CLASS NOTES ON LINEAR ALGEBRA

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1 CLASS NOTES ON LINEAR ALGEBRA STEVEN DALE CUTKOSKY 1. Matrices Suppose that F is a field. F n will denote the set of n 1 column vectors with coefficients in F, and F m will denote the set of 1 m row vectors with coefficients in F. M m,n (F ) will denote the set of m n matrices A = (a ij ) with coefficients in F. We will sometimes denote the zero vector of F n by 0 n, the zero vector of F m by 0 m and the zero vector of M m,n by 0 m,n. We will also sometimes abbreviate these different zeros as 0 or 0. We will let e i F n be the column vector whose entries are all zero, except for a 1 in the i-the row. If A M m,n (F ) then A t M n,m (F ) will denote the transpose of A. We will write A = (A 1, A 2,..., A n ) where A i F m are the columns of A, and A = where A j F n are the rows of A. For x = (x 1,..., x n ) t F n, we have the formula A 1 A 2. A m Ax = x 1 A 1 + x 2 A x n A n. In particular, Ae i = A i for 1 i n and (e i ) t Ae j = a ij for 1 i m, 1 j n. We also have A 1 x A 2 x Ax =. A m x = (A 1 ) t x (A 2 ) t x. (A m ) t x where for x = (x 1,..., x n ) t, y = (y 1,..., y n ) F n, x y = x 1 y 1 + x 2 y x n y n is the dot product on F n. Lemma 1.1. The following are equivalent for two matrices A, B M m,n (F ). 1. A = B. 2. Ax = Bx for all x F n. 3. Ae i = Be i for 1 i n. N will denote the natural numbers {0, 1, 2...}. Z + is the set of positive integers {1, 2, 3,...}. 1

2 2. Span, Linear Dependence, Basis and Dimension In this section we suppose that V is a vector space over a field F. We will denote the zero vector in V by 0 V. Sometimes, we will abbreviate, and write the zero vector as 0 or 0. Definition 2.1. Suppose that S is a nonempty subset of V. Then Span(S) = {c 1 v c n v n n Z +, v 1,..., v n S and c 1,..., c n F }. This definition is valid even when S is an infinite set. We define Span{ } = { 0}. Lemma 2.2. Suppose that S is a subset of V. Then Span(S) is a subspace of V. Definition 2.3. Suppose that S is a subset of V. S is a linearly dependent set if there exists n Z +, v 1,..., v n S and c 1,..., c n F such that c 1,..., c n are not all zero, and c 1 v 1 + c 2 v c n v n = 0. S is a linearly independent set if it is not linearly dependent. Observe that is a linearly independent set. Definition 2.4. Suppose that S is subset of V which is a linearly independent set and Span(S) = V. Then S is a basis of V. Theorem 2.5. Suppose that V is a vector space over a field F. Then V has a basis. We have that the empty set is a basis of the 0 vector space { 0}. Theorem 2.6. (Extension Theorem) Suppose that V is a vector space over a field F and S is a subset of V which is a linearly independent set. Then there exists a basis of V which contains S. Theorem 2.7. Suppose that V is a vector space over a field F and S 1 and S 2 are two bases of V. Then S 1 and S 2 have the same cardinality. This theorem allows us to make the following definition. Definition 2.8. Suppose that V is a vector space over a field F. Then the dimension of V is the cardinality of a basis of V. The dimension of { 0} is 0. Definition 2.9. V is called a finite dimensional vector space if V has a finite basis. For the most part, we will consider finite dimensional vector spaces. If V is finite dimensional of dimension n, a basis is considered as an ordered set β = {v 1,..., v n }. Lemma Suppose that V is a finite dimensional vector space and W is a subspace of V. Then dim W dim V, and dim W = dim V if and only if V = W. Lemma Suppose that V is a finite dimensional vector space and β = {v 1, v 2,..., v n } is a basis of V. Suppose that v V. Then v has a unique expression with c 1,..., c n F. v = c 1 v 1 + c 2 v c n v n Lemma Let {v 1,..., v n } be a set of generators of a vector space V. Let {v 1,..., v r } be a maximal subset of linearly independent elements. Then {v 1,..., v r } is a basis of V. 2

3 3. Direct Sums Suppose that V is a vector space over a field F, and W 1, W 2,..., W m are subspaces of V. Then the sum W W m = Span(W 1 W m ) = {w 1 + w w m w i W i for 1 i m} is the subspace of V spanned by W 1, W 2,..., W m. The sum W W m is called a direct sum, denoted by W 1 W2 Wm, if every element v W 1 + W W m has a unique expression v = w w m with w i W i for 1 i m. We have the following useful criterion. Lemma 3.1. The sum W 1 + W W m is a direct sum if and only if 0 = w 1 + w w m with w i W i for 1 i m implies w i = 0 for 1 i m. In the case when V is finite dimensional, the direct sum is characterized by the following equivalent conditions. Lemma 3.2. Suppose that V is a finite dimensional vector space. Let n = dim (W W m ) and n i = dim W i for 1 i m. The following conditions are equivalent 1. If v W 1 + W W m then v has a unique expression v = w w m with w i W i for 1 i m. 2. Suppose that β i = {w i,1,..., w i,ni } are bases of W i for 1 i m. Then β = {w 1,1,..., w 1,n1, w 2,1,..., w m,nm } is a basis of W W m. 3. n = n n m. The following lemma gives a useful criterion. Lemma 3.3. Suppose that U and W are subspaces of a vector space V. Then V = U W if and only if V = U + W and U W = { 0}. 4. Linear Maps In this section, we will suppose that all vector spaces are over a fixed field F. We recall the definition of equality of maps, which we will use repeatedly to show that two maps are the same. Suppose that f : V W and g : V W are maps (functions). Then f = g if and only if f(v) = g(v) for all v V. If f : U V and g : V W, we will usually write the composition g f : U W as gf. If x U, we will sometimes write fx for f(x), and gfx for (g f)(x). Definition 4.1. Suppose that V and W are vector spaces. A map L : V W is linear if L(v 1 + v 2 ) = L(v 1 ) + L(v 2 ) for v 1, v 2 V and L(cv) = cl(v) for v V and c F. Lemma 4.2. Suppose that L : V W is a linear map. Then 1. L(0 V ) = 0 W. 2. L( v) = L(v) for v V. Lemma 4.3. Suppose that L : V W is a linear map and T : W U is a linear map. Then the composition T L : V U is a linear map. If S and T are sets, a map f : S T is 1-1 and onto if and only if there exists a map g : T S such that g f = I S and f g = I T, where I S is the identity map of S and I T is 3

