Math 19b Linear Algebra with Probability Oliver Knill Spring 2011 The pivot columns of A span the image of A Proof You can see this by deleting the other columns The new matrix B still allows to solve Bx b if Ax b could be solved Lecture 13 Image and Kernel 5 Find the image of The image of a matrix If T Rm Rn is a linear transformation then T x x Rm is called the image of T If T x A x where A is a matrix then the image of T is also called the image of A We write im A or im T 1 If T x y cos x sin y sin x cos y is a rotation in the plane then the image of T is the whole plane 3 The averaging map T x y z x y z 3 from R3 to R has as image the entire real axes R The span of vectors v1 vk in Rn is the set of all linear combinations c1 v1 ck vk 4 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 The kernel of a matrix The map T x y z x y 0 maps the three dimensional space into itself It is linear x 1 0 0 x because we can find a matrix A for which T x A y 0 1 0 y The image z 0 0 0 z of T is the xy plane 2 1 1 2 3 4 5 5 2 A 3 4 If T Rm Rn is a linear transformation then the set x T x 0 is called the kernel of T These are all vectors which are annihilated by the transformation If T x A x then the kernel of T is also called the kernel of A The kernel of A are all solutions to the linear system Ax 0 We write ker A or ker T 6 The kernel of T x y z x y 0 is the z axes Every vector 0 0 z is mapped to 0 7 The kernel of a rotation in the plane consists only of the zero point 8 The kernel of the averaging map consists of all vector x y z for which x y z 0 The kernel is a plane In the language of random variables the kernel of T consists of the centered random variables The span of the standard basis vectors e1 e2 is the xy plane Also the kernel of a matrix A is a linear space n A subset V of R is called a linear space if it is closed under addition scalar multiplication and contains 0 The image of a linear transformation x 7 A x is the span of the column vectors of A The image is a linear space kernel domain How do we compute the kernel Just solve the linear system of equations A x 0 Form rref A For every column without leading 1 we can introduce a free variable si If x is P the solution to A xi 0 where all sj are zero except si 1 then x j sj xj is a general vector in the kernel image codomain 9 A column vector of A is called a pivot column if it contains a leading one after row reduction The other columns are called redundant columns 3 6 9 2 6 0 5 1 0 Gauss Jordan elimination gives B rref A 1 3 0 0 1 There are two pivot columns and one redundant column The equation B x 0 0 0 0 0 0 is equivalent to the system x 3y 0 z 0 After fixing z 0 can chose y t freely and 3 obtain from the first equation x 3t Therefore the kernel consists of vectors t 1 0 How do we compute the image If we are given a matrix for the transformation then the image is the span of the column vectors But we do not need all of them in general 1 2 Find the kernel of A 3 0 0 Step II Transmission Now add an error by switching one entry in each vector Homework due March 2 2011 1 A 2 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 0 0 0 0 1 1 0 0 A 0 0 0 0 1 0 4 0 1 0 3 0 1 0 2 0 6 0 1 0 3 0 0 0 1 0 4 0 0 0 1 0 0 0 0 0 1 1 0 0 1 0 1 1 Hu 0 1 0 1 1 0 1 0 0 1 1 1 1 0 B 0 0 0 1 0 0 1 0 E 0 0 0 1 0 1 1 0 H 0 0 0 1 1 0 1 0 K 0 0 0 1 1 1 1 0 N 0 0 1 0 0 0 1 0 Q 0 0 1 0 0 1 1 0 T 0 0 1 0 1 0 1 0 W 0 0 1 0 1 1 1 0 Z 0 0 1 1 1 0 1 0 0 0 1 1 1 0 1 0 C 0 0 0 1 0 0 1 1 F 0 0 0 1 0 1 1 1 I 0 0 0 1 1 0 1 1 L 0 0 0 1 1 1 1 1 O 0 0 1 0 0 0 1 1 R 0 0 1 0 0 1 1 1 U 0 0 1 0 1 0 1 1 X 0 0 1 0 1 1 1 1 0 0 1 1 1 0 1 1 0 0 1 1 1 0 1 1 the matrix multiplications to build 0 1 1 0 1 0 0 1 0 1 0 0 1 0 1 1 0 0 0 0 1 1 1 1 My 1 0 0 0 0 1 1 0 1 0 0 1 0 1 0 0 1 0 Now look in which column Hu or Hv is Assume this column is k Place 0 s everywhere in the vectors e except at the k th entry where you put 1 For example if Hu is the second e column then put a 1 at the second place We obtain vectors e and f 1 1 0 0 0 0 1 Pu f Step IV Decode the message x1 x2 x4 x x 3 5 x Use P to determine P e P P f P 4 x6 x5 x7 x 6 x7 an error free transmission P u P v would give the right result back Now Choose a letter to get a pair of vectors x y …