Don t forget to think of a matrix as a map (a.k.a. A Linear Algebra Primer)
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1 Don t forget to think of a matrix as a map (aka A Linear Algebra Primer) Thomas Yu February 5, Matrix Recall that a matrix is an array of numbers: a 11 a 1n A a m1 a mn In this note, we focus on the case of m n 2, although everything in this section has a natural generalization to general m and n (the concept of determinant, however, is only meaningful in the case of m n) It is important to not just think of a matrix as a boring array of number Matrix originated from the study of linear equations: a 11 x 1 + a 12 x a 1n x n b 1 a m1 x 1 + a m2 x a mn x n b m The first use of matrix is to write the above in the following compact form where A is a matrix, x and b are column vectors Ax b, The key point of this note is to urge you to think of a matrix as a function (aka a mapping/map) The study of various properties of linear solution urged mathematicians to think about the mapping: x [x 1, x 2 ] T Ax [a 11 x 1 + a 12 x 2, a 21 x 1 + a 22 x 2 ] T (11) As such, we think of a 2 2 matrix as a function from R 2 to R 2 (ie a function that maps two numbers to two numbers) Of course, there are many, many functions from R 2 to R 2 that are not of the form (11), those functions are called nonlinear functions The functions that are of the form (11) are called linear functions Linear functions are very special, and many basic properties of a linear function (eg 1-1, onto properties) can be determined easily The properties of linear maps are what you learn in a course of linear algebra You may wonder: why spend so much effort just to learn about one kind of functions? Besides the fact that there is much to learn about linear maps, an important point is that any nonlinear function can almost always be locally approximated by a linear function: if f : R 2 R 2 is differentiable near a point x R 2, then f1 f(x) f(x ) + x 1 (x f ) 1 x 2 (x ) f 2 x 1 (x f ) 2 x 2 (x (x x ), when x is near x ) 1
2 This has many important applications Once you start thinking matrices as functions, standard concepts pertaining to functions (function composition, 1-1, onto properties, etc etc) apply to matrices Composing two functions is something very basic, so one may ask: If we have two 2 2 matrices A and B, and we view them as functions from R 2 to R 2, then these functions can be composed Is the composed function still linear? If so, what is the new 2 2 matrix that represents the composite function? I put the answers to these two questions in the form of a theorem: Theorem 11 The composition of two linear functions is still linear The matrix that represents the composite function is exactly the matrix obtained by multiplying the two matrices in the way you were taught Proof: For clarity, we use the symbol f A : R 2 R 2 to denote the linear function represented by the matrix A, so if x R 2, f A (x) Ax [a 11 x 1 + a 12 x 2, a 21 x 1 + a 22 x 2 ] T Similarly we use the notation f B : R 2 R 2 to denote the linear function represented by the matrix B, so f B (x) Bx We seek to prove that the composite function is also linear Notice that f B f A : R 2 R 2 f B f A (x) f B (f A (x)) (def of function composition) B(Ax) a11 x B 1 + a 12 x 2 a 21 x 1 + a 22 x 2 b11 (a 11 x 1 + a 12 x 2 ) + b 12 (a 21 x 1 + a 22 x 2 ) b 21 (a 11 x 1 + a 12 x 2 ) + b 22 (a 21 x 1 + a 22 x 2 ) (b11 a 11 + b 12 a 21 )x 1 + (b 11 a 12 + b 12 a 22 )x 2 (b 21 a 11 + b 22 a 21 )x 1 + (b 21 a 12 + b 22 a 22 )x 2 b11 a 11 + b 12 a 21 b 11 a 12 + b 12 a 22 x1 b 21 a 11 + b 22 a 21 b 21 a 12 + b 22 a 22 x 2 So f B f A (x) is indeed a linear function and is represented by the matrix b11 a 11 + b 12 a 21 b 11 a 12 + b 12 a 22 ; b 21 a 11 + b 22 a 21 b 21 a 12 + b 22 a 22 moreover this matrix is exactly BA under the standard definition of matrix multiplication The main point here is that the definition of matrix multiplication which may look quite unmotivated as first glance is exactly the operation for composing two linear maps Given the above interpretation of matrix multiplication, it is hardly surprising that matrix multiplication is not commutative, ie AB is not always equal to BA Composing two maps in two different orders typically lead to very different composite maps (A non-mathematical example: Compare (i) you put clothes into a washing machine, take them out, and then put them into a dryer; versus (ii) you put clothes into a drying machine, take them out, and then put them into a washing machine You get totally different results!) Exercise: Consider the square Let A 2 3, B Calculate P AB, Q BA S {[x, y] T : x 1, y 1} Now, determine the 4 corners of each of the following parallelograms A(S), B(S), P (S), Q(S), and draw these parallelograms 2
3 (Again, we view a matrix as a function from R 2 to R 2, so if A is a matrix and S R 2, A(S) is the image of the set S under the function A as defined in the book, ie A(S) {Ax : x S}, which is also a subset of R 2 in this case) Notice that P (S) and Q(S) are two different parallelograms; compare their areas 2 Determinant So now you understand that we should think of a matrix as a transformation on R 2 Different matrices have different actions on R 2 Indeed, some matrices can transform the plane quite radically, as illustrated by 1 2 A (22) 1 2 This matrix maps every point on the plane onto the line {[x, y] T : x y} (see it?), it collapses any region with a positive area into a region with zero area The determinant of a 2 2 matrix is given by the following simple formula: a11 a det( 12 ) a a 21 a 11 a 22 a 12 a The simplicity of this formula is deceptive Where does this formula come from? What does this magical number say about the matrix itself? Below I give the answers to these questions, with proofs omitted Let S R 2, f : R 2 R 2, and I want to measure: area(f(s)) area(s) Here, for a technical