590 Notes July 8, Theorem (1.11). Let W be a subspace of a finite-dimensional vector space V. Then

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1 59 Notes July 8, 214 Theorem (111) Let W be a subspace of a finite-dimensional vector space V Then 1 W is finite dimensional 2 If dim(w ) = dim(v ), Then W = V Proof of theorem 111 Let dim(v ) = n If W = {}, then dim(w ) = n Otherwise, W contains a nonzero vector v 1, and {v 1 } is a linearly independent subset of W We continue choosing vectors v 2, v 3,, v k, if possible, so that {v 1, v 2,, v k } is linearly independent, and adjoining another vector from V results in a linearly dependent set Since no linearly independent subset of V has more than n vectors, this process stops at k n Then {v 1, v 2,, v k } is a basis for W, so dim(w ) dim(v ) If dim(w ) = n, then a basis for W is a linearly independent subset of V with exactly n vectors, and thus is a basis V, so W = V Example Let W be the set of diagonal n n matrices as a subset of M n n (F ) Then for E pq = A M n n (F ) with (A) ij = 1 if i = p, j = q, and A ij = otherwise, the set {E 11, E 22,, E nn } is a basis for W Therefore, dim(w ) = n Corollary If W is a subspace of a finite-dimensional vector space V Then Any basis for W can be extended to a basis for V Proof Since a basis for W is a linearly independent subset of V, c) of corollary 2 of the replacement theorem gives the result The Lagrange Interpolation Formula Let c, c 1,, c n be distinct scalars (ie no pair are equal) in an infinite field Define the polynomials f (x), f 1 (x),, f n (x) in P (F ) via: f i (x) = (x c )(x c 1 ) (x c i 1 )(x c i+1 ) (x c n ) (c i c )(c i c 1 ) (c i c i 1 )(c i c i+1 ) (c i c n ) = x c k c i c k k i

2 note that these polynomials are degree n For examlpe if our field is R and c = 1, c 1 = 2, c 2 = 4, we get f = (x 2)(x 4) (1 2)(1 4) = x2 6x = x2 3 2x { 1 if i = j Note that f i (c j ) = if i j Thus if we take a linear combination of the f i s and get zero: Then necessarily a i f i (x) = (The zero polynomial in P (F )) i= a i f i (c j ) = for every j =, 1, n i= This is only possible if a i = for i =, 1,, n Therefore, the set {f, f 1,, f n } is a set of n + 1 linearly independent vectors in P n (F ) Since dim(p n (F )) = n + 1, we conclude that {f, f 1,, f n } is a basis of P n (F ) Suppose g(x) P n (F ) What does g(x) look like in this basis? Well: Therefore if g(x) = a i f i (c j ) = a j for every j =, 1, n i= a i f i (x), g(c i ) = a i f i (c j ) = a j for every j =, 1, n i= Therefore, for any g(x) P n (F ): g(x) = g(c i )f i (x) (1) i= We call (1) the Lagrange Interpolation Formula We now state, without proof, a result of section 17: Theorem Every vector space has a basis On to chapter 2: Linear Transformations and matrices Page 2

3 * Section 21 Linear Transformations, Null Spaces and Ranges A few words on notation: 1 Suppose a function f has domain A and codomain B Then we write f : A B If for x A and y B, f(x) = y, we write f : x y 2 We use the symbol to denote for all or for every For example, v V reads for all v in V Definition Let V and W be vector spaces over a field F We call a function T : V W a linear transformation from V to W If x, y V and c F : 1 T(x+y)=T(x)+T(y) 2 T(cx)=cT(x) Facts: 1 T is linear T () = 2 T is linear T (cx + y) = ct (x) + T (y) 3 T is linear T (x y) = T (x) T (y) Examples of linear operators on R 2 : projection, rotation, reflection More explicitly, take reflection about the x-axis: T : R 2 R 2 via T : (a, b) (a, b) T is linear because: T (c(a 1, b 1 ) + (a 2, b 2 )) = T (ca 1 + a 2, cb 1 + b 2 ) = (ca 1 + a 2, cb 1 b 2 ) = c(a 1, b 1 ) + (a 2, b 2 ) = ct (a 1, b 1 ) + T (a 2, b 2 ) Some other classic linear operators are differentiation on P n (R) and integration over an interval on C(R) Two particular linear transformations appear often and get their own names For vector spaces V and W over the field F, we define: The identity transformation I V is defined by: I V : V V and I V (x) = x x V The zero transformation T defined by T : V W and T (x) = x V Page 3

4 Definitions Let V and W be vector spaces, and let T : V W be linear We define the null space or kernel N(T ) of T to be the set of all vectors in x V such that T (x) = We define the range or image R(T ) to be the set of all y W such that y = T (x) for some x V, that is, R(T ) = {T (x) : x V } Theorem (21) Let V and W be vector spaces and T : V W be linear Then N(T ) is a subspace of V and R(T ) is a subspace of W Proof Let V denote the zero vector in V and W denote the zero vector in W Since T is linear, T ( V ) = W so V N(T ) If x, y N(T ), then T (x + y) = T (x) + T (y) = W + W = W and if c F, T (cx) = ct (x) = c W = W, so N(T ) is a subspace of V Similarly, since T ( V ) = W, W R(T ) If y 1, y 2 R(T ) then there exist x 1, x 2 V such that y 1 + y 2 = T (x 1 ) + T (x 2 ) = T (x 1 + x 2 ) and if c F, cy 1 = ct (x 1 ) = T (cx 1 ) so R(T ) is a subspace of W Theorem (22) Let V and W be vector spaces, and let T : V W be linear If β = {v 1, v 2,, v n } is a basis for V then R(T ) = span(t (β)) = span({t (v 1 ), T (v 2 ),, T (v n )}) Proof We see that T (v i ) R(T ) for all v i β by the definition of R(T ) Since R(T ) is a subspace, R(T ) contains span({t (v 1 ), T (v 2 ),, T (v n )}) = span(t (β)) Let y R(T ) Then x V such that y = T (x) Write x in terms of the basis β, so x = n a iv i Then y = T (x) = T ( n a iv i ) = n a it (v i ) Therefore R(T ) span({t (v 1 ), T (v 2 ),, T (v n )}) = span(t (β)) Example Define the linear transformation T : P 2 (R) M 2 2 (R) via: ( ) f() f(1) T (f(x)) = f(2) Since β = {1, x, x 2 } is a basis for P 2 (R): R(T ) = span(t (β)) = span({t (1), T (x), T (x 2 )}) ({( ) ( ) ( )}) 1 1 = span,, ({( ) ( )}) 1 = span, 1 We have found a basis for R(T ) and we see that dim(r(t )) = 2 Page 4