4 the identity map of T. When this happens, g is uniquely determined. We write g = f 1 and say that g is the inverse of f. Definition 4.4. Suppose that L : V W is a linear map. L is an isomorphism if there exists a linear map T : W V such that T L = I V is the identity map of V, and LT = I W is the identity map of W. The map T in the above definition is unique, if it exists, by the following lemma. If L is an isomorphism, we write L 1 = T, and call T the inverse of L. Lemma 4.5. Suppose that L : V W is a map. Suppose that T : W V and S : W V are maps which satisfy T L = I V, LT = I W and SL = I V, LS = I W. Then S = T. Lemma 4.6. a linear map L : V W is an isomorphism if and only if L is 1-1 and onto. Lemma 4.7. Suppose that L : V W is a linear map. Then 1. Image L = {L(v) v V } is a subspace of W. 2. Kernel L = {v V L(v) = 0} is a subspace of V. Lemma 4.8. Suppose that L : V W is a linear map. Then L is 1-1 if and only if Kernel L = {0}. Definition 4.9. Suppose that F is a field, and V, W are vector spaces over F. L F (V, W ) be the set of linear maps from V to W. We will sometimes write L(V, W ) = L F (V, W ) if the field is understood to be F. Lemma Suppose that F is a field, and V, W are vector spaces over F. L F (V, W ) is a vector space. Let Then Theorem (Dimension Theorem) Suppose that L : V W is a homomorphism of finite dimensional vector spaces. Then dim Kernel L + dim Image L = dim V. Proof. Let {v 1,..., v r } be a basis of Kernel L. Extend this to a basis {v 1,..., v r, v r+1,..., v n } of V. Let w i = L(v i ) for r + 1 i n. We will show that {w r+1,..., w n } is a basis of Image L. Since L is linear, {w r+1,..., w n } span Image L. We must show that {w r+1,..., w n } are linearly independent. Suppose that we have a relation for some c r+1,..., c n F. Then c r+1 w r c n w n = 0 L(c r+1 v r c n v n ) = c r+1 L(v r+1 ) + + c n L(v n ) = c r+1 w r c n w n = 0. Thus c r+1 v r+1 + c n v n Kernel L, so that we have an expansion c r+1 v r+1 + c n v n = d 1 v d r v r for some d 1,..., d r F, which we can rewrite as d 1 v d r v r c r+1 v r+1 c n v n = 0. Since {v 1,..., v n } are linearly independent, we have that d 1 = = d r = c r+1 = = c n = 0. Thus {w r+1,..., w n } are linearly independent. 4

5 Corollary Suppose that V is a finite dimensional vector space and Φ : V V and Ψ : V V are linear maps such that ΨΦ = I V. Then Φ and Ψ are isomorphisms and Ψ = Φ 1. Theorem (Universal Property of Vector Spaces) Suppose that V and W are finite dimensional vector spaces, β = {v 1,..., v n } is a basis of V, and w 1,..., w n W. Then there exists a unique linear map L : V W such that L(v i ) = w i for 1 i n. Proof. We first prove existence. For v V, define L(v) = c 1 w 1 + c 2 w c n w n if v = c 1 v c n v n with c 1,..., c n F. L is a well defined map since every v V has a unique expression v = c 1 v c n v n with c 1,..., c n F, as β is a basis of V. We have that L(v i ) = w i for 1 i n. We leave the verification that L is linear to the reader. Now we prove uniqueness. Suppose that T : V W is a linear map such that T (v i ) = w i for 1 i n. Suppose that v V. Then v = c 1 v c n v n for some c 1,..., c n F. We have that T (v) = T (c 1 v 1 + c n v n ) = c 1 T (v 1 ) + + c n T (v n ) = c 1 w c n w n = L(v). Since T (v) = L(v) for all v V, we have that T = L. Lemma Suppose that A M m,n (F ). Then the map L A : F n F m defined by L A (x) = Ax for x F n is linear. Lemma Suppose that A = (A 1,..., A n ) M m,n (F ). Then Image L A = Span{A 1,..., A n }. Kernel L A = {x F n Ax = 0} is the solution space to Ax = 0. Lemma Suppose that A, B M m,n (F ). Then L A+B = L A + L B. Suppose that c F. Then cl A = L ca. Suppose that A M m,n (F ) and C M l,m. Then the composition of maps L C L A = L CA. Lemma The map Φ : M m,n L F (F n, F m ) defined by Φ(A) = L A for A M m,n (F ) is an isomorphism. Proof. By Lemma 4.16, Φ is a linear map. We will now show that Kernel Φ is {0}. Let A Kernel Φ. Then L A = 0. Thus Ax = 0 for all x F n. By Lemma 1.1, we have that A = 0. Thus Kernel Φ = {0} and Φ is 1-1. Suppose that L L F (F n, F m ). Let w i = L(e i ) for 1 i n. Let A = (w 1,..., w n ) M m,n (F ). For 1 i n, we have that L A (e i ) = Ae i = w i. By Theorem 4.13, L = L A. Thus Φ is onto, and Φ is an isomorphism. Definition Suppose that V is a finite dimensional vector space. Suppose that β = {v 1,..., v n } is a basis of V. Define the coordinate vector (v) β F n of v respect to β by (v) β = (c 1,..., c n ) t, where c 1,..., c n F are the unique elements of F such that v = c 1 v c n v n. Lemma Suppose that V is a finite dimensional vector space. Suppose that β = {v 1,..., v n } is a basis of V. Suppose that u 1, u 2 V and c F. Then (u 1 + u 2 ) β = (u 1 ) β + (u 2 ) β and (cu 1 ) β = c(u 1 ) β. Theorem Suppose that V is a finite dimensional vector space. Let β = {v 1,..., v n } be a basis of V. Then the map Φ : V F n defined by Φ(v) = (v) β is an isomorphism. 5