reason, I have to assume that S is nice in a certain sense 1 in order to talk about its area, a measurable set S can be quite complicated and needs not be of a familiar shape such as a polygon or an ellipse In general, the above ratio depends on both the function f and the subset S: a general nonlinear function f : R 2 R 2 can distort area mildly at some places but very wildly at some other places However, a linear function cannot do that; we have: Theorem 21 (I) The above ratio does not depend on S when f is a linear mapping, ie a function of the form (11) (II) So the ratio is only dependent on the matrix A and this ratio is exactly det(a) I will skip the proof of the above important fact For one thing, I did not clarify what I mean by S being nice, and when S is nice, how do I exactly define the area of S However, the above theorem is not too hard to prove if I assume S to be just rectangles; and it is a very good exercise to do Now I have explained what the magnitude of a matrix measures about the matrix itself What, then, does the sign of the determinant of a matrix say about the matrix? It is not hard to see that a matrix preserves orientation if and only if its determinant is positive; equivalently, a matrix reverses orientation if and only if its determinant is negative You may remember learning about the following fact: Theorem 22 A is invertible if and only if det(a) 0 The above theorem is actually closely related to Theorem 21 (although here I am not attempting to prove either one of the two theorems using the other): Recall that any function is invertible if and only if 1 The technical term for niceness here is measurable, a concept developed in a subject called measure theory 3
4 it is 1-1 and onto In general, 1-1 and onto are two totally different concepts; however, any linear function from R n to R n is 1-1 if and only if it is onto So a matrix A is invertible iff the associated linear map is onto When is a function onto? Answer: exactly when the linear map does not collapse any subset with a positive area into a subset with zero area, ie when the determinant is not zero For example, the matrix in (22) has a zero determinant, it is non-invertible, and, as a map, it collapses any region with a positive area to a set with zero area Theorem 23 det(ab) det(a) det(b) det(ba) (although in general AB BA) Proof: Let S be any nice subset of R 2 with a positive area, then det(ab) T hm21 area(ab(s)) area(s) area(a(b(s))) area(b(s)) T hm21 det(a) det(b) area(b(s)) area(s) 3 Orthogonal matrices and matrices with determinant ±1 To wrap up this crash course on linear algebra, we compare here matrices which (i) are orthogonal (Definition: an n n matrix is orthogonal if A 1 A T ) and (ii) have a determinant equals to 1 or 1 Anyone seeing the condition A 1 A T the first time must ask: what does this condition really say about the linear map defined by the matrix A? The concept of orthogonal matrices come from the following fact: Theorem 31 A linear map preserves Euclidean distance if and only if it is one represented by an orthogonal matrix (As such, an orthogonal matrix represents a linear map that is like a rotation, a reflection, or a combination of the two) Proof: ( ) Recall that the Euclidean distance x of a vector x R n is: x n x 2 i x T x i1 Therefore if A T A I, then Ax 2 (Ax) T (Ax) x T A T Ax x T (I)x x T x x 2 In other words, the the action of A to any vector can not change the length of the vector ( ) (The proof of this part can be skipped) First of all, notice that if A preserves length, it must be invertible: assume not, then a very basic result in linear algebra says that Ax 0 for some non-zero vector x; but this immediately violates the length preserving assumption Let r 1,, r n be the rows of the n n matrix A, then r1 T,, r n are column vectors Note that no row is identically zero (otherwise A is singular), so ri T > 0 for all i Next, consider r 1 r T r1 T 1 Ar1 T r 2 r T 1 r n r1 T r n r1 T Since A preserves length, Ar1 T r1 T ; this also means r 2 r1 T r n r1 T 0 Similarly, we can show that r i rj T 0, i j So all together we have shown that all the off-diagonal entries of AA T are zeros, and all the diagonal entries of AA T are positive This also means that the inverse of AA T has the same structure 4
5 To finish the proof, we need to show that the diagonal entries of (AA T ) 1 are all equal to one Note that A 1 also preserves length: A 1 y A(A 1 y) y for any y So y T A T A 1 y y T y for any y By choosing y e i (ie the column vector with zeros in all but the ith position, and equals 1 at the ith position), we have (A T A 1 ) i,i e T i (A T A 1 )e i e T i e i 1 Note: the ( ) part of the above theorem is the same as Property 732 on Page 398 of DeGroot and Schervish Next we want to show that a an orthogonal matrix must have determinant equals to 1 The proof of this relies on the algebraic fact that det(a T ) det(a) for any square matrix A: 1 det(i) det(a T A) det(a T ) det(a) det(a) 2 det(a) ±1 By our earlier interpretation of the determinant (Theorem 21), the above results mean that a length preserving linear map must also preserve area/volume/measure (when n 2/n 3/general n) You may now wonder: is it true that any (say 2 2) area preserving matrix must also preserve length? The answer is negative, consider the following shearing map: 1 1 A 0 1 It has determinant equals to 1, meaning that it preserves area In fact, A shears any square into a parallelogram of the same area (Why? Please draw[ some [ pictures) ] [ However, A T A I, and a shearing map clearly does not preserve length; for instance, 0 1] 1 1] 4 Conclusion Many concepts in linear algebra which at first glance look rather un-motivating have simple and elegant geometric interpretations The key is not to forget that a matrix should be interpreted as a linear map 5
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