5 Definitions Let V and W be vector spaces and T : V W be linear If the nullspace N(T ) is finite dimensional, we define the nullity of T, denoted nullity(t ), to be the dimension of N(T ) Similarly, If the Range R(T ) is finite dimensional, we define the rank of T, denoted rank(t ), to be the dimension of R(T ) Theorem (Dimension Theorem) Let V and W be vector spaces and let T : V W be linear If V is finite dimensional then nullity(t ) + rank(t ) = dim(v ) Proof Suppose dim(v ) = n N(T ) is a subspace of V, so dim(n(t )) = k for some k n Let {v 1, v 2,, v k } be a basis for N(T ) Now extend this basis to a basis {v 1, v 2,, v k, v k+1,, v n } We will prove that {T (v k+1 ), T (v k+2 ),, T (v n )} is a basis for R(T ) Let y R(T ) Then there exists x V such that y = T (x) Writing x as a linear combination of basis vectors, say x = a i v i, we see that ( ) y = T (x) = T a i v i = a i T (v i ) = Therefore {T (v k+1 ), T (v k+2 ),, T (v n )} spans R(T ) i=k+1 a i T (v i ) Now we prove {T (v k+1 ), T (v k+2 ),, T (v n )} is ( a linearly independent set Suppose that a i T (v i ) = Since a i T (v i ) = T a i v i ), we have a i v i N(T ) Thus i=k+1 i=k+1 a i v i = i=k+1 k b i v i for some b 1, b 2,, b k F, ie i=k+1 a i v i i=k+1 k b i v i = But {v 1, v 2,, v k, v k+1,, v n } is a basis, so all the a i s (and b i s) are zero Therefore {T (v k+1 ), T (v k+2 ),, T (v n )} is a linearly independent set, and the T (v i ) s are distinct, so dim(r(t )) = n k Theorem (24) Let V and W be vector spaces, and let T : V W be linear Then T is one-to-one N(T ) = {} Proof ) Suppose N(T ) = and x, y V Then T (x) = T (y) T (x) T (y) = T (x y) = x y N(T ) x y = x = y Therefore T is one-to-one Page 5

6 ) If x N(T ), T (x) = but also necessarily T () = Since T is one-to-one, we conclude x = So N(T ) = {} Theorem (25) Let V and W be vector spaces of equal (finite) dimension, and let T : V W be linear Then the following are equivalent a) T is one-to-one b) T is onto c) rank(t ) = dim(v ) Proof Outline of the proof: a) c) b) From the dimension theorem, we know nullity(t ) + rank(t ) = dim(v ) We know by theorem 24 that a) implies N(T ) = which implies that nullity(t ) =, thus a) implies c) If we assume c), then by the dimension theorem and theorem 24, we get that a) is true Therefore c) implies a) Lastly, we notice that T is onto if and only if dim(r(t )) = dim(w ) dim(v ) = dim(w ), we see that b) is true if and only if c) is Since Theorem (26) Let V and W be vector spaces over F, and suppose that {v 1, v 2,, v n } is a basis for V For w 1, w 2,, w n in W, there exists exactly one linear transformation T : V W such that T (v i ) = w i for i = 1, 2,, n Proof Let x V Then x = a i v i Where a 1, a 2,, a n are unique scalars We define T : V W by T (x) = n a iw i (We are sending v i to T (v i ) = w i for each i, and extending linearly All that is left to do is prove that this map is linear and unique) 1 T is linear: Let u, v V, with u = n b iv i, v = n c iv i, for some scalars b 1, b 2,, b n and c 1, c 2,, c n It follows that, for any d F, ( ) ( ) T (du + v) = T d b i v i + c i v i = T (db i + c i )v i Page 6

7 By the definiton of T, we get: ( ) T (db i + c i )v i = (db i + c i )w i and finally (db i + c i )w i = d b i w i + c i w i = dt (u) + T (v) 2 Suppose that U : V W is linear and U(v i ) = w i for i = 1, 2,, n Then ( ) U(x) = U a i v i = a i U(v i ) = a i w i = T (x) Corollary Let V and W be vector spaces, and suppose that V has a finite basis {v 1, v 2,, v n } If U, T : V W are linear and U(v i ) = T (v i ) for i = 1, 2,, n, then U = T Example Consider the map (transformation) T : R 2 R 2 defined by: T (a 1, a 2, a 3 ) = (a 1 a 2, 2a 3 ) Let x = (x 1, x 2, x 3 ), y = (y 1, y 2, y 3 ) R 3 and let c R Then T (cx + y) = T (cx 1 + y 1, cx 2 + y 2, cx 3 + y 3 ) = (cx 1 + y 1 (cx 2 + y 2 ), 2(cx 3 + y 3 ) = c(x 1 x 2, 2x 3 ) + (y 1 y 2, 2y 3 ) = ct (x) + T (y), so T is linear The kernel of T, N(T ), is the solution set to the equation a 1 a 2 =, 2a 3 =, ie the set {a 1, a 2, a 3 ) R 3 : a 1 = a 2, a 3 = } which we can span with the vector (1, 1, ) Therefore N(T ) has dimension one and {(1, 1, )} is a basis for N(T ) By the dimension theorem, we know the image must have dimension 2 Let s confirm this by finding a basis for the image R(T ) By theorem 22, we need only compute the vectors T (1,, ), T (, 1, ), T (,, 1) which are (1, ), ( 1, ), (, 2) The vectors (1, ), ( 1, ), (, 2) clearly span R 2, therefore we can confirm dim(r(t )) = 2 Finally, we see that because the nullity of T is one, T is not one-to-one, and since the rank of T is 2, T is onto * 22 The Matrix Representation of A Linear Transformation Definition Let V be a finite-dimensional vector space An ordered basis for V is a basis for V with a specific order; that is, an ordered basis for V is a finite sequence of linearly independent vectors in V that generate V Page 7

8 Example If V = F 3, then β 1 = {e 1, e 2, e 3 } and β 2 = {e 3, e 1, e 2 } are both ordered bases for V but β 1 β 2 as ordered bases Definition Let β = {u 1, u 2,, u n } be an ordered basis for a finite-dimensional vector space V For x V, let a 1, a 2,, a n be the unique scalars such that x = a i u i We define the coordinate vector of x relative to β, denoted by [x] β, by [x] β = Note that [u i ] β = e i In fact the map x [x] β is a special kind of linear map, called an isomorphism We will study this later Example Let V = P 2 (R) and let {1, x, x 2 } be the standard ordered basis for V Then, for example, if f(x) = 1 + 5x + 3x 2, 1 [f] β = 5 3 Now we will see how to represent a linear transformation as a matrix Suppose that V and W are finite dimensional vector spaces with ordered bases β = {v 1, v 2,, v n } and γ = {w 1, w 2,, w m } respectively Let T : V W be linear Then for each j, 1 j n, there exist unique scalars a ij F, 1 i m, such that a 1 a 2 a n T (v j ) = m a ij w i for 1 j n Definition For the a ij s as above, we call the m n matrix A defined by (A) ij = a ij the matrix representation of T in the ordered bases β and γ and write A = [T ] γ β If V = W and β = γ, we write A = [T ] β Note that the j-th column of A is just [T (v j )] γ Page 8

