Chapter 19. General Matrices. An n m matrix is an array. a 11 a 12 a 1m a 21 a 22 a 2m A = a n1 a n2 a nm. The matrix A has n row vectors


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1 Chapter 9. General Matrices An n m matrix is an array a a a m a a a m... = [a ij]. a n a n a nm The matrix A has n row vectors and m column vectors row i (A) = [a i, a i,..., a im ] R m a j a j a nj col j (A) =. Rn. The ij entry a ij is the number in row i, column j. If [a ij ] is n m and B = [b ij ] is m k, then the product matrix AB is defined by the formula ij entry of AB = m a il b lj = a i b j +a i b j + +a im b mj = row i (A), col j (B). l= Note that the number m of columns of A must equal the number of rows of B, otherwise, the product AB is not defined. The inner dimensions must match. The product then has the outer dimensions. Mnemonic : (n m)(m k) = n k. For example, we can multiply our n m matrix A times an m matrix (which is a vector in R m ), and we get an n matrix (which is a vector in R n ). In other words, A is a function A : R m R n which has the linearity property: A(su + tv) = sau + tav, for all u, v in R m and s, t in R. Any linear function from R m to R n is multiplication by some matrix A.
2 If u j = col j (A), and x = (x,..., x m ), then Ax = x u + + x m u m R n. In particular, if e j is the vector with in the j th spot and zeros in all the other spots, then we have the familiar fact that the columns of A are given by Ae j = col j (A). () As before, the kernel of A is ker {x R m : Ax = 0} R m. If n =, then [ a a a m ] and ker A is the hyperplane a x + a x + + a m x m = 0. The kernel of a general n m matrix A is the intersection of the row hyperplanes The image of A is a i x + a i x + + a im x m = 0, i =,,..., n. im {Ax R n : x R m } R n. That is, im A consists of the vectors in R n that come from R m via A. From equation (), it follows that im A is the span of the columns of A. The kernel and image are related to each other by the KernelImage Theorem: For any n m matrix A, we have dim ker A + dim im m. For small n, m we have an ad hoc understanding of dimension. The general definition of dimension will be given in the next chapter. Intuitively, you can think of A as a messenger, carrying information. In this imaginary viewpoint, the kernel of A is the information lost by A and im A is the information retained by A. The startingplace R m is the total information. Then the KernelImage Theorem says: Information lost + information retained = total information.
3 Example : Here A : R R 3, sending e u = (, 3, 5), e v = (, 4, 6), and A sends a general vector (x, y) R to the linear combination (x, y) x u + x v. Thus, A is an embedding of a plane in R 3, namely the plane spanned by u, v. Recall that we can find the equation of this plane, take the cross product u v = (, 4, ), which we can scale to (,, ). Thus the equation of the plane is x y + z = 0. This plane is called the image of A. In general, the image of a matrix A : R m R n is the set of all vectors in R n of the form Au, for some u R m. Example : Here we don t get a plane because the columns are proportional. Instead, every vector of the form Au is a scalar multiple of u = (,, 3), so the image is the line Ru. In this case, A is squashing R into a line in R 3. Example 3: [ 0 ] sends e e, e e, e 3 0. An arbitrary vector u = (x, y, z) R 3 is sent by A to Au = (x, y). Every vector in R 3 is squashed vertically onto the floor, which is R. The image of A is the whole R. In particular, the cube whose vertices are the vectors with entries 0, is squashed straight down into the square in R whose vertices have entries 0,. 3
4 Example 4: [ 0 ] x 0 y For some choice of (x, y). This sends the 0, cube to an image of the cube in R that can be seen by Miserable Dimensional Creatures. For example, A sends (, 0, ) to the vector (, 0) + (x, y), and the edge from (, 0, 0) to (, 0, ) to the line segment from (, 0) to (, 0) + (x, y). As the MDC varies (x, y), it gets different views of the cube, but with the angles deformed. The same thing happens when we try to draw a cube on paper. The price of cramming a 3d object into d is loss of information, in this case the angles and relative lengths of the sides. But we can at least see the relation between the vertices and edges and faces. Exercise 9. Compute the product, if can be computed 0 [ ] 0 (a) 3 0 (b) [ ] (c) Exercise 9. Let [a ij ] be an arbitrary n n matrix, and let D be a diagonal matrix d d 0 D = d n a) What are the ij entries of DA and AD? b) Suppose the d i s are distinct, and that AD = DA. What can you say about A? Exercise 9.3 MDCs, hence M3DCs like us, can see images of the hypercube in R 4, by using a matrix A : R 4 R. First of all, the hypercube has 6 corners (x, x, x 3, x 4 ), x i = 0 or. Two corners have an edge between them if they differ in only one component. For example, there is an edge between (0) and (00), but no edge between (0) and (0). a) Choose two random vectors u = (u, u ) and v = (v, v ) in R, and let [ ] 0 u v. 0 u v 4
5 (To get a good view, make sure no u i or v i is 0 or.) b) Plot the image under A of the 6 corners of the hypercube. You will plot 6 points in the plane R, but label them by the coordinates of the corresponding points in R 4, to keep track of them. For example, the corner (00) goes via A to (0, ) + (v, v ) = (v, + v ). Label this point by (00). c) Connect pairs of your 6 points in R by line segments if their 4tuple labels differ in only one component. d) Behold the hypercube! 5
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