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2.3 The Inverse Of a Linear Transforma- tion Definition. A function T

an m × n matrix the transformation is invert- ible if the linear system A x = y has a unique solution. 1. Case 1: m < n The system A x = y has either no 



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2 jan. 2012 Inverse Linear Transformations. ? A matrix operator T. A. :Rn. ?Rn is one-to-one if and only if the matrix A is invertible.



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[PDF] 23 The Inverse Of a Linear Transforma- tion Definition A function T

an m × n matrix the transformation is invert- ible if the linear system A x = y has a unique solution 1 Case 1: m < n The system A x = y has either no 



[PDF] Which Linear Transformations are Invertible - University of Lethbridge

We have mentioned taking inverses of linear transformations A linear transformation is invertible if and only if it is injective and surjective



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  • What is the inverse of a linear transformation?

    T is said to be invertible if there is a linear transformation S:W?V such that S(T(x))=x for all x?V. S is called the inverse of T. In casual terms, S undoes whatever T does to an input x. In fact, under the assumptions at the beginning, T is invertible if and only if T is bijective.

  • How to do inverse transformations?

    A general method for simulating a random variable having a continuous distribution—called the inverse transformation method—is based on the following proposition. then the random variable X has distribution function F . ( F - 1 ( u ) is defined to equal that value x for which F ( x ) = u .)
  • Let L: V ? W be a linear transformation. Then L is an invertible linear transformation if and only if there is a function M: W ? V such that (M ° L)(v) = v, for all v ? V , and (L ° M)(w) = w, for all w ? W . Such a function M is called an inverse of L.

Which Linear Transformations are Invertible

We have mentioned taking inverses of linear transformations. But when can we do this?

Theorem

A linear transformation is invertible if and only if it is injective and surjective.

This is a theorem about functions.Theorem

A linear transformationL:U!Vis invertible if and only ifker(L) =f~0gand

Im(L) =V.

This follows from our characterizations of injective and surjective.Theorem A linear transformationL:U!Vis invertible if and only if whenever~e1;:::;~enis a

basis forUthe collectionL(~e1);:::;L(~en) is a basis forV.Proof:)-direction. We assumeLis bijective.ThenLis injective, soKer(L) =f~0gand so

Ker(L)\Span(~e1;:::;~en) =f~0g

so by the assignment,L(~e1);:::;L(~en) are linearly independent.BecauseLis surjective we knowIm(L) =V, and as~e1;:::;~enare a basis forUthey are

a generating set, and so so by the assignment,L(~e1);:::;L(~en) are a generating set for

Im(L) =V.We concludeL(~e1);:::;L(~en) are a basis.(-directionthis is on the assi gnmentMath 3410 (University of Lethbridge)Spring 2018 1 / 7

Finite Dimensional Case

Theorem(Rank-Nullity Theorem)

SupposeL:U!Vis a linear transformation between nite dimensional vector spaces thennull(L) +rank(L) =dim(U). We will eventually give two (dierent) proofs of this.Theorem SupposeUandVare nite dimensional vector spaces a linear transformationL:U!V

is invertible if and only ifrank(L) =dim(V) andnull(L) = 0.Proof IdeaThis is just checking surjectivity and injectivity by looking at the dimensions

of the image and kernel.Theorem SupposeUandVare nite dimensional vector spaces a linear transformationL:U!V is invertible if and only ifdim(U) =dim(V) andrank(L) =dim(V).Theorem SupposeUandVare nite dimensional vector spaces a linear transformationL:U!V

is invertible if and only ifdim(U) =dim(V) andnull(L) = 0.Proof IdeaThese last two results just playing games with the equalities in the above

theorems. Math 3410 (University of Lethbridge)Spring 2018 2 / 7

Finite Dimensional Case - Matrix

Recall:The following result just says that we can check invertibility by looking at the matrix.