9 Example Let T : P 3 (R) P 2 (R) be defined by T (f(x)) = f (x) Let β and γ be the standard ordered bases for P 3 (R) and P 2 (R) respectively Then: T (1) = = 1 + x + x 2 [T (1)] γ = 1 T (x) = 1 = x + x 2 [T (x)] γ = T (x 2 ) = 2x = x + x 2 [T (x 2 )] γ = 2 T (x 3 ) = 3x 2 = 1 + x + 3 x 2 [T (x 3 )] γ = 3 and therefore [T ] γ β = Definition Let T, U : V W be arbitrary functions, where V and W are vector spaces over F, and let a F We define T + U : V W via (T + U)(x) = T (x) + U(x) for all x V and at : V W via (at )(x) = at (x) for all x V Theorem (27) Let V and W be vector spaces over F and let T, U : V W be linear a) for all a F, at + U is linear b) Using the addition an scalar multiplication defined above, the set of all linear transformations from V to W is a vector space over F Proof A tedius but easy exercise Write this proof for extra credit (due this Friday, June 2) if you are inclined Definitions Let V and W be vector spaces over F We denote the vector space of all linear transformations from V into W by L(V, W ) In the case V = W, we write L(V ) instead of L(V, V ) Theorem (28) Let V and W be finite-dimensional vector spaces with ordered bases β and γ, respectively, and let T, U : V W be linear transformations Then a) [T + U] γ β = [T ]γ β + [U]γ β Page 9

10 b) [at ] γ β = a [T ]γ β for all scalars a Proof We prove part a) The proof of part b) is similar Let β = {v 1, v 2,, v n } and γ = {w 1, w 2,, w m } Then there exit scalars a ij and b ij ( 1 i m, 1 j n) such that: m m T (v j ) = a ij w i and U(v j ) = b ij w i therefore (U + T )(v j ) = m a ij w i + so ( ) [U + T ] γ β ij m b ij w i = = a ij + b ij = m (a ij + b ij )w i ( ) [T ] γ β + [U]γ β ij (If this proof looks funny just keep in mind all we are doing is showing two matrices are equal) * 23 Composition of Linear Transformations and Matrix Multiplication Compositions of linear maps are linear: Theorem (29) Let V, W and Z be vector spaces over the same field F and let T : V W and U : W Z be linear Then UT : V Z is linear Proof Let x, y V, c F Then UT (cx + y) = U(cT (x) + T (y)) = cut (x) + UT (y) Furthermore: Theorem (21) Let V be a vector space Let T, U 1, U 2 L(V ) Then: a) T (U 1 + U 2 ) = T U 1 + T U 2 and (U 1 + U 2 )T = U 1 T + U 2 T b) T (U 1 U 2 ) = (T U 1 )U 2 c) T I = IT = T d) a(u 1 U 2 ) = (au 1 )U 2 = U 1 (au 2 ) for all scalars a Page 1

11 Proof This is a mind-numbing exercise of applying the definition of function composition Let s just prove b): Let x V Then on the one hand: T (U 1 U 2 )(x) = T (U 1 (U 2 )(x)) = T (U 1 (U 2 (x))) and on the other hand: (T U 1 )U 2 (x) = (T U 1 )(U 2 (x)) = T (U 1 (U 2 (x))) Therefore, T (U 1 U 2 ) = (T U 1 )U 2 (Note that we didn t even use linearity! Or the fact that V and W are vector spaces! Only a) and d) use those facts b) and c) are true in much more generality) Suppose we have these three vector spaces: Vector space Ordered Basis V α = {v 1, v 2,, v n } W β = {w 1, w 2,, w m } Z γ = {z 1, z 2,, z p } and suppose T : V W and U : W Z are linear maps Let B = [T ] β α and A = [U] γ β Then ( m ) ( m ) (UT )(v j ) =U(T (v j )) = U (B) kj w k = (B) kj U(w k ) k=1 ( p ( m )) ( m = (B) kj (A) ik z i = k=1 k=1 k=1 ) p (A) ik (B) kj z i ( p ( m ) ) ( p ) = (A) ik (B) kj z i = (C) ik z i k=1 For this reason we define the product of an m n matrix A with an m p matrix B, denoted AB, to be the m p matrix: (AB) ij = (A) ik (B) kj for 1 i m, 1 j p k=1 The way we defined matrix multiplication gives us a theorem for free: Theorem (211) Let V, W and Z be finite-dimensional vector spaces with ordered bases α, β, γ respectively Let T : V W and U : W Z be linear transformations Then [UT ] γ α = [U]γ β [T ]β α Corollary Let V be a finite-dimensional vector space with an ordered basis β Let T, U L(V ) Then [UT ] β = [U] β [T ] β Page 11

12 Definitions We define the Kronecker delta δ ij by { 1 if i = j δ ij = if i j The n n identity matrix I n is defined by (I n ) ij = δ ij Theorem (212) Let A M m n (F ), B, C M n p (F ) and D, E M q m (F ) Then a) A(B + C) = AB + AC and (D + E)A = DA + EA b) a(ab) = (aa)b = A(aB) for any scalar a F c) I m A = A = AI n d) If V is an n-dimensional vector space with an ordered basis β, then [I V ] β = I n Proof Proofs of a) and c) are in the book We ll prove b) and d): proof of b): ( ) ( ) a(ab) ij = a (A) ik (B) kj = (a(a) ik )(B) kj = (aa)b ij k=1 k=1 and ( ) A(aB) ij = (A) ik (a(b) kj ) = k=1 (a(a) ik )(B) kj = (aa)b ij k=1 proof of d): Letβ = {v 1, v 2,, v n } Then by definition, I V (v i ) = v i for i = 1, 2,, n and 1 1 [v 1 ] β =, [v 2 ] β = and so on Therefore [I V ] β = I n Corollary Let A be an m n matrix, B 1, B 2,, B k be n p matrices, C 1, C 2,, C k be q m matrices, and a 1, a 2,, a k be scalars Then A ( k a ib i ) = k a iab i and ( k a ic i ) A = k a ic i A Proof Exercise Hint: use induction and theorem 212 Page 12

13 Note that this corollary is just generalizing parts a) and b) of theorem 212 Theorem (213) Let A be an m n matrix and B be an n p matrix For each j, (1 j p), let u j and v j denote the j-th columns of AB and B respectively Then a) u j = Av j b) v j = Be j Proof Exercise Note that this theorem just says that AB = [ Av 1 Av 2 Av n ] = [ ABe1 ABe 2 ABe n ] Theorem (214) Let V and W be finite-dimensional vector spaces having ordered bases β and γ respectively, and let T : V W be linear Then, for each u V, we have: [T (u)] γ = [T ] γ β [u] β Proof The idea of the proof is that Theorem 211 says that [T U] γ α = [T ]γ β [U]β α, so we look for an appropriate map U and an ordered basis α so that we can apply Theorem 211 We will also need a map g to make α appear as desired To this end We define the linear map f : F V by f(a) = au and g : F W by g(a) = at (u), for all a F We let α = {1} be the standard ordered basis for F Then g = T f and [T (u)] γ = [g(1)] γ = [g] γ α = [T f]γ α = [T ]γ β [f]β α = [T ]γ β [f(1)] β = [T ]γ β [u] β Next we make the bridge between matrices and linear maps more explicit We do this by giving the name L A to the linear map that corresponds to left multiplication by the matrix A Definition Leta A be an m n matrix with entries in the field F We denote by L A the mapping L A : F n F m defined by L A (x) = Ax for each column vector x F n We call L A a left-multiplication-transformation Theorem (215) Let A be an m n matrix with entries from F Then the left multiplication transformation L A : F n F m is linear Furthermore, if B is any other m n matrix (with entries in F ) and β and γ are the standard ordered bases for F n and F m respectively, then we have the following properties: Page 13