6 Proof. Φ is a linear map by Lemma Note that Φ(v i ) = e i for 1 i n. By Theorem 4.13, there exists a unique linear map Ψ : F n V such that Ψ(e i ) = v i for 1 i n (where {e 1,..., e n } is the standard basis of F n ). Now ΨΦ : V V is a linear map which satisfies ΨΦ(v i ) = v i for 1 i n. By Theorem 4.13, there is a unique linear map from V V which takes v i to v i for 1 i n. Since the identity map I V of V has this property, we have that ΨΦ = I V. By a similar calculation, ΦΨ = I F n. Thus Φ is an isomorphism (with inverse Ψ). Definition Suppose that V and W are finite dimensional vector spaces. Suppose that β = {v 1,..., v n } is a basis of V and β = {w 1,..., w m } is a basis of W. Suppose that L : V W is a linear map. Define the matrix M β β (L) M m,n (F ) of L with respect to the bases β and β to be the matrix M β β (L) = ((L(v 1 ) β, (L(v 2 )) β,..., (L(v n )) β ). The matrix M β β (L) of L with respect to β and β has the following important property: (1) M β β (L)(v) β = (L(v) β for all v V. In fact, M β β (L) is the unique matrix A such that A(v) β = (L(v)) β for all v V. Lemma Suppose that V and W are finite dimensional vector spaces. Suppose that β = {v 1,..., v n } is a basis of V and β = {w 1,..., w m } is a basis of W. Suppose that L 1, L 2 L F (V, W ) and c F. Then M β β (L 1 + L 2 ) = M β β (L 1 ) + M β β (L 2 ) and M β β (cl 1 ) = cm β β (L 1 ). Theorem Suppose that V and W are finite dimensional vector spaces. Suppose that β = {v 1,..., v n } is a basis of V and β = {w 1,..., w m } is a basis of W. Then the map Λ : L F (V, W ) M m,n (F ) defined by Λ(L) = M β β (L) is an isomorphism. Proof. Λ is a linear map by Lemma It remains to verify that Λ is 1-1 and onto, from which it will follow that Λ is an isomorphism Suppose that L 1, L 2 L F (V, W ) and Λ(L 1 ) = Λ(L 2 ). Then M β β (L 1 ) = M β β (L 2 ) so that (L 1 (v i )) β = (L 2 (v i )) β for 1 i n. Thus L 1 (v i ) = L 2 (v i ) for 1 i n. Since β is a basis of V, L 1 = L 2 by the uniqueness statement of Theorem Thus Λ is 1-1. Suppose that A = (a ij ) M m,n (F ). Write A = (A 1,..., A n ) where A i = (a 1,i,..., a m,i ) t F m are the columns of A. Let z i = m j=1 a jiw j W for 1 i n. By Theorem 4.13, there exists a linear map L : V W such that L(v i ) = z i for 1 i n. We have that (L(v i )) β = (z i ) β = A i for 1 i n. We have that Λ(L) = M β β (L) = (A 1,..., A n ) = A, and thus Λ is onto. Corollary Suppose that V is a vector space of dimension n and W is a vector space of dimension m. Then L F (V, W ) is a vector space of dimension mn. Lemma Suppose that U, V and W are finite dimensional vector spaces, with respective bases β 1, β 2, β 3. Suppose that L 1 L F (U, V ) and L 2 L F (V, W ). Then M β 1 β 3 (L 2 L 1 ) = M β 2 β 3 (L 2 )M β 1 β 2 (L 1 ). Theorem Suppose that V is a finite dimensional vector space, and β is a basis of V. Then the map Λ : L F (V, V ) M n,n (F ) defined by Λ(L) = M β β (L) is a ring isomorphism. 6

7 5. Bilinear Forms Let U, V, W be vector spaces over a field F. Let Bil(U V, W ) be the set of bilinear maps from U V to W. Bil(U V, W ) is a vector space over F. We call an element of Bil F (V V, F ) a bilinear form. A bilinear form is usually written as a pairing < v, w > F for v, w V. A bilinear form <, > is symmetric if < v, w >=< w, v > for all v, w V. A symmetric bilinear form is nondegenerate if whenever v V is such that < v, w >= 0 for all w V then v = 0. Theorem 5.1. Let F be a field and A M m,n (F ). Let g A : F m F n F be defined by g A (v, w) = v t Aw. Then g A Bil F (F m F n, F ). Further, the map Ψ : M m,n (F ) Bil F (F m F n, F ) defined by Ψ(A) = g A is an isomorphism of F -vector spaces. Also, in the case when m = n, 1. Ψ gives a 1-1 correspondence between n n matrices and bilinear forms on F n. 2. Ψ gives a 1-1 correspondence between n n symmetric matrices and symmetric forms on F n 3. Ψ gives a 1-1 correspondence between invertible n n symmetric matrices and nondegenerate symmetric forms From now on in this section, suppose that V is a vector space over a field F, and <, > is a nondegenerate symmetric form on V. An important example is the dot product on F m where F is a field. We begin by stating the identity (2) < v, w >= 1 [< v + w, v + w > < v, v > < w, w >] 2 for v, w V. Define v, w V to be orthogonal (or perpendicular) if < v, w >= 0. Suppose that S V is a subset. Define S = {v V < v, w >= 0 for all w S}. Lemma 5.2. Suppose that S is a subset of V. Then 1. S is a subspace of V. 2. If U is the subspace Span(S) of V, then U = S. Lemma 5.3. Suppose A = A 1. A m M m,n (F ). Let U = Span{A t 1,..., At m} F n, which is the column space of A t. Then U = Kernel L A. A basis β = {v 1,..., v n } of V is called an orthogonal basis if < v i, v j >= 0 whenever i j. Theorem 5.4. Suppose that V is a finite dimensional vector space with a symmetric nondegenerate bilinear form <, >. Then V has an orthogonal basis. Proof. We prove the theorem by induction on n = dim V. If n = 1, the theorem is true since any basis is an orthogonal basis. Assume that dim V = n > 1, and that any subspace of V of dimension less than n has an orthogonal basis. There are two cases. The first case is when every v V satisfies 7

8 < v, v >= 0. Then by (2), we have < v, w >= 0 for every v, w V. Thus any basis of V is an orthogonal basis. Now assume that there exists v 1 V such that < v 1, v 1 > 0. Let U = Span{v 1 }, a one dimensional subspace of V. Suppose that v V. Let c = < v, v 1 > < v 1, v 1 > F. Then v cv 1 U, and thus v = cv 1 + (v cv 1 ) U + U. It follows that V = U + U. We have that U U = { 0}, since < v 1, v 1 > 0. Thus V = U U. Since U has dimension n 1 by Lemma 3.2, we have by induction that U has an orthogonal basis {v 2,..., v n }. Thus {v 1, v 2,..., v n } is an orthogonal basis of V. Theorem 5.5. Let V be a finite dimensional vector space over a field F, with a symmetric bilinear form <, >. Assume dim V > 0. Let V 0 be the subspace of V defined by V 0 = V = {v V < v, w >= 0 for all w V }. Let {v 1,..., v n } be an orthogonal basis of V. < v i, v i >= 0 is equal to the dimension of V 0. Then the number of integers i such that Proof. Order the {v 1,..., v n } so that < v i, v i >= 0 for 1 i r and < v i, v i > 0 for r < i n. Suppose that w V. Then w = c 1 v c n v n for some c 1,..., c n F. n < v i, w >= c i < v i, v j >= c i < v i, v i >= 0 j=1 for 1 i r. Thus v 1,..., v r V 0. Now suppose that v V 0. We have v = d 1 v d n w n for some d 1,..., d n F. We have n 0 =< v i, v >= d j < v i, v j >= d i < v i, v i > j=1 for 1 i n. Since < v i, v i > 0 for r < i n we have d i = 0 for r < i n. Thus v = d 1 v d r v r Span{v 1,..., v r }. Thus V 0 = Span{v 1,..., v r }. Since v 1,..., v r are linearly independent, they are a basis of V 0, and thus dim V 0 = r. The dimension of V 0 in Theorem 5.5 is called the index of nullity of the form. Theorem 5.6. (Sylvester s Theorem) Let V be a finite dimensional vector space over R with a symmetric bilinear form. There exists an integer r with the following property. If {v 1,..., v n } is an orthogonal basis of V, then there are precisely r integers i such that < v i, v i >> 0. Proof. Let {v 1,..., v n } and {w 1,..., w n } be orthogonal bases of V. We may suppose that their elements are arranged so that < v i, v i >> 0 is 1 i r, < v i, v i >< 0 if r+1 i s and < v i, v i >= 0 if s + 1 i n. Similarly, < w i, w i >> 0 is 1 i r, < w i, w i >< 0 if r + 1 i s and < w i, w i >= 0 if s + 1 i n. We first prove that v 1,..., v r, w r +1,..., w n are linearly independent. Suppose we have a relation x 1 v x r v r + y r +1w r +1 + y n w n = 0 for some x 1,..., x r, y r;+1,..., y n F. Then x 1 v x r v r = (y r +1w r y n w n ). 8

