Linear Algebra. Grinshpan

Transcription

1 Linear Algebra Grinshpan Saturday class, 2/23/9 This lecture involves topics from Sections 3-34 Subspaces associated to a matrix Given an n m matrix A, we have three subspaces associated to it The column space (image, range) of A is the span of its columns The column space is a subspace of R n, it is spanned by the non-redundant columns From the reduced row-echelon form, we see that the dimension of the column space is ranka The row space of A is the span of its rows The row space is a subspace of R m (row version), it is spanned by the non-redundant rows From the reduced row-echelon form, we see that the dimension of the row space is ranka The row space and column space dimensions are equal Any matrix has as many linearly independent rows as it has linearly independent columns Note: the number of redundant columns of a matrix is not necessarily the same as the number of its redundant rows The kernel of A is also known as its null space The null space is a subspace of R m The null space has dimension m ranka Every nontrivial relation between columns gives rise to a kernel vector, and vice versa Note that the kernel vectors and the row space vectors are mutually orthogonal Thus the kernel and the row space are orthogonal subspaces of R m of complementary dimension Proposition Let V be a subspace of R n and let v,, v k be vectors in V Then any two of the following conditions imply the third: v,, v k span V v,, v k are linearly independent V has dimension k For instance, if V has dimension 3 and we know three linearly independent vectors in V, then these vectors must form a basis for V Similarly, if we know four linearly independent vectors in V, then the dimension of V must be at least 4 Proof of proposition Let V be spanned by k linearly independent vectors These vectors then form a basis for V So dim V = k, by definition Let V have dimension k and let it be spanned by k vectors If any of the vectors were redundant, we could remove them from the list and obtain a list of less than k vectors which are linearly independent and still span V But this would mean that dim V < k Let V have dimension k and let v,, v k be linearly independent vectors in V If v,, v k failed to span V, there would be a vector w in V which is not their linear combination This would give k+ linearly independent vectors in V : v,, v k, w But this would mean that dim V > k

2 Example Consider the matrix A = We will determine its image and kernel by inspection The columns 2, 4, 5 are redundant: ( ) C 2 = 3C, 2C 4 = 3C + C 3, C 5 = 2C + C 3 The image of A is therefore spanned by the non-redundant columns, 3 ) ( ) We have im A = span(, 2 = span, Thus im A is a two-dimensional subspace of R 3 Note that im A is the plane x = x 3 and, form a basis for it Now ker A has dimension 5 2 = 3, it is spanned by three linearly independent vectors We may read them off the column relations ( ): 3 3 ker A = span,, 2 2 The arrangement of s confirms that the spanning vectors are linearly independent Thus ker A is a three-dimensional subspace of R 5 and we may take 3 3 2,, 2 as a basis for it 2

3 Coordinates with respect to a basis Let V be a subspace of R n and let B = (v,, v k ) be a basis for V Then, for each x in V, the linear system c v v k = x c k c has a unique solution So each x in V can be uniquely written the form c k x = c v + + c k v k The coefficients c,, c k are the coordinates of x with respect to the basis B Notation: x B = c c k Example Let V = R 2 and B = So x B = x x 2 Example Let V = R 2, B = So x = B 3 x = (, ) x = x x + x 2 2 (standard basis) Then are the standard coordinates of x (with respect to the standard basis) (, 3, and x = Then ) 2 x = are the coordinates of x = 2 with respect to B Example Let V be the plane x + x 2 + x 3 = in R 3 ( ) Then dim V = 2 and B =, is a basis for V The vector x = 2 3 So the coordinates of x = 2 3 belongs to V and can be written as x = 2 with respect to B are x 2 = B

4 Linearity of coordinates The operation of taking coordinates with respect to a given basis is a linear transformation: x + y = x + y B B B cx = c x B B Matrix of basis elements The n k matrix S = v v k with columns given by the vectors of B = (v,, v k ) transforms the coordinates of x with respect to B back to the standard coordinates of x S x B = x Honeycomb tiling Consider a tiling of R 2 with regular hexagons I have labeled some of the points with their coordinates relative to the basis B = (v, w) Each point with integer B-coordinates is either a corner or a center Is (3, 7) a corner or a center? How about (2, 52)? Find a good way to tell Matrix with respect to a given basis Let B = (v,, v n ) be a basis for R n and let T be a linear transformation of R n By the matrix of T with respect to B is meant the n n matrix B which transforms the B-coordinates of x into the B-coordinates of T (x), B x = T (x) B B Notation: T B = B 4

5 With respect to the standard basis of R n, the preceding relation takes the form Ax = T (x), where A = T (e ) T (e n ) is the standard matrix of T Example Let T (x, x 2 ) = (3x, 5x 2 ) be a change of scale in R 2 3, so that A = 5 ( ) Let B =, (this basis is not standard) Then T v = 5v and T v 2 = 3v 2 So x = c v + c 2 v 2 is transformed into T (x) = 5c v + 3c 2 v 2 c In B-coordinates then, 5c is transformed into So B = c 2 3c Column-by-column formula We can write down the matrix B column by column Note that v = v + v v n, v 2 = v + v v n, v n = v + + v n + v n Therefore vj B = e j, j =,, n Consequently, Be j = B v j B = T (v j ) B is the jth column of B Our formula is ready: B = T (v ) B T (vn ) B Example Let T be the reflection about the line x 2 = 2x in R 2 ( ) 2 Let B =, (these vectors go well with the line of reflection) 2 Then T v = v and T v 2 = v 2 We have v = and v B 2 = B So B = T (v ) B T (v2 ) B = v B v2 5 B =

6 Similarity relation What is the relation between the matrices A and B? Observe that B x = Ax B B The n n matrix S = v v n is invertible and its inverse transforms the standard coordinates into the B-coordinates, S x = x B Therefore, BS x = S Ax for every x in R n This gives BS = S A or AS = SB This relation is known as similarity or intertwining We can also write B = S AS and A = SBS ( ) 2 Example Let T be the reflection about x 2 = 2x and B =, 2 2 Then S = and S 2 = 2 So 5 2 A = SBS = = 2 2 3/5 4/5 4/5 3/5 This is indeed the standard matrix of reflection about the span of the unit vector as above 5 2