14 a) [L A ] γ β = A b) L A = L B if and only if A = B c) L A+B = L A + L B and L aa = al A for all a F d) If T : F n F m is linear, then there exists a unique m n matrix C such that T = L C In fact, C = [T ] γ β e) If E is an n p matrix, then L AE = L A L E f) If m = n, then L In = I F n Proof Parts a),b),d),e) are in the book so we will prove c) and f) proof of c): Let x F n Then and if a F, L A+B (x) = [L A+B ] γ β [x] β = (A + B) [x] β = A [x] β + B [x] β = [L A ] γ β [x] β + [L B] γ β [x] β = L A(x) + L B (x) L aa (x) = [L aa ] γ β [x] β = (aa) [x] β = a(a [x] β ) = a([l A] γ β [x] β ) = al A(x) proof of f): If m = n then L In : F n F n and for any x F n, L In (x) = [L In ] γ β [x] β = I n [x] β = [x] β = x We can use this correspondence between matrices and linear maps to carry our theorem about associativity of linear maps over to a theorem about associativity of matrix multiplication Theorem (216) Let A, B, and C be matrices such that A(BC) is defined Then (AB)C is also defined and A(BC) = (AB)C Proof Notice that for A(BC) to be defined these must be matrices over the same field and the number of rows of C must match the number of columns of B and the number of rows of B must match the number of columns of A These are the same conditions we need to ensure that (AB)C is defined So (AB)C is defined Associativity of the matrix product now follows from the associativity of function composition: L A(BC) = L A L BC = L A (L B L C ) = (L A L B )L C = L AB L C = L (AB)C Page 14

15 24 Invertibility and Isomorphisms Definition Let V and W be vector spaces, and let T : V W be linear A function U : W V is said to be the inverse of T if T U = I W and UT = I V If T has an inverse, T is said to be invertible The inverse is unique and is denoted by T 1 Uniqueness follows from this argument: Assume T U = I W and UT = I V and also T U = I W and U T = I V Then U = UI W = U(T U ) = (UT )U = I V U = U Facts: For all invertible functions T and U, 1 (T U) 1 = U 1 T 1 2 (T 1 ) 1 = T ; so in particular T 1 is invertible Luckily, the inverse of a linear map is also a linear map: Theorem (217) Let V and W be vector spaces, and let T : V W be linear and invertible Then T 1 : W V is linear Proof Let y 1, y 2 W and c F Since T is invertible, T is one-to-one and onto, hence there exist x 1, x 2 V such that T (x 1 ) = y 1, T (x 2 ) = y 2 and x 1 = T 1 (y 1 ), x 2 = T 1 (y 2 ) Then T 1 (cy 1 + y 2 ) =T 1 (ct (x 1 ) + T (x 2 )) = T 1 (T (cx 1 + x 2 )) = cx 1 + x 2 =ct 1 (y 1 ) + T 1 (y 2 ) Definition Let A be an n n matrix Then A is invertible if there exists an n n matrix B such that AB = BA = I Only square matrices are invertible This suggests that in the set of linear maps between finite dimensional vector spaces V and W, we should only find invertible maps if the dimensions of V and W match Lemma Let T be an invertible linear transformation from V to W Then V is finite-dimensional if and only if W is finite-dimensional In this case, dim(v ) = dim(w ) Proof Suppose V is finite-dimensional Then since T is one-to-one, N(T ) = {} so by the dimension theorem rank T = dim(v ) Since T is onto, dim(w ) = dim(r(t )) = dim(v ), and W is finite-dimensional On the other hand, if W is finite dimensional, making the above argument with T 1 instead shows that V is finite-dimensional Page 15

16 Fact: A function T is invertible if and only if T is one-to-one and onto Also note that Theorem 25 says the if V, W are vector spaces of equal and finite dimension and T : V W is linear, then tfae (the following are equivalent): a) T is one-to-one b) T is onto c) rank(t ) = dim(v ) Therefore, we can reinterpret the conclusion of Theorem 25 to read T is invertible if and only if rank(t ) = dim(v ) Theorem (218) Let V and W be finite-dimensional vector spaces with ordered bases β and γ respectively Let T : V W be linear Then T is invertible if and only if [T ] γ β is invertible Furthermore, [T 1 ] β γ = ([T ] γ β) 1 Proof Suppose T is invertible Then V and W have the same dimension, call it n So [T ] γ β is an n n matrix and T T 1 = I W and T 1 T = I V Therefore, for I n the n n identity matrix, I n = [I V ] β = [ T 1 T ] β = [ T 1 ] β γ [T ]γ β and similiarly [T ] γ β [T 1 ] β γ = I n So [T ] γ β is invertible with inverse [T 1 ] β γ Suppose A = [T ] γ β is invertible Then there exists an n n matrix B such that AB = BA = I n By theorem 26 there exists U L(W, V ) such that U(w j ) = B ij v i for j = 1, 2,, n, where γ = {w 1, w 2,, w n } and β = {v 1, v 2,, v n } It follows that [U] β γ = B Furthermore [UT ] β = [U] β γ [T ]γ β = BA = I n = [I V ] β So UT = I V and similarly T U = I W Question: Why didn t we choose U = L B in the proof above? Corollary (1) Let V be a finite-dimensional vector space with an ordered basis β, and let T : V V be linear Then T is invertible if and only if [T ] β is invertible Furthermore, [T 1 ] β = ( [T ] β ) 1 Corollary (2) Let A be an n n matrix Then A is invertible if and only if L A is invertible Furthermore, (L 1 ) 1 = L A 1 Page 16

17 Definitions Let V and W be vector spaces We say V is isomorphic to W if there exists an invertible linear transformation T : V W Such a linear transformation is called an isomorphism from V onto W Isomorphism is an equivalence relation Essentially, if two vector spaces V and W are isomorphic, then as vectors, the vectors in V are just the vectors in W just given different names, and vice-versa Examples The following linear maps are isomorphisms [ ] a b 1 T : F 3 M 2 2 (F ), T (a, b, c, d) = c d [ ] a b 2 T : M 2 2 (F ) P 3 (F ), T : a + bx + cx c d 2 + dx 3 [ ] f(2) f(4) 3 T : P 3 (R) M 2 2 (R), T (f) = f(6) f(8) At this point, we may wonder, for each positive integer n, is each vector space of dimension n isomorphic to all other vector spaces of dimension n over the same field? Theorem (219) Let V and W be finite-dimensional vetor spaces (over the same field) Then V is isomorphic to W if and only if dim V = dim W Proof Suppose V is isomorphic to W Then there exists an isomorphism T : V W Since T is invertible, by the lemma above dim(v ) = dim(w ) On the other hand, if dim(v ) = dim(w ) then let their common dimension be n Then let β = {v 1, v 2,, v n } be a basis for V and γ = {w 1, w 2,, w n } be a basis for W, respectively We claim the linear map T such that T (v i ) = w i for i = 1, 2,, n is an isomorphism Firstly, T ( n a iv i ) = n a it (v i ) = n a iw i = and since γ is a basis for W, we see that N(T ) = {} On the other hand R(T ) = span(t (β)) = span(γ) = W Since T is one-to-one and onto, T is invertible Therefore T is an isomorphism Corollary Let V be a vector space over F Then V is isomorphic to F n if and only if dim(v ) = n So in terms of just the vector space structure, every finite dimensional vector space V with dimension n is a re-labeling of F n Page 17