9 Let c i =< v i, v i > and d i =< w i, w i > for all i. Taking the product of each side of the preceding equation with itself, we have c 1 x c r x 2 r = d r +1y 2 r d s y2 s. Since the left hand side is 0 and the right hand side is 0, we must have that both sides of this equation are zero. Thus x 1 = = x r = 0. Since w r +1,..., w n are linearly independent, y r +1 = = y n = 0. Since dim V = n, we have that r + n r n, so that r r. Interchanging the roles of the two bases in the proof, we also obtain r r, so that r = r. The integer r in the proof of Sylvester s theorem is call the index of positivity of the form. 6. Exercises 1. Let V be an n-dimensional vector space over a field F. The dual space of V is the vector space V = L(V, F ). Elements of V are called functionals. i) Suppose that β = {v 1,..., v n } is a basis of V. For 1 i n, define vi V by vi (v) = c i if v = c 1 v c n v n V (so that vi (v j) = δ ij ). Prove that the vi are elements of V and that they form a basis of V. (This basis is called the dual basis to β.) ii) Suppose that V has a nondegenerate symmetric bilinear form <, >. For v V define L v : V F by L v (w) =< v, w > for w V. Show that L v V, and that the map Φ : V V defined by Φ(v) = L v is an isomorphism of vector spaces. 2. Suppose that L : V W is a linear map of finite dimensional vector spaces. Define ˆL : W V by ˆL(ϕ) = ϕl for ϕ W. i) Show that ˆL is a linear map. ii) Suppose that L is 1-1. Is ˆL 1-1? Is ˆL onto? Prove your answers. iii) Suppose that L is onto. Is ˆL 1-1? Is ˆL onto? Prove your answers. iv) Suppose that A M m,n (F ) where F is a field, and L = L A : F n F m. Let {e 1,..., e n } be the standard basis of F n, and {f 1,..., f m } be the standard basis of F m. Let β = {e 1,..., e n} be the dual basis of (F n ) and let β = {f1,..., f m} be the dual basis of (F m ). Compute M β β (ˆL). 3. Let V be a vector space of dimension n, and W be a subspace of V. Define Perp V (W ) = {ϕ V ϕ(w) = 0 for all w W }. i) Prove that dim W + dim Perp V (W ) = n. ii) Suppose that V has a nondegenerate symmetric bilinear form <, >. Prove that the map v L v of Problem 1.ii) induces an isomorphism of W = {v V < v, w >= 0 for all w W } with Perp V (W ). Conclude that dim W + dim W = n. iii) Use the conclusions of Problem 3.ii) to prove that the column rank of a matrix is equal to the row rank of a matrix (see Definition 7.4). This gives a different proof of part of Theorem 7.5. iv) (rank-nullity theorem) Suppose that A M m,n (F ) is a matrix. Show that the dimension of the solution space {x F n Ax = 0} 9

10 is equal to n rank A. Thus with the notation of Lemma 5.3, dim U + rank A = n. v) Consider the complex vector space C 2 with the dot product as nondegenerate symmetric bilinear form. Give an example of a subspace W of C 2 such that C 2 W W. Verify for your example that dim W +dim W = dim C 2 = Suppose that V is a finite dimensional vector space with a nondegenerate symmetric bilinear form <, >. An operator Φ on V is a linear map Φ : V V. If Φ is an operator on V, show that there exists a unique operator Ψ : V V such that < Φ(v), w >=< v, Ψ(w) > for all v, w V. Ψ is called the transpose of Φ, and we write Ψ = Φ t. Hint: Define a map Ψ : V V as follows. For w V, Let L : V F be the map defined by L(v) =< Φ(v), w > for v V. Verify that L is linear, so that L V. By Problem 1.ii), there exists a unique w V such that for all v V, we have L(v) =< v, w >. Define Ψ(w) = w. Prove that Ψ is linear, and that < Φ(v), w >=< v, Ψ(w) > for all v, w V. Don t forget to prove that Ψ is unique! 5. Let V = F n with the dot product as nondegenerate symmetric bilinear form. Suppose that L : V V is an operator. Show that L t = L A t if A M n n (F ) is the n n matrix such that L = L A. 6. Give a down to earth (but coordinate dependent ) proof of Problem 4, starting like this. Let β = {v 1,..., v n } be a basis of V. The linear map Φ β : V F n defined by Φ β (v) = (v) β is an isomorphism, with inverse Ψ : F n V defined by Ψ((x 1,..., x n ) t ) = x 1 v x n v n. Define a product [, ] on F n by [x, y] =< Ψ(x), Ψ(y) >. Verify that [, ] is a nondegenerate symmetric bilinear form on F n. By our classification of nondegenerate symmetric bilinear forms on F n, we know that [x, y] = x t By for some symmetric invertible matrix B M n n (F ). 7. Determinants Theorem 7.1. Suppose that F is a field, and n is a positive integer. Then there exists a unique function Det : M n,n (F ) F which satisfies the following properties: 1. Det is multilinear in the columns of matrices in M n,n (F ). 2. Det is alternating in the columns of matrices in M n,n (F ); that is Det(A) = 0 whenever two columns of A are equal. 3. Det(I n ) = 1. Existence can be proven by defining a function on A = (a ij ) M n,n (F ) by Det(A) = σ S n sgn(σ)a σ(1),1 a σ(2),2 a σ(n),n where for a permutation σ S n, { 1 if σ is even sgn(σ) = 1 if σ is odd. It can be verified that Det is multilinear, alternating and Det(I n ) = 1. Uniqueness is proven by showing that if a function Φ : M n,n (F ) F is multilinear and alternating on columns and satisfies Φ(I n ) = 1 then Φ = Det. Lemma 7.2. Suppose that A, B M n,n (F ). Then 10

