Linear Maps, Matrices and Linear Sytems

Transcription

1 Linear Maps, Matrices and Linear Sytems Let T : U V be linear. We defined kert x x U, Tx and imt y Tx y for some x U. According to the dimension equality we have that dimkert dimimt dimu n. We have that the linear map T is one-one or injective iff kert. If U V then T is injective iff T is surjective (onto). This is an easy consequence of the dimension equality. For any linear map T we have a matrix representation. The matrix A of T depends on the chosen bases and of U and V, repectively: MatT;,,, n ;,, m a ij where T j m i a ij i, j, n;i, m The matrix for T is an m n matrix where dimv m, dimu n. Each of the n columns of A contain the m components of T j with respect to the basis j. Let x be a vector in U. If x n j where Thus x j j, then Tx T n j a a a n a a a n a m a m a mn x j j n j x j T j n j x j m i a ij i i y i n j a ij x j x x x n y y y m m n j a ij x j i m j y i i Thus the map Tx y has a coordinate representation as Ax y. with this in mind, we see that dimimt dimcolumnspace A s, dimkert dimsolution space for Ax n r where r is the row rank of A. By the dimension equality we have that s n r n which is s r that is row rankcolumn rank. Example : Let U 3, V 4 then the matrix A stands for the linear map T where

2 T 5, T where we have chosen the unit vectors as bases for U 3 and V 4. We have that , T has rank 3. This means that kert, the map is injective and maps the 3 unit vectors to three linearly independent vectors. The image of the vector x 3 is a vector in 4 : x y z x 8y 3z x 4y z 5x y z x 3y 8z 3 kert is the solution space for the homogeneous system which as we already saw consists only of the zero-vector of 3. The map T : 3 4 is injective. We can find a linear map S : 4 3 such that S T : 3 3 is the identity. Because Te, Te, Te 3 are linearly independent, we can find a vector 4 such that Te, Te, 3 Te 3, 4 form a basis of 4. We define S on this basis by e, e, 3 e 3, 4 where is any vector in 3. For example 3 is fine. Then STe S e and similarly for the other unit vectors of 3. That is STe i e i. That is, the composition S T is the identity on the unit vectors of 3 and therefore the identity on 3. Because of S T id 3 we have for general reasons that S is surjective and T injective. Something we showed for arbitrary maps. We also see that S is not uniquely determined by T. First 4 is not unique and if we have found some 4 we can assign any vector in 3 as its image under S. The map S is unique only on ims by assigning to T, the vector. Example. This is a simple example which makes the logic quite transparent. Let T : 3, e e 3, e e 3. The matrix of T is third unit vector e 3 3 to e 3 Te, e 3 Te and define the map S as Se 3 e, Se 3 e, Se 3 a 3,a, b arbitrarily chosen. Then b 8 For a basis of 3, we add the MatS;e 3, e 3, e 3 3 ;e, e a b

3 and MatS T;e, e ;e, e a b This is an example where the product of two non-square matrices is square and invertible. Notice that a b a b is also square but not the identity. Matrices and linear maps can be identified. If A is an m n matrix, then L A : n m, X AX Y is linear and MatL A A. This is because Ae j n A j where A j is the j th column vector of A. Example 3. The matrix 3 4 stands for a mapa from 3 into. (The book denotes it as L A ). The map is onto (why?), and therefore kera 3, which gives us a one-dimensional null-space. How can we compute the kernel, that is find a basis for it? We have for 3 4 as row echelon form: kera is the span of the vector 7 7 This is x z, y 7z or We have that the matrix of the composition of maps corresponds to the product of the matrices If S : U V, T : V W, A MatS;,,, n ;,, m, B MatT;,, m ;,,, l then MatT S;,,, n ;,,, l BA In particular, MatL BA BA : L BA X BAX BAX L B L A X. Thus L BA L B L A and therefore MatL BA MatL B L A _ MatL B MatL A BA The linear map T : U V is invertible if there is a linear map S : V U such that S T id U 3

4 and T S id V. For a linear map T : U V to have an inverse, T, it is necessary that dimu dimv. (Proof?) If A MatT;,,, n ;,, n and B MatT ;,, n ;,,, n then B A where AA A A I n, the identity n n matrix. We have the following important result: For an n n matrix the following are equivalent: Ax has only the trivial solution; Ax y has for every y exactly one solution x A has an inverse. All of this follows from the theorem that a linear map on a finite dimensional vector space is injective if an only if it is surjective. Every elementary row operation on an m, n-matrix A can be expressed as as a product of an "elementary" m, m matrix E and A. If E stands for the operation "multiply the first row of A by c " then the matrix for this operation is just what you get if you apply this operation on the m, m unit matrix I m. That is: c a a a n ca ca ca n a a a n a a a n a m a m a mn a m a m a mn Similarly for the elementary operation where you add a multiple of one row to another row. The elementary matrix that "add the c-multiple of the second row to the first row" is what you get if you apply this operation on the m, m unit matrix I m. That is, it is the matrix c If E i, i,, k are k elementary operations successively applied to the matrix A transform A into its row-echelon form REFA then E k E A REFA where now the E i are the corresponding elementary matrices. For an invertible matrix n, n matrix A, the n, n identity matrix I n is its row-echelon form. And B E k E is its inverse. Let A I n be the matrix A augmented by the n columns of unit vectors e, e,, e n. Then using the elementary row operations transforms A into I n and I n into A. A I n elementary row operations I n A This is a celebrated algorithm for finding the inverse of a matrix where you apply simultaneously elementary row operations to the left A portion as well to the right I portion of the A I n block of two matrices. As a corollary of this process we get the following 4

5 TheoremThe n, n-matrix A is invertible if and only if it is a product of elementary matrices. Proof. A is invertible if and only if its row echelon form is the identity matrix. But then E k E A I and then A has inverse E k E. This is E k E A and A E E k. Each elementary operation can be reversed by an elementaty operation. For example, "multiply the first row by c " has the converse operation "multiply the first row by c. Similarly for the other elementary opertation. That is, an elementary matrix is invertible and its inverse is also elementary. Finding the inverse of an n, n matrix A means finding the n, n matrix X such that AX I n. This amounts to simultaneously solving the n equational systems Ax e i n, i,, n. This amounts to the just described algorithm. See Example on page of the book. 5