18 Theorem (22) Let V and W be finite-dimensional vector spaces over F of dimensions m and n, respectively, and let β and γ be ordered bases for V and W respectively Then the function Φ : L(V, W ) M m n (F ), defined by Φ(T ) = [T ] γ β for T L(V, W ), is an isomorphism Proof By Theorem 28, Φ is linear (ie [cu + T ] γ β = c [U]γ β +[T ]γ β ) Thus we need to show T is one-to-one and onto Let β = {v 1, v 2,, v n } and γ = {w 1, w 2,, w m } To show that T is onto, we can show that for each A M m n (F ), there is a map T L(V, W ) such that Φ(T ) = A Well, by Theorem 26, there is a unique map T defined by the two properties 1) T is linear and 2) T (v j ) = m A ijw i, for 1 j n The second property says that A = [T ] γ β Since this map is T is unique, we have also shown that Φ is one-to-one Now that the relationship between matrices and linear maps is becoming clearer, we rename the transformation x [x] β Definition Let β be an ordered basis for an n-dimensional vector space V over a field F The standard representation of V with respect to β is the function φ β, φ β : V F n defined by φ β (x) = [x] β for every x V Theorem (221) For any finite-dimensional vector space V with ordered basis β, φ β is an isomorphism Proof Exercise 12 of section 24 With Theorem 221 in hand we can construct the following commutative diagram: V T W φ β φ γ F n L A F m By commutative diagram we mean that you can start with any x V in the top left and either apply T and then φ γ or first apply φ β and then L A, and either way the result is the same, ie φ γ (T (x)) = L A (φ β (x)) Page 18

19 * 25 The Change of Coordinate Matrix In calculus 2 we claim that the level sets of degree two polynomials in two variables with coefficients in R give conic sections For example, the set 4x 2 + 9y 2 = 36 is easily seen to be a parabola (for example, use the parameterization x = 3 cos t, y = 2 sin t But what about something like 2x 2 4xy + 5y 2 = 1? Well, using the coordinates x, y defined by x = 2 5 x 1 5 y y = 1 5 x y we turn the equation 2x 2 4xy + 5y 2 = 1 into the equation (x ) 2 + 6(y ) 2 = 1, which is much more familiar Geometrically, we have rotated our frame of reference In other words, we have changed the basis of unit vectors whose linear combinations we use to express any point P in R 2 The unit vectors along the x -axis and y -axis form the ordered basis { ( ) ( )} 1 2 β 1 1 = 5, {( 1 In other words, if we let β be the standard ordered basis ( ) ( x x changed variables from [P ] β = to [P ] y β = y new coordinates are related by the equation ( ) x = 1 ( ) ( ) 2 1 x y y The matrix Q = 1 ( ) is exactly [I] β β, and we have [v] β = Q [v] β, for all v R 2 ) (, 1 )}, we have ) Notice that the old and Theorem (222) Let β and β be two ordered bases for a finite-dimensional vector space V, and let Q = [I V ] β β Then a) Q is invertible b) For any v V, [v] β = Q [v] β Proof a) Since the identity map I V : V V is invertible, so is Q by theorem 218 b) [v] β = [I V (v)] β = [I V ] β β [v] β = Q [v] β Page 19

20 With this theorem in hand we are ready to change coordinates at will We call the matrix Q = [I V ] β β a change of coordinates matrix In particular we say that Q changes β -coordinates into β-coordinates For the remainder of this section we consider only linear maps from V into itself We call such maps linear operators on V Given a linear operator T : V V, with ordered bases β and β of V, we know we can represent T via both [T ] β and [T ] β How are these matrices related? Theorem (223) Let T be a linear operator on a finite dimensioinal vector space V, and let β and β be ordered bases for V Suppose that Q is the change of coordinate matrix that changes β -coordinates into β-coordinates Then Proof We show Q [T ] β = [T ] β Q Let I = I V Then T = IT = T I, so [T ] β = Q 1 [T ] β Q Q [T ] β = [I] β β [T ] β β = [IT ] β β = [T I] β β = [T ] β β [I]β β = [T ] β Q Example Let T be the linear operator on R 2 defined by ( ) ( ) a 2a + b T = b a 3b {( ) ( )} {( ) ( )} and let β =,, β =, ( ) ( ) ( ) Then since T (e 1 ) = and T (e 1 2 ) =, we see that [T ] 3 β = 1 3 We can also see that the change of coordinate matrix that changes β -coordinates into β-coordinates is Q = ( ) Now suppose we are handed the fact that ( Then by Theorem 223, we get: ( 2 1 [T ] β = Q 1 [T ] β Q = 1 1 ) 1 = ( ) ( ) ) ( Did we get the right matrix? Well, the first column of [T ] β is ( ) ( ) ( ) that for v =, [T (v)] 1 β = [T ] β =, ie 5 ) ( ) 8 13 = 5 9 ( 8 5 ), so we claim Page 2

21 ( 1 T 1 out too ) ( 1 = 8 1 ) Similarly, we claim T ( ) ( ) 1 2 ( ) 3 = This checks out 2 ( ) ( ) 1 1 = 13 9 = 1 2 ( 4 5 ) This checks Since our map is clearly linear (as it is given by matrix multiplication) and agrees with T on a basis for the domain (R 2 ) of T, our map is the desired map Example 3 starting on page 113 is a good advertisement for this section, you should work through this example Corollary Let A M n n (F ), and let γ be an ordered basis for F n Then [L A ] γ = Q 1 AQ, where Q is the n n matrix whose j-th column is the j-th vector of γ Definition Let A and B be matrices in M m n (F ) We say B is similar to A if there exists and invertible matrix Q such that B = Q 1 AQ Note that similarity is an equivalence relation (see exercise 9 in this section) Things we need from chapters 3 and 4 Definitions Let A be an m n matrix Any of the following three operations on the rows[columns] of A is called an elementary row[column] operation: 1 interchanging any two rows[columns] of A 2 multiplying any row[column] of A by a nonzero scalar 3 adding a scalar multiple of a row[column] of A to another row[column] Any of these three operations is called an elementary operation Elementary operations are of type 1, type 2, or type 3 according to the list above We also define elementary matrices Definition An n n elementary matrix is a matrix obtained by performing an elementary operation on I n The elementary matrix is said to by of type 1, type 2 or type 3 according to which type of elementary operation was applied to I n Example The matrix is an elementary matrix of type 1 The next theorem connects elementary matrices to elementary operations: Page 21