11 1. Det(AB) = Det(A)Det(B). 2. Det(A t ) = Det(A). Lemma 7.3. Suppose that A M n,n (F ) Then the following are equivalent 1. Det(A) 0 2. The columns {A 1,..., A n } of A are linearly independent. 3. The solution space to Ax = 0 is (0). Proof. 2. is equivalent to 3. by the dimension formula, applied to the linear map L A : F n F n. We will now establish that 1. is equivalent to 2. Suppose that {A 1,..., A n } are linearly dependent. Then, after possibly permuting the columns of A, there exist c 1,..., c n 1 F such that A n = c 1 A c n 1 A n 1. Thus Det(A) = Det(A 1,..., A n 1, A n ) = Det(A 1,..., A n 1, c 1 A c n 1 A n 1 ) = c 1 Det(A 1,..., A n 1, A 1 ) + + c n 1 Det(A 1,..., A n 1, A n 1 ) = 0. Now suppose that {A 1,..., A n } are linearly independent. Then dim Image L A = n. By the dimension theorem, dim Kernel L A = 0. Thus L A is an isomorphism, so A is invertible with an inverse B. We have that 1 = Det(I n ) = Det(AB) = Det(A)Det(B). Thus Det(A) 0. Definition 7.4. Suppose that A M m,n (F ). 1. The column rank of A is the dimension of the column space Span{A 1,..., A n }, which is a subspace of F m. 2. The row rank of A is the dimension of the row space Span{A 1,..., A m }, which is a subspace of F n. 3. The rank of A is the largest integer r such that there exists an r r submatrix B of A such that Det(B) 0. Theorem 7.5. Suppose that A M m,n (F ). Then the column rank of A, the row rank of A and the rank of A are all equal. Proof. Let s be the column rank of A, and let r be the rank of A. We will show that s = r. There exists an r r submatrix B of A such that Det(B) 0. There exist 1 i 1 < i 2 < < i r m and 1 j 1 < j 2 < < j r n such that B is obtained from A by deleting the rows A i from A such that i {i 1,..., i r } and deleting the columns A j such that j {j 1,..., j r }. We have that the columns of B are linearly independent, so the columns A j 1,..., A jr of A are linearly independent. Thus s r. We now will prove that r s be induction on n, from which it follows that r = s. The case n = 1 is easily verified; r and s are either 0 or 1, depending on if A is zero or not. Assume that the induction statement is true for matrices of size less than n. After permuting the columns of A, we may assume that the first s columns of A are linearly independent. Since the column space is a subspace of F m, we have that s m. Let B the matrix consisting of the first s rows of A. if s < n, then by induction we have that the rank of B is s, so that A has rank r s. We have reduced to the case where s = n m. Let C be the (n 1) m submatrix of A consisting of the first n 1 columns of A. By induction, there exists an (n 1) (n 1) submatrix E of C whose determinant is non zero. After possibly interchanging rows of A, we may assume that E consists of the first n 1 rows of C. For 1 i n, let E i be the 11

12 column vector consisting of the first n 1 rows of the i-th column of A, and for 1 i n and n j m, let E j i be the column vector consisting of the first n 1 rows of the i-th column of A, followed by the j-th row of the i-th column of A. Since Det(E) 0, {E 1,..., E n 1 } are linearly independent, and are thus a basis of F n 1. Thus there are unique elements a 1,..., a n 1 F such that (3) a 1 E a n 1 E n 1 = E n. Suppose that all determinants of n n submatrices of A are zero. Then {E j 1,..., Ej n} are linearly dependent for n j m. By uniqueness of the relation (3), we must then have that for n j m. Thus a 1 E j a n 1E j n 1 E n = 0 a 1 A a n 1 A n 1 A n = 0, a contradiction to our assumption that A has column rank n. Thus some n n submatrix of A has nonzero determinant, and thus r s. Taking the transpose of A, the above argument shows that the row rank of A is equal to the rank of A. 8. Rings A set R is a ring if it has an addition operation under which A is an abelian group, and an associative multiplication operation such that a(b + c) = ab + ac and (b + c)a = ba + ca for all a, b, c A. A further has a multiplicative identity 1 A (written 1 when there is no danger of confusion). A ring A is a commutative ring if ab = ba for all a, b A. Suppose that A and B are rings. A ring homomorphism ϕ : A B is a mapping such that ϕ is a homomorphism of abelian groups, ϕ(ab) = ϕ(a)ϕ(b) for a, b A and ϕ(1 A ) = 1 B. Suppose that B is a commutative ring, A is a subring of B and S is a subset of B. The subring of B generated by S and A is the intersection of all subrings T of B which contain A and S. The subring of B generated by S and A is denoted by A[S]. It should be verified that A[S] is in fact a subring of B, and that A[S] = { n i 1,i 2,...,i r=0 and let R i = R for i N. Let a ii,...,i r s i 1 1 s ir r r N, n N, a i1,...,i r A, s 1,..., s r S}. T = {{a i } i N R i a i = 0 for i 0}. We can also write a sequence {a i } T as (a 0, a 1, a 2,..., a r, 0, 0,...) for some r N. T is a ring with addition {a i } + {b i } = {a i + b i } and {a i }{b j } = {c k } where c k = i+j=k a ib k. The zero element 0 T is the sequence all of whose terms are 0, and 1 T is the sequence all of whose terms are zero expect the zero th term which is 1. That is, 0 T = (0, 0,..., ) and 1 T = (1, 0, 0,...). Let x be the sequence whose first term is 1 and all other terms are zero; that is, x = (0, 1, 0, 0,...). Then x i is the sequence whose i-th term is 1 and all other terms are zero. The natural map of R into T which maps a R to the sequence whose zero-th term is a and all other terms are zero identifies R with a subring of T. Suppose 12