Theorem

SupposeUandVare nite dimensional vector spaces a linear transformationL:U!V

is invertible if and only if either equivalentlyFor some choice of basis forUandVthe matrix associated toLis invertible.For any choice of basis forUandVthe matrix associated toLis invertible.Proof

)-directionassuming Linvertible letMbe its inverse, then we have the formulas

LM=IdVandML=IdU

thus for any choice of basis, ifAis the matrix forLandBis the matrix forMwe know that

AB=IdandBA=Id

because the matrix forIdVandIdUare always the identity matrix.This proves the matricies are always invertible.

(-directionFix any basis in which the ma trixA, associated toLis invertible.In the same basis, letMbe the matrix associated toA1.ThenLMandMLare respectively the transformations associated to

AA

1=IdandA1A=Idin particular they areIdVandIdU.Math 3410 (University of Lethbridge)Spring 2018 3 / 7

Invertibility of a Matrix

Theorem

A (sqaure) matrixAis invertible if and only if the determinant is non-zero.There are lots of dierent ways to prove this, depending on what you know about

determinants. For some other approaches see the notes on the determinant on Moodle or check in your textbook. If the determinant is non-zero then we can check directly that

1det(A)Adj(A)

A=Id=A1det(A)Adj(A)

by using the denition ofAdj(A) (if you forget what this is ask me about it later, we will never use it for anything else

) and properties of the determinant.Conversely ifAhas an inverse then by multiplicativity of the determinant

det(A)det(A1) =det(AA1) =det(Id) = 1 and so if there is an inverse, the determinant can't be zero. Math 3410 (University of Lethbridge)Spring 2018 4 / 7 Invertibility of a Matrix - Other Characterizations

Theorem

SupposeAis annbyn(so square) matrix then the following are equivalent:1Ais invertible.2det(A) is non-zero.See p reviousslide 3A

tis invertible.on assignment 1 4The reduced row echelon form ofAis the identity matrix.(algo rithmto nd inverse) 5Ahas rankn,rank is numb erof lead 1s in RREF 6the columns ofAspanRn,rank is dim of span of columns 7A

~x=~balways has a solution,denition of columns spanning 8the columns ofAare a basis forRn,generating set of size n must b eLI 9the columns ofAare linearly independent.basis is LI 10whenA~x=~bhas a solution it is unique,LI implies unique rep resentations11The kernel ofAisf~0gWhat you check when you check LI12Ahas nullity 0,Denition of nullit y13the rows ofAspanRn,Apply ab oveto At14the rows ofAare a basis forRn,Appl yab oveto At15the rows ofAare linearly independent.Apply ab oveto AtMath 3410 (University of Lethbridge)Spring 2018 5 / 7

Isomorphisms

Recall

We call an invertible linear transformation between vector spacesUandVan isomorphism. We say that vector spaces areUandVareisomorphicif there exists an isomorphism between them, so if there exists a bijectiveL:U!V.Theorem Vector spacesUandVare isomorphic if and only ifdim(U) =dim(V).Proof: )-directionrecall that if Lis bijective, andBis a basis forU, thenL(B) is a basis forV, hence both have a basis of the same size.(-direction Iff~eigis a basis forUandf~figis a basis forV, and both have the same size then we can dene maps

L:U!VandM:V!U

by

L(~ei) =~fiandM(~fi) =~ei

notice why we need the bases to have the same sizeIt is clear that these maps are inverses, thus give the desired isomorphism.

Math 3410 (University of Lethbridge)Spring 2018 6 / 7

Natural Questions About Isomorphisms and Inverses

Given some description of a linear transformationL:Rn!Rn, is it an isomorphism?

does it have an inverse? and if yes, what is a description for the inverse?There are a lot of conditions you could check,and it is not alw aysobvi ouswhich one

is easiest.

T ond the inverse y oup rettymuch alw aysuse g aussianelimination. Given some description of a linear transformationL:V!W, is it an isomorphism?

does it have an inverse? and if yes, what is a description for the inverse?

There are a lot of conditions you could check,

and it is not alw aysobvi ouswhich one is easiest. Math 3410 (University of Lethbridge)Spring 2018 7 / 7quotesdbs_dbs20.pdfusesText_26
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