22 Theorem (31) Let A M m n (F ), and suppose that B is obtained from A by performing an elementary row[column] operation Then there exists an m m [n n] elementary matrix E such that B = EA[B = AE] In fact, E is obtained from I m [I n ] by performing the same elementary row[column] operation as that was performed on A to obtain B Conversely, if E is an elementary m m [n n] matrix, then EA[AE] is the matrix obtained from A by performing the same elementary row[column] operation that produces E from I m [I n ] To prove this theorem you have to verify it for each operation, which is super tedious We ll skip it The point is, you can produce row operations by left multiplication by an elemenatry matrix and you can produce column operations by right multiplication by an elementary matrix Theorem (32) Elementary matrices are invertible, and the inverse of an elementary matrix is an elementary matrix of the same type Proof Let E be an elementary n n matrix Then E can be obtained by an elementary row operation on I n By reversing the steps applied to transform I n into E, we can transform E back into I n So I n can be obtained from E by an elementary row operation of the same type By Theorem 31, there is an elementary matrix Ẽ such that ẼE = I n Therefore, E is invertible and E 1 = Ẽ Definition If A M m n (F ), we define the rank of A, denoted rank(a) to the the rank of the linear map L A : F n F m Note that an n n matrix is invertible if and only if its rank is n Theorem (34) Let A be an m n matrix If P and Q are invertible m m and n n matrices, respectively, then a) rank(aq) = rank(a) b) rank(p A) = rank(a) and therefore b) rank(p AQ) = rank(a) Proof R(L AQ ) = R(L A L Q ) = L A L Q (F n ) = L A (L Q (F n )) = L A (F n ) = R(L A ) since L Q is onto So rank(aq) = dim(r(l AQ )) = dim(r(l A )) = rank(a) The proof of b) is similar With a) and b) in hand, we have rank(p AQ) = rank(aq) = rank(a) Page 22

23 Corollary Elementary row and column operations on a matrix are rank-preserving Theorem (36) Let A be an m n matrix of rank r Then r m, r n, and, by means of a finite number of elementary row and column operations, A can be transformed into the matrix ( ) Ir D = 1, 2 3 where 1, 2, and 3 are zero matrices So D ii = 1 for i r and D ij = otherwise The proof of this theorem is not hard but it is long details See pages for Corollary Every invertible matrix is a product of elementary matrices Proof If A is an invertible n n matrix, then rank(a) = n Hence the matrix D is I n and there exists invertible matrices (compositions of elementary matrices corresponding to row and column operations) B and C such that I n = BAC Let B = E p E p 1 E 1, C = G 1 G 2 G q Then A = B 1 I n C 1 = B 1 C 1, that is, A = E 1 1 E 1 2 E 1 p G 1 q G 1 q 1 G 1 1 Now we are ready to talk about determinants ( ) a b Definition If A = is a 2 2 matrix over a field F, we define the c d determinant of A, denoted by det(a) or A to be the scalar (ad bc) Definition If A M n n (F ), we define the determinant of A as follows: 1 If n = 1, A = A 11 and det(a) = A 11 2 if n 2 we define det(a) recursively : det(a) = ( 1) j+1 A 1j det(ã1j) i=j where Ãij is the (n 1) (n 1) matrix obtained by removing the i-th row and j-th column of A We call the scalar ( 1) n+1 det(ã1j) the cofactor of A ij We call the formula above cofactor expansion along the first row of A In fact you can compute a determinant by cofactor expansion along any row: Theorem Let A M n n (F ) Then for any i {1, 2,, n}, det(a) = n j=1 ( 1)j+i A ij det(ãij) Page 23

24 Adapted from Robert A Beezer s book A First Course in Linear Algebra We use induction The n = 1 case is trivial When n = 2 we just compute expansion along the second row: det(a) = ( 1) 2+1 A 12 det(a 21 ) + ( 1) 2+2 A 22 det(a 11 ) = A 11 A 22 A 21 A 12 For the general case we ll introduce some notation Let A ij,kl denote the matrix obtained from A be deleting the i-th and j-th rows and k-th and l-th columns As we sum up to a missing column, { our index will bump up by one To make l < j this easier to deal with, we define ɛ lj = Let i {2, 3,, n} 1 l j Now: det(a) = = = = = = = ( 1) j+1 A 1j det(ã1j) j=1 ( 1) j+1 A 1j ( 1) i 1+l ɛ lj A il det(a 1i,jl ) j=1 l j ( 1) j+i+l ɛ lj A 1j A il det(a 1i,jl ) j=1 l j ( 1) j+i+l ɛ lj A 1j A il det(a 1i,jl ) l=1 j l ( 1) i+l A il ( 1) j ɛ lj A 1j det(a 1i,jl ) l=1 j l ( 1) i+l A il ( 1) ɛlj+j A 1j det(a i1,lj ) l=1 j l ( 1) i+l A il det(ãil) l=1 Theorem (43) The determinant of an n n matrix is a linear function of each row then the remaining rows are held fixed In other words, for 1 r n we have: det a 1 a r 1 u + kv = det a r+1 a n a 1 a r 1 u a r+1 a n + k det a 1 a r 1 v a r+1 a n Page 24

25 Proof We induct on n The n = 1 case: det(u + kv) = u + kv = det(u) + k det(v) Now assume n 2 and the theorem holds for the n 1 case Let A be an n n matrix with rows a 1, a 2,, a n, respectively, and suppose that for some r, 1 r n, we have a r = u + kv, for some u, v F n and k F Let u = (b 1, b 2,, b n ) and v = (c 1, c 2,, c n ) Let B and C denote the matrices obtained from A by replacing row r of A by u and v respectively Then we want to show det(a) = det(b) + k det(c) Firstly, if r = 1, we get det(a) = = = ( 1) j+1 A 1j det(ã1j) = ( 1) n+1 (b j ) det(ã1j) + j=1 ( 1) n+1 (b j + kc j ) det(ã1j) j=1 ( 1) n+1 (b j ) det(ã1j) + k j=1 = det(b) + k det(c) ( 1) n+1 (kc j ) det(ã1j) j=1 ( 1) n+1 (c j ) det(ã1j) Now for r > 1 and 1 j n the rows of Ã1j, B 1j, C 1j are the same except for row r 1 Row r 1 of Ã1j is: (b 1 + kc 1, b 2 + kc 2,, b j 1 + kc j 1, b j+1 + kc j+1,, b n + kc n ) which is the sum of row r 1 of B 1j and k times row r 1 of C 1j By the induction hypothesis, det(ãij) = det( B 1j ) + k det( C 1j ) Since A 1j = B 1j = C 1j for each j, we finally have: det(a) = = = ( 1) j+1 A 1j det(ã1j) j=1 ( 1) j+1 A 1j j=1 j=1 ( det( B 1j ) + k det( C ) 1j ) ( 1) j+1 A 1j det( B 1j ) + k j=1 = det(b) + k det(c) ( 1) j+1 A 1j det( C 1j ) j=1 Corollary If A M n n (F ) has a row of all zeros, then det(a) = Proof Suppose every entry of the r-th row of A M n n (F ) is zero Then the r-th row of A is equal to the r-th row of A plus itself By theorem 43, det(a) = det(a) + det(a), which is only possible if det(a) = Page 25