13 that {a i } T. Then there is some natural number r such that a i = 0 for all i > r. We have that r {a i } = (a 0, a 1,...,, a r, 0, 0,...) = a i x i. Thus the subring R[x] of T generated by x and R is equal to T. R[x] is called a polynomial ring. Lemma 8.1. Suppose that f(x), g(x) R[x] have expansions f(x) = a 0 + a 1 x + a r x r R[x] and g(x) = b 0 +b 1 x+ +b r x r R[x] with a 0,..., a r R and b 0,..., b r R Suppose that f(x) = g(x). Then a i = b i for 0 i r. Proof. We have f(x) = (a 0, a 1,..., a r, 0,...) and g(x) = (b 0, b 1,..., b r, 0,...) as elements of i N R i. Since two sequences in i N R i are equal if and only if their coefficients are equal, we have that a i = b i for all i. Theorem 8.2. (Universal Property of Polynomial Rings) Suppose that A and B are commutative rings and ϕ : A B is a ring homomorphism. Suppose that b B. Then there exists a unique ring homomorphism ϕ : A[x] B such that ϕ(a) = ϕ(a) for a A and ϕ(x) = b. Proof. We first prove existence. Define ϕ : A B by ϕ(f(x)) = ϕ(a 0 ) + ϕ(a 1 )b + + ϕ(a r )b r for (4) f(x) = a 0 + a 1 x + + a r x r R[x] with a 0,..., a r R. This map is well defined, since the expansion (4) is unique by Lemma 8.1. It should be verified that ϕ is a ring homomorphism with the desired properties. We will now prove uniqueness. Now suppose that Ψ : A[x] B is a ring homomorphism such that Ψ(a) = ϕ(a) for a A and Ψ(x) = b. Suppose that f(x) A[x]. Write f(x) = a 0 + a 1 x + + a r x r with a 0,..., a r A. We have Ψ(f(x)) = Ψ(a 0 + a 1 x + + a r x r ) = Ψ(a 0 ) + Ψ(a 1 )Ψ(x) + + Ψ(a r )Ψ(x) r = ϕ(a 0 ) + ϕ(a 1 )b + + ϕ(a r )b r = ϕ(a 0 + a 1 x + + a r x r ) = ϕ(f(x)). Since this identity is true for all f(x) A[x], we have that Ψ = ϕ. Let F be a field, and F [x] be a polynomial ring over F. Suppose that f(x) = a 0 + a 1 x + + a n x n F [x] with a 0, a n F and a n 0. The degree of f(x) is defined to be deg f(x) = n. If f(x) = 0 define deg f(x) =. For f, g F [x] we have and deg fg = deg f + deg g i=0 deg f + g max {deg f, deg g}. f F [x] is a unit (has a multiplicative inverse in F [x]) if and only if f is a nonzero element of F, if and only if deg f = 0 (this follows from the formulas on degree). F [x] is a domain; that is it has no zero divisors. A basic result is Euclidean division. Theorem 8.3. Suppose that f, g F [x] with f 0. Then there exist unique polynomials q, r F [x] such that g = qf + r with deg r < deg f. Corollary 8.4. Suppose that f(x) F [x] is nonzero. distinct roots in F. 13 Then f(x) has at most deg f

14 Corollary 8.5. F [x] is a principal ideal domain; that is every ideal in F [x] is generated by a single element. An element f(x) F [x] is defined to be irreducible if it has degree 1, and f cannot be written as a product f = gh with f, g F. A polynomial f(x) F [x] of positive degree n is monic if f(x) = a 0 + a 1 x + + a n 1 x n 1 + x n for some a 0,..., a n 1 F. Theorem 8.6. (Unique Factorization) Suppose that f(x) F [x] has positive degree. Then there is a factorization (5) f(x) = cf n 1 1 fr nr where r is a positive integer, f 1,..., f r F [x] are monic irreducible, 0 c F and n 1,..., n r are positive integers. This factorization is unique, in the sense that any other factorization of f(x) as a product of monic irreducibles is obtained from (6) by permuting the f i. If F is an algebraically closed field (such as the complex numbers C) the monic irreducibles are exactly the linear polynomials x α for α F. Theorem 8.7. Suppose that F is an algebraically closed field and f(x) F [x] has positive degree. Then there is a (unique) factorization (6) f(x) = c(x α 1 ) n1 (x α r ) nr where r is a positive integer, α 1,..., α r F are distinct, 0 c F and n 1,..., n r are positive integers. Suppose that f(x), g(x) F [x] are not both zero. A common divisor of f and g in F [x] is an element h F [x] such that h divides f and h divides g in F [x]. A greatest common divisor of f and g in F [x] is an element d F [x] such that d is a common divisor of f and g in F [x] and d divides every other common divisor of f and g in F [x]. Theorem 8.8. Suppose that f, g F [x] are not both zero. Then 1. there exists a unique monic greatest common divisor d(x) = gcd(f, g) of f and g in F [x]. 2. There exist p, q F [x] such that d = pf + qg. 3. Suppose that K is a field containing F. Then d is a greatest common divisor of f and g in K[x]. Suppose that F is a field. 9. Eigenvalues and Eigenvectors Definition 9.1. Suppose that V is a vector space over a field F and L : V V is a linear map. An element λ F is an eigenvalue of L if there exists a nonzero element v V such that L(v) = λv. A nonzero vector v V such that L(v) = λv is called an eigenvector of L with eigenvalue λ. If λ F is an eigenvalue of L, then is the eigenspace of λ for L. E(λ) = {v V L(v) = λv} Lemma 9.2. Suppose that V is a vector space over a field F, L : V V is a linear map, and λ F is an eigenvalue of L. Then E(λ) is a subspace of V of positive dimension. 14

15 Theorem 9.3. Suppose that V is a vector space over a field F, and L : V V is a linear map. Let λ 1,..., λ r F be distinct eigenvalues of L. Then the subspace of V spanned by E(λ 1 ),..., E(λ r ) is a direct sum; that is E(λ 1 ) + E(λ 2 ) + + E(λ r ) = E(λ 1 ) E(λ 2 ) E(λ r ). Proof. We prove the theorem by induction on r. The case r = 1 follows from the definition of direct sum. Now assume that the theorem is true for r 1 eigenvalues. Suppose that v i E(λ i ) for 1 i r, w i E(λ i ) for 1 i r and (7) v 1 + v v r = w 1 + w w r. We must show that v i = w i for 1 i r. Multiplying equation (7) by λ r, we obtain (8) λ r v 1 + λ r v λ r v r = λ r w 1 + λ r w λ r w r. Applying L to equation (7) we obtain (9) λ 1 v 1 + λ 2 v λ r v r = λ 1 w 1 + λ 2 w λ r w r. Subtracting equation (8) from (9), we obtain (λ 1 λ r )v (λ r 1 λ r )v r 1 = (λ 1 λ r )w (λ r 1 λ r )w r 1. Now by induction on r, and since λ i λ r 0 for 1 i r 1, we get that v i = w i for 1 i r 1. Substituting into equation (7), we then obtain that v r = w r, and we see that the sum is direct. Corollary 9.4. Suppose that V is a finite dimensional vector space over a field F, and L : V V is linear. Then L has at most dim(v ) distinct eigenvalues. 10. Characteristic Polynomials Suppose that F is a field. We let F [t] be a polynomial ring over F. Definition Suppose that F is a field, and A M n,n (F ). The characteristic polynomial of A is the polynomial p A (t) = Det(tI n A) F [t]. p A (t) is a monic polynomial of degree n. Lemma Suppose that V is an n-dimensional vector space over a field F, and L : V V is a linear map. Let β, β be two bases of V. Then Proof. We have that p M β β (L)(t) = p M β β (L)(t). M β β (L) = M β β (I)M β β β (L)Mβ (I). Let A = M β β β (L), B = Mβ (L) and C = M β β (I), so that B = C 1 AC. Then p M β β (L)(t) = Det(tI n B) = Det(tI n C 1 AC) = Det(C 1 (ti n A)C) = Det(C) 1 Det(tI n A)Det(C) = Det(tI n A) = p M β β (L)(t). 15

