Linear Algebra Notes


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1 Linear Algebra Notes Chapter 19 KERNEL AND IMAGE OF A MATRIX Take an n m matrix a 11 a 12 a 1m a 21 a 22 a 2m a n1 a n2 a nm and think of it as a function A : R m R n The kernel of A is defined as Note that ker A lives in R m The image of A is ker set of all x in R m such that Ax = 0 im set of all vectors in R n which are Ax for some x R m Note that im A lives in R n Many calculations in linear algebra boil down to the computation of kernels and images of matrices Here are some different ways of thinking about ker A and im A In terms of equations, ker A is the set of solution vectors x = (x 1,, x m ) in R m of the n equations a 11 x 1 + a 12 x a 1m x m = 0 a 21 x 1 + a 22 x a 2m x m = 0 a n1 x 1 + a n2 x a nm x m = 0, (19a) and im A consists of those vectors y = (y 1, y n ) in R n for which the system a 11 x 1 + a 12 x a 1m x m = y 1 a 21 x 1 + a 22 x a 2m x m = y 2 a n1 x 1 + a n2 x a nm x m = y n, 1 (19b)
2 2 has a solution x = (x 1,, x m ) A single equation a i1 x 1 + a i2 x a im x m = y i is called a hyperplane in R m (So a line is a hyperplane in R 2, and a plane is a hyperplane in R 3 ) Geometrically, ker A is the intersection of hyperplanes (19a), and im A is the set of vectors (y 1,, y n ) R n for which the hyperplanes (19b) intersect in at least one point If x and x are two solutions of (19b) for the same y, then A(x x ) = Ax Ax = y y = 0, so x x belongs to the kernel of A If ker 0 (ie, consists just of the zero vector) then there can be at most one solution In general, the bigger the kernel, the more solutions there are to a given equation that has at least one solution Thus, if x is one solution, then all other solutions are obtained from x by adding a vector from ker A However, we will see that there is a certain conservation principle at work here, which implies that the bigger the kernel, the smaller the image, so the less likely it is that there will be even one solution Intuitively, you can think of vectors in R m as representing information Then ker A is the information lost by A, while im A is the information retained by A We can also describe im A as the span (=set of linear combinations of) the columns of A That is, if u 1,, u m are the columns of A, then im A consists of the vectors Ax = x 1 u 1 + x 2 u x m u m R n, with all possible choices of scalars x 1,, x m Example 1: (2 3 case) 12 a 13 a 21 a 22 a 23 Suppose first that A is the zero matrix (all a ij = 0) Then ker R 3 and im A consists only of the zero vector Suppose then that A is not the zero matrix Then ker A is the intersection of two planes through (0, 0, 0) a 11 x 1 + a 12 x 2 + a 13 x 3 = 0 a 21 x 1 + a 22 x 2 + a 23 x 3 = 0 Each plane corresponds to a row vector of A, whereby the row vector is the normal vector to the plane If the row vectors of A are not proportional, then the planes are distinct In this case the planes intersect in a line, and ker A is this line If the rows of A are proportional, then the two equations determine just one plane, and ker A is this plane For example, ker is the line R(1, 2, 1),
3 3 while ker is the plane x + 2y + 3z = 0 What about im A? Since im A R 2 and is nonzero (because we re assuming A 0), the image of A is either a line or all of R 2 How to tell? Recall that im A is spanned by the three column vectors 12 a13 u 1 =, u a 2 =, u 21 a 3 = 22 a 23 The image of A will be a line l exactly when these three column vectors all live on the same line l If, say, u 1 0, and the image is a line, then there are scalars s, t such that u 2 = su 1 and u 3 = tu 1 This would mean that a11 sa 11 ta 11 a 21 sa 21 ta 21 But look, this means the rows are proportional They are both proportional to (1, s, t) By what we saw before, this means the kernel is a plane In summary: Recall also the case 0: If im line then ker plane (19a) If im plane then ker line (19b) If im 0 then ker R 3 (19c) We can summarize (19ac) in a table of dimensions: 2 3 matrix expected, rows not proportional 2 1 rows proportional A only Note that for any 2 3 matrix A, we have dim(ker A) + dim(im A) = 3 As you vary A, the quantity dim(ker A) can vary from 1 to 3, and the quantity dim(im A) can vary from 0 to 2, but the sum dim(ker A) + dim(im A) remains constant at 3 For 3 2 matrices the table is a 11 a 12 a 21 a 22 a 31 a 32 : R 2 R 3,
4 4 2 3 matrix 0 2 expected, columns not proportional 1 1 columns proportional, A only Again dim(ker A) + dim(im A) = 3 You can think of this as conservation of information It is a general fact: KernelImage Theorem Let A be an n m matrix Then dim(ker A) + dim(im A) = m The corresponding table depends on whether n or m is bigger n m, n m m n n expected, rows linearly independent m n + 1 n 1 m 0 0 only n m, n m 0 m expected, columns linearly independent 1 m 1 m 0 0 only A set of vectors is linearly independent if none of them is a linear combination of the others Note that the expected situation has minimal kernel As you go down the rows in the tables, there are more and more conditions to be satisfied, hence each row is less likely than the one above, until finally only 0 satisfies all the conditions of the last row The conditions can be expressed as certain determinants being zero, as follows Let µ be the smaller of n or m Each table has µ + 1 rows Number the rows 0, 1,, µ starting at the top row Then a matrix satisfies the conditions for row 0 (the expected case) if some µ µ subdeterminant of A is nonzero The conditions for some lower row p are that some p p subdeterminant of A is nonzero, but all (p+1) (p+1) subdeterminants are zero This is illustrated in the exercises Intuitively, the KernelImage Theorem says the amount of information lost plus the amount of information retained equals the amount of information you started with However, to really understand the KernelImage Theorem, we have to understand dimension inside R n for any n
5 Exercise 191 Determine the kernel and image, and the dimensions of these, for the following matrices (A line is described by giving a nonzero vector on the line, and a plane can by described by giving two nonproportional vectors in the plane) (a) (b) 3 0 (d) (e) (f) (g) (h) a b Exericise 192 Let c d (c) (a) Suppose A is the zero matrix What are ker A and im A? (b) Suppose A 0, but det 0 What are ker A and im A? (c) Suppose det A 0 What are ker A and im A? Exercise 193 Let u = (u 1, u 2 ) and v = (v 1, v 2 ) be vectors in R 2, and let 1 0 u1 v u 2 v 2 (This is the sort of matrix you used to map the hypercube in R 4 into R 2 ) (a) Describe the kernel of A in terms of the vectors u and v (b) What is the image of A? Exercise 194 A 2 3 matrix 12 a 13 a 21 a 22 a 23 has three subdeterminants det 12, det 13 a 21 a 22 a 21 a 23 a12 a, det 13 a 22 a 23 Assume A is non zero Explain how these subdeterminants determine the dimensions of the kernel and image of A (Study the 2 3 analysis given above ) Exercise 195 Explain how to use subdeterminants to determine the dimensions of the image and kernel of a 2 m matrix 12 a 1m a 21 a 22 a 2m 5 Exercise 196 Make the tables of dimensions of kernels and images of 3 4 and 4 3 matrices, and find a matrix for each row of each table
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