26 51 Eigenvalues and Eigenvectors Definitions A linear operator T on a finite-dimensional vector space V is called diagonalizable if there is an ordered basis β for V such that [T ] β is a diagonal matrix A square matrix A is called diagonalizable if L A is diagonalizable Definitions Let T be a linear operator on a vector space V A nonzero vector v V is called an eigenvector of T if there exists a scalar λ such that T (v) = λv The scalar λ is called the eigenvalue corresponding to the eigenvector v Let A M n n (F ) A nonzero vector v F n is called an eigenvector of A if v is an eigenvector of L A, ie Av = λv for some scalar λ The scalar λ is called an eigenvalue of A corresponding to the eigenvector v Theorem (51) A linear operator T on a finite-dimensional vector space V is diagonalizable if and only if there exists an ordered basis β for V consisting of eigenvectors of T Furthermore, if T is diagonalizable, β = {v 1, v 2,, v n } is an ordered basis of eigenvectors of T, and D = [T ] β, then D is a diagonal matrix and D jj is the eigenvalue corresponding to v j for 1 j n Proof A direct consequence of the definitions above To diagonalize a matrix or linear operator is to find a basis of eigenvectors and corresponding eigenvalues A few more tidbits from chapter 4 We begin with a corollary to Theorem 36 Recall that Theorem 36 states: Let A be an m n matrix of rank r Then r m, r n, and, by means of a finite number of elementary row and column operations, A can be transformed into the matrix ( ) Ir D = 1, 2 3 where 1, 2, and 3 are zero matrices So D ii = 1 for i r and D ij = otherwise The corollary is: Corollary Let A be an m n matrix Then rank(a t ) = rank(a) Proof A consequence of Theorem 36 is that there exist invertible matrices (products of elementary matrices) B and C such that D = BAC for D as in Theorem 36 Now D t = (BAC) t = C t A t B t and since B and C are invertible, so are their transposes (by homework exercise 5 of section 24) Thus rank(a t ) = rank(c t A t B t ) = rank(d t ) Page 26

27 But D t has the same rank as D, and rank(d) = rank(a) Some corollaries of Theorem that a determinant can be computed by cofactor expansion on any row 1 If A M n n has two identical rows, then det(a) = : Proof By induction The 2 2 case is clear Otherwise, n 3 and suppose row r of A is the same as row s of A, r s The we can take the determinant of A by cofactor expansion on a row that is not row r or row s, say row i Then det(a) = ( 1) i+j A ij det(ãij) j=1 and by the induction hypotheses, each det(ãij) = for each j {1, 2,, n} 2 If A is a triangular matrix, then det(a) is the product of the diagonal entries Proof (Really a sketch of the proof): Suppose T is upper triangular Use induction and evaluate the determinant by successively cofactor expanding first on the last row, then the second to last row, and so on Theorem If A M n n (F ) and B is a matrix obtained from A by interchanging two rows of A, then det(b) = det(a) Proof Let A M n n (F ) have rows a 1, a 2,, a n, and suppose B M n n is A but with rows r and s switched (say r < s) Then A = a 1 a r a s a n, B = a 1 a s a r a n Page 27

28 Now = det a 1 a r + a s a r + a s a n = det a 1 a r a r + a s a n + det a 1 a s a r + a s a n = det a 1 a r a r a n + det a 1 a r a s a n + det a 1 a s a r a n + det a 1 a s a s a n = + det(a) + det(b) + Very similarly we can prove Theorem If A M n n (F ) and B is obtained by adding a multiple of one row of A to another row of A, then det(a) = det(b) Proof Using the same terminology as before: det a 1 a r + ka s a s a n = det a 1 a r a s a n + k det a 1 a s a s a n Page 28

29 that is, det(b) = det(a) + = det(a) Corollary If A M n n (F ) has rank less than n, then det(a) = Proof If A has rank less than n, then the rows of A are linearly dependent because the columns of A t are linearly dependent, and these columns as column vectors span R(L A ) So we may write one row as a linear combination of the others Using linearity of the determinant on this row, we can write det(a) as a sum of scalar multiples of determinants of matrices each of which has at least one row duplicated (say row i equals row j) These determinants are all zero thus det(a) is zero Finally we can prove: Theorem (47) For any A, B M n n (F ), det(ab) = det(a) det(b) Proof Firstly we establish the result for elementary matrices For example, if A is obtained by multiplying the r-th row of I by the nonzero scalar k, then A is diagonal with all diagonal entries one expect for the entry in the r-th row, so det(a) = k det(i) If AB is the matrix obtained by multiplying the r-th row of B by k, then det(ab) = k det(b) Similarly one shows the result for type 1 or type 3 matrices If A is an n n matrix with rank less than n, then det(a) = So the result holds in this case In the remaing case, A has rank n and so is invertible and hence is a product of elementary matrices Therefore, A = E m E 2 E 1 and thus det(ab) = det(e m E 2 E 1 B) = det(e m ) det(e m 1 E 2 E 1 B) = det(e m ) det(e m 1 ) det(e m 2 E 2 E 1 B) = det(e m ) det(e m 1 ) det(e 1 ) det(b) = det(e m E 2 E 1 ) det(b) = det(a) det(b) Corollary A matrix A M n n (F ) is invertible if and only if det(a) Furthermore, if A is invertible, then det(a 1 1 ) = det(a) Proof ) We show the contrapositive: If A is not invertible, then the rank of A is less than n, so det(a) = ) On the other hand, if A is invertible, and det(a) det(a) det(a 1 ) = det(aa 1 ) = det(i) = 1 det(a 1 ) = 1 det(a) Page 29

30 Theorem (48) For any A M n n (F ), det(a t ) = det(a) Proof If A is not invertible, then rank(a) < n but rank(a) = rank(a t ) so A t is not invertible either On the other hand, if A is invertible, it is a product of elementary matrices Each elementary matrix E satisfies det(e) = det(e t ) (you can check this) So A = E m E 2 E 1 and thus det(a t ) = det(e t 1E t 2 E t m) = det(e t 1) det(e t 2) det(e t m) = det(e 1 ) det(e 2 ) det(e m ) = det(e m ) det(e m 1 ) det(e 1 ) = det(e m E m 1 E 1 ) = det(a) Back to chapter 5: ( ) ( ) 5 1 Notice that A = has eigenvector x = with corresponding eigenvalue λ = 5 because Ax = λx However a matrix (or linear operator) need not have any eigenvalues: ( ) 1 The matrix A = is [T ] 1 β in the standard ordered basis where T is rotation by the angle π, thus neither T nor A have any real eigenvalues 2 Theorem (52) Let A M n n (F ) Then a scalar λ is an eigenvalue of A if and only if det(a λi n ) = Proof If λ is an eigenvalue of A, then for some nonzero vector x V, Ax = λx, ie (A λi n )x = Thus A λi n is not invertible, so det(a λi n ) = On the other hand, if det(a λi n ) =, then A λi n is not invertible, so there exists a nonzero vector x N(A λi n ), ie λ is an eigenvalue of A Definition Let A M n n (F ) The polynomial f(t) = det(a ti n ) is called the characteristic polynomial of A ( ) 5 Looking back at A = we see that the charateristic polynomial of A is: 4 1 ( ) 5 t det(a ti) = det = t 2 6t + 5 = (t 5)(t 1) 4 1 t It turns out (exercise 12) that similar matrices have the same characteristic polynomial For this reason we define characteristic polynomials for linear operators as follows: Page 3