16 Definition Suppose that V is an n-dimensional vector space over a field F, and L : V V is a linear map. Let β be a basis of V. Define the characteristic polynomial p L (t) of L to be p L (t) = p M β β (L)(t). Lemma 10.2 shows that the definition of characteristic polynomial of L is well defined; it is independent of choice of basis of V. Theorem Suppose that V is an n-dimensional vector space over a field F, and L : V V is a linear map. Then the eigenvalues of L are the roots of p L (t) = 0 which are in F. In particular, we have another proof that L has at most n distinct eigenvalues. Proof. Let β be a basis of V, and suppose that λ F. Then the following are equivalent: 1. λ is an eigenvalue of L. 2. There exists a nonzero vector v V such that Lv = λv. 3. the Kernel of λi L : V V is nonzero. 4. The solution space to M β β (λi L)y = 0 is non zero. 5. The solution space to (λi n M β β (L))y = 0 is non zero. 6. Det(λI n M β β (L)) = 0 7. p L (λ) = 0. Definition Suppose that V is an n-dimensional vector space over a field F, and L : V V is a linear map. L is diagonalizable if there exists a basis β of V such that M β β (L) is a diagonal matrix. Theorem Suppose that V is an n-dimensional vector space over a field F, and L : V V is a linear map. Then the following are equivalent: 1. L is diagonalizable. 2. There exists a basis of V consisting of eigenvectors of L. 3. Let λ 1,..., λ r be the distinct eigenvalues of L. Then dim E(λ 1 ) + + dim E(λ r ) = n. Proof. We first prove 1 implies 2. By assumption, there exists a basis β = {v 1,..., v n } of V such that λ M β β (L) = ((L(v 0 λ 2 0 1)) β,..., (L(v n )) β ) =. 0 0 λ n is a diagonal matrix. Thus L(v i ) = λ i v i for 1 i n and so β is a basis of V consisting of eigenvectors of L. We now prove 2 implies 3. By assumption, there exists a basis β = {v 1,..., v n } of V consisting of eigenvectors of L. Thus if λ 1,..., λ r are the distinct eigenvalues of L, then V E(λ 1 ) + + E(λ r ). Thus by Theorem 9.3, so V = E(λ 1 ) + + E(λ r ) = E(λ 1 ) E(λ r ) n = dim V = dim E(λ 1 ) + + dim E(λ r ). 16

17 We now prove 3 implies 1. We have that E(λ 1 ) + + E(λ r ) = E(λ 1 ) E(λ r ) by Theorem 9.3 so that E(λ 1 ) E(λ r ) is a subspace of V which has dimension n = dim V. Thus V = E(λ 1 ) E(λ r ). Let {v i 1,..., vi s i } be a basis of E(λ i ) for 1 i r. Then β = {v 1 1,..., v 1 s 1, v 2 1,..., v 2 s 2,..., v r 1,..., v r s r } is a basis of V = E(λ 1 ) E(λ r ). We have that M β β (L) = ((L(v1 1 ) β,..., (L(vs r r )) is a diagonal matrix. Lemma Suppose that A M n,n (F ) is a matrix. Then L A is diagonalizable if and only if A is similar to a diagonal matrix (there exists an invertible matrix B M n,n (F ) and a diagonal matrix D M n,n (F ) such that B 1 AB = D). Proof. Suppose that L A is diagonalizable. Then there exists a basis β = {v 1,..., v n } of K n consisting of eigenvectors of L A. Let λ i be the eigenvalue of v i, so that Av i = λ i v i for 1 i n. We have that D = M β β (L A) = ((L(v 1 )) β,..., (L(v n )) β ) = λ λ λ n is a diagonal matrix. Let β = {E 1,..., E n } be the standard basis of F n. D = M β β (L A) = M β β (id β F n)mβ (L A )M β β (id F n) = B 1 AB where B = M β β (id K n) = (v 1, v 2,..., v n ). Now suppose that B 1 AB = D is a diagonal matrix. β = {B 1,..., B n }, a basis of F n since B is invertible. is a diagonal matrix, M β β (L A) = M β β (id β F n)mβ (L A )M β β (id F n) = B 1 AB = D D = λ λ λ n.. Let B = (B 1,..., B n ) and so L A (B i ) = λ i B i for 1 i n, and so {B 1,..., B n } is a basis of eigenvectors of L A, and so L A is diagonalizable. 11. Triangulation Let V be a finite dimensional vector space over a field F, of positive dimension n 1. Suppose that L : V V is a linear map. A subspace W of V is L-invariant if L(W ) W. A fan {V 1,..., V n } of L in V is a finite sequence of L-invariant subspaces V 1 V 2 V n = V such that dim V i = i for 1 i n. A fan basis (of the fan {V 1,..., V n }) for L is a basis {v 1,..., v n } of V such that {v 1,..., v i } is a basis of V i for 1 i n. Given a fan, a fan basis always exists. 17

18 Theorem Let β = {v 1,..., v n } be a fan basis for L. Then the matrix M β β (L) of L with respect to the basis β is an upper triangular matrix Proof. Let {V 1,..., V n } be the fan corresponding to the fan basis. L(V i ) V i for all i implies there exist a ij F such that Thus is upper triangular. L(v 1 ) = a 11 v 1 L(v 2 ) = a 21 v 1 + a 22 v 2. L(v n ) = a n1 v a nn v n. M β β (L) = (L(v 1) β, L(v 2 ) β,, L(v n ) β ) = (a ij ) Definition A linear map L : V V is triangular if there exists a basis β of V such that M β β (L) is upper triangular. A matrix A M nn(f ) is triangulizable if there exists an invertible matrix B M nn (F ) such that B 1 AB is upper triangular. A matrix A is triangulizable if and only if the linear map L A : F n F n is triangulizable; if β is a fan basis for L A, then M β β (L A) = Mβ st st (I)Mst (L A )M β st (I) = B 1 AB where I : V V is the identity map and st denotes the standard basis of F n, and B = M β st (I). Suppose that V is a vector space, and W 1, W 2 are subspaces such that V = W 1 W2. Then every element v V has a unique expression v = w 1 + w 2 with w 1 W 1 and w 2 W 2 We may thus define a projection π 1 : V W 1 by π 1 (v) = w 1 if v = w 1 + w 2 with w 1 W 1 and w 2 W 2, and define a projection π 2 : V W 1 by π 1 (v) = w 2 for v = w 1 + w 2 V. These projections are linear maps, which depend on both W 1 and W 2. Composing with the inclusion W 1 V, we can view π 1 as a map from V to V. Composing with the inclusion W 2 V, we can view π 2 as a map from V to V. Then π 1 + π 2 = I V. Theorem Let V be a non zero finite dimensional vector space over a field F, and let L : V V be a linear map. Suppose that the characteristic polynomial p L (t) factors into linear factors in F [t] (this will always happen if F is algebraically closed). Then there exists a fan of L in V, so that L is triangulizable. Proof. We prove the theorem by induction on n = dim V. If n = 1, then any basis of V is a fan basis for V. Assume that the theorem is true for linear maps of vector spaces W over F of dimension less than n = dim V. Since p L (t) splits into linear factors in F [t], and p L (t) has degree n > 0, p L (t) has a root λ 1 in F, which is thus an eigenvalue of L. Let v 1 V be an eigenvector of L with the eigenvalue λ 1. Let V 1 be the one dimensional subspace of V generated by {v 1 }. Extend {v 1 } to a basis β = {v 1, v 2,... v n } of V. Let W = Span{v 2,..., v n }. β = {v 2,..., v n } is a basis of W. V is the direct sum V = V 1 W. Let π1 : V V 1 and π 2 : V W be the projections. The composed linear map is π 2 L : W W. From ( ) M β λ1 β (L) =, 0 M β β (π 2 L) 18

