk.

ThenA(BC) = (AB)C.

5

MATH 2121 | Linear algebra (Fall 2017) Lecture 7

Proof.Use our rst denition of matrix multiplication. By this denition,ABandBCare the unique matrices such that (AB)x=A(Bx) and (BC)x=B(Cx). In turn,A(BC) is the unique matrix such that (A(BC))x=A((BC)x) =A(B(Cx)). But (AB)Cis also the unique matrix such that ((AB)C)x= (AB)(Cx) =A(B(Cx)). ThereforeA(BC) = (AB)C.Here are some easier properties. SupposeA;B;Care matrices andr2R.

IfAismnandB;CarenlthenA(B+C) =AB+AC.

IfA;BaremnandCisnlthen (A+B)C=AC+BC.

IfAismnandBisnlthenr(AB) = (rA)B=A(rB).

IfAismnthenImA=AIn=A.

3 Pathologies of matrix multiplication

SupposeAandBare matrices.

Four important observations:

1. The productABis dened only if the number of columns ofAis the number of rows ofB.

2. Even ifABandBAare both dened, it typically happens thatAB6=BA.

3.AB=ACdoes not implyB=C.

4. It can happen thatAB= 0 =2

6 400
003 7

5even if bothA6= 0 andB6= 0.

Example.We have

1 0 1 0 0 0 1 1 = 1 0 1 0 0 1 1 0 1 0 0 1 =0 0 0 0 while 0 0 1 1 1 0 1 0 = 0 0 1 1 1 1 0 0 1 1 0 0 =0 0 2 0 IfAandBare bothsquarematrices of the same size (meaning they have the same number of rows and columns), andAB=BA, then we say thatAandBcommutes.

4 Matrix transpose

Thetransposeof anmnmatrixAis thenmmatrixATwhose columns are the rows ofA. Ifaijis the entry in rowiand columnjofA, then this is the entry in rowjand columniofAT.

For example, ifA=a b c

d e f andAT=2 4a d b e c f3 5

The transpose ofAis given by

ippingAacross the main diagonal, in order to interchange rows/columns. 6

MATH 2121 | Linear algebra (Fall 2017) Lecture 7

Another example: ifC=2

41 1 1 1

3 52 7

0 0 1 03

5 thenCT=2 6

6413 0

1 5 0 12 1

1 7 03

7 75.
We nish this lecture by noting some basic properties of the transpose operation: (AT)T=Asince ipping twice does nothing.

IfAandBhave the same size the (A+B)T=AT+BT.

Ifc2Rthen (cA)T=c(AT).

IfAis ankmmatrix andBis andmnmatrix then (AB)T=BTAT.

To prove the last property, use our earlier results to compute the entries inith row andjth column of

the matrices on either side (in terms of the entries ofAandB), and check that these are equal.

Question for later: how is the linear transformation with standard matrixArelated to the linear trans-

formation with standard matrixAT? What is the transpose as an operation on linear transformations? 7quotesdbs_dbs7.pdfusesText_13

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