31 Definition Let T be a linear operator on an n-dimensional vector space V with ordered basis β We define the characteristic polynomial f(t) of T to be the characteristic polynomial of [T ] β Using induction we can easily prove: Theorem (53) Let A M n n (F ) Then a) The characteristic polynomial of A is a polynomial of degree n with leading coefficient ( 1) n b) A has at most n distinct eigenvalues This theorem tells us how to find the eigenvector(s) corresponding to an eigenvalue: Theorem (54) Let T be a linear operator on a vector space V, and let λ be an eigenvalue of T A vector v V is an eigenvector of T corresponding to λ if and only if v and v N(T λi) Proof Exercise Recall the commutative diagram for Theorem 221: V T W φ β φ γ F n L A F m In the special case where V = W, β = γ, we get: V T W φ β φ β F n L A F m Page 31

32 Now suppose V is a finite dimensional vector space and T : V V is a linear operator with eigenvector v V and corresponding eigenvalue λ In other words, assume T (v) = λv Choose an ordered basis β for V and let A = [T ] β Then using the diagram immediately above we see Aφ β (v) = L A (φ β (v)) = φ β (T (v)) = φ β (λv) = λφ β (v) That is, the matrix A has eigenvector φ β (v) with corresponding eigenvalue λ Since the φ β s are isomorphisms, we can reverse the argument and show that if φ β (v) is an eigenvector of the matrix A with corresponding eigenvalue λ, then v is an eigenvector of T with corresponding eigenvalue λ The point is, we can find the eigenvalues and eigenvectors of T by finding the eigenvalues and eigenvectors of A = [T ] β, for any (ordered) basis β of V 2 3 Example Let A = and F = R Let s find the eigenvalues (if any) of A, and the corresponding eigenvectors To find the eigenvalues, we find the characteristic polynomial of A and then find it s zeroes, which are the eigenvalues of A: det(a λi) = det λ λ λ = (λ 3)(λ 2)(λ 1) = λ 3 + 6λ 2 11λ + 6 Now let s find an eigenvector corresponding to λ = 1 To accomplish this we find the kernel of L A 1I : A 1I = *We take the rref of A 1I and get: This matrix tells us that if nonzero vector in span eigenvalue λ = 1 a b c N(L A 1I ) then a = b = c Accordingly, any is an eigenvector of A corresponding to the Page 32

33 1 Similarly we can find that is a basis for the kernel of A 3I and is a basis for the kernel of A 2I These three vectors form a basis for R 3, so L A is diagonalizable and in the ordered basis β = 1,, [L A ] β = 2 3 Note (*) rref-ing A 1I is applying a finite sequence of row operations to A 1I, ie multiplying A 1I on the left by an product of elementary matrices, which is a product of invertible matrices, ie which is an invertible matrix, call it E In other words, if rref(a 1I) = B, then B = E(A 1I), so if (A 1I)x = then E(A 1I)x = and on the other hand, if E(A 1I)x = we have (A 1I)x = E 1 = The point is, (A 1I) and it s rref have the same kernel as left multiplication transformations, so we are justified in this procedure, 52 Diagonalizability Theorem (55) Let T be a linear operator on a vector space V, and let λ 1, λ 2,, λ k be distinct eigenvalues of T If v 1, v 2, v k are eigenvectors of T such that λ i corresponds to v i, (1 i k), then {v 1, v 2,, v k } is linearly independent Proof We use induction on k If k = 1, {v 1 } is a set of one nonzero vector and is therefore linearly independent Now assume the theorem holds for k 1 distinct eigenvalues, where k 1 1, and we have k distinct eigenvalues {λ 1, λ 2,, λ k } with corresponding eigenvectors {v 1, v 2,, v k } We shall prove the contrapositive for the n = k case, that is, assume that {v 1, v 2, v k } is linearly dependent Then by the inductive assumption, this means we can write v k as a linear combination of v 1, v 2,, v k 1 Say v k = k 1 a iv i Then Which gives k 1 k 1 a i v i = λ k v k = Av k = A( a i v i ) = a i λ i v i λ k k 1 k 1 a i (λ i λ k )v i = Page 33

34 Since the v i s are linearly independent for 1 i k 1, this is only possible if a i (λ i λ k ) = for i = 1, 2,, k 1 The λ s are all distinct by hypothesis, so we conclude that the a i s are all zero, which says that v k is zero, which contradicts the assumption that v k is an eigenvector We conclude that the set {v 1, v 2,, v k } is linearly independent Corollary Let T be a linear operator on an n-dimensional vector space V If T has n distinct eigenvalues, then T is diagonalizable This if is not an if and only if For example, any diagonal matrix with a repeated entry on the diagonal is both diagonalizable and does not have distinct eigenvalues More concretely, such an example would be the identity matrix Definition A polynomial f(t) in P (F ) splits over F if there are scalars c, a 1, a 2,, a n such that f(t) = c(t a 1 )(t a 2 ) (t a n ) So, for example, t 2 2 splits over R but not over Q, the field of rationals Similarly, t splits over C but not over R If f(t) is a characteristic polynomial of a linear operator or matrix over a field F, then the statement f(t) splits is understood to mean f(t) splits over F Theorem (56) The characteristic polynomial of any diagonalizable linear operator splits Proof If T is a diagonalizable linear operator on an n-dimensional vector space V, then there exists a basis β = {v 1, v 2,, v n } of V composed of eigenvectors, with corresponding eigenvalues λ 1, λ 2,, λ n Then [T ] β = λ 1 λ 2 λ n and [T ] β ti = λ 1 t λ 2 t λ n t so det([t ] β ti) = (λ 1 t)(λ 2 t) (λ n t) = ( 1) n (t λ 1 )(t λ 2 ) (t λ n ) On the other hand, If the characteristic polynomial T splits, T need not be diagonalizable Definition Let λ be an eigenvalue of a linear operator or matrix with a characteristic polynomial f(t) The (algebraic) multiplicity of λ is the largest positive integer k such that (t λ) k is a factor of f(t) 1 For example, If A = 2 3 then the characteristic polynomial of A is 5 3 f(t) = (t 1)(t 3) 2 So λ = 1 is an eigenvalue of A with multiplicity 1 and λ = 3 Page 34