19 we calculate p L (t) = (t λ 1 )p π2 L(t). Since p L (t) factors into linear factors in F [t], p π2 L(t) also factors into linear factors in F [t]. By induction, there exists a fan of π 2 L in W, say {W 1,..., W n 1 }. Let V i = V 1 + W i 1 for 2 i n. The subspaces V i form a chain V 1 V 2 V n = V. Let {u 1,..., u n 1 } be a fan basis for π 2 L, so that {u 1,..., u j } is a basis of W j for 1 j n 1. Then {v 1, u 1,..., u i 1 } is a basis of V i. We will show that {V 1,..., V n } is a fan of L in V, with fan basis {v 1, u 1,..., u n 1 }. We have that L = IL = (π 1 + π 2 )L = π 1 L + π 2 L. Let v V i. We have an expression v = cv 1 + w i 1 with c F and w i 1 W i 1. π 1 L(v) = π 1 (L(v)) V 1 V i. π 2 L(v) = π 2 L(cv 1 ) + π 2 L(w i 1 ). Now π 2 L(cv 1 ) = cπ 2 L(v 1 ) = cπ 2 (λ 1 v 1 ) = 0 and π 2 L(w i 1 ) W i 1 since {w 1,..., w n 1 } is a fan basis for π 2 L. Thus π 2 L(v) W i 1 and so L(v) V i. Thus V i is L-invariant, and we have shown that {V 1,..., V n } is a fan of L in V. 12. The minimal polynomial of a linear operator Suppose that L : V V is a linear map, where V is a finite dimensional vector space over a field F. Let L F (V, V ) be the F -vector space of linear maps from V to V. L F (V, V ) is a ring with multiplication given by composition of maps. Let I : V V be the identity map, which is the multiplicative identity of the ring L F (V, V ). There is a natural inclusion of the field F as a subring (and subvector space) of L F (V, V ), obtained by mapping λ F to λi L F (V, V ). Let F [L] be the subring of L F (V, V ) generated by F and L. F [L] = {a 0 I + a 1 L + + a r L r r N and a 0,..., a r F }. F [L] is a commutative ring. By the universal property of polynomial rings, there is a surjective ring homomorphism ϕ from the polynomial ring F [t] onto F [L], obtained by mapping f(t) F [t] to f(l). ϕ is also a vector space homomorphism. Since F [L] is a subspace of L F (V, V ), which is an F -vector space of dimension n 2, F [L] is a finite dimensional vector space. Thus the kernel of ϕ is nonzero. Let m L (t) be the monic generator of the kernel of ϕ. We have an isomorphism F [L] = F [t]/(m L (t)). m L (t) is the minimal polynomial of L. We point out here that if g(t), h(t) F [t] are polynomials, then since F [L] is a commutative ring, we have equality of composition of operators g(l) h(l) = h(l) g(l). We generally write g(l)h(l) = h(l)g(l) to denote the composition of operators. If v V, we will also write Lv for L(v) and f(l)v for f(l)(v). The above theory also works for matrices. The n n matrices M n,n (F ) is a vector space over F, and is also a ring with matrix multiplication. F is embedded as a subring by the map λ λi n, where I n denotes the n n identity matrix. Given A M n,n (F ), F [A] is a commutative subring, and there is a natural surjective ring homomorphism F [t] F [A]. The kernel is generated by a monic polynomial m A (t) which is the minimal polynomial of A. Now suppose that V is a finite dimensional vector space over F, and β is a basis of V. Then the map L F (V, V ) M n,n (F ) defined by ϕ M β β (ϕ) for ϕ L F (V, V ) is a vector space isomorphism, and a ring isomorphism. 19

20 Suppose that L : V V is a linear map, and let A = M β β (L). Then we have an induced isomorphism of F [L] with F [A]. For a polynomial f(x) F [x], we have that the image of f(l) in F [A] is M β β (f(l)) = f(m β β (L)) = f(a). Thus we have that m A (t) = m L (t). Lemma Suppose that A is an n n matrix with coefficients in a field F, and K is an extension field of F. Let f(x) be the minimal polynomial of A over F, and let g(x) be the minimal polynomial of A over K. Then f(x) = g(x). Proof. Let a 1, a 2,..., a r K be elements of K which are linearly independent over F. We will use the following observation. Suppose that A 1,..., A r M nn (F ) and a 1 A a r A r = 0 in M nn (K). Then since each coefficient of a 1 A a r A r is zero, we have that A 1 = A 2 = = A r = 0. Write g(x) = x e + b e 1 x e b 0 with b i K. Let {a 1 = 1,..., a r } be a basis of the subspace of K (recall that K is a vector space over F ) spanned by {1, b e 1,..., b 0 }. Expand b i = t j=1 c ija j with c ij F. Let f 1 (x) = x e + c e 1,1 x e c 0,1 and f j (x) = c e 1,j x e c 0,j for j 2. f i (x) F [x] for all i, and g(x) = r j=1 a jf j (x). We have that 0 = g(a) = t i=1 a if i (A), which implies that f i (A) = 0 for all i, as observed above. Thus f(x) divides f i (x) in F [x] for all i, and then f(x) divides g(x) in K[x]. Since f(a) = 0, g(x) divides f(x) in K[x] so f and g generate the same ideal, so f = g is the unique monic generator. 13. The theorem of Hamilton-Cayley Theorem Let V be a non zero finite dimensional vector space, over an algebraically closed field F, and let L : V V be a linear map. Let p L (t) be its characteristic polynomial. Then p L (L) = 0. Here p L (L) = 0 means that p L (L) is the zero map on V. Proof. There exists a fan {V 1,..., V n } of L in V, with associated fan basis β = {v 1,..., v n }. The matrix M β β (L) M n,n(f ) is upper triangular. Write M β β (L) = (a ij). Then p L (t) = Det(tI n M β β (L)) = (t a 11)(t a 22 ) (t a nn ). We shall prove by induction on i that (L a 11 I) (L a ii I)v = 0 for all v V i. We first prove this in the case i = 1. Then (L a 11 I)v = Lv a 11 v = 0 for v V 1, since V 1 is in the eigenspace for the eigenvalue a 11. Now assume that i > 1, and that (L a 11 I) (L a i 1,i 1 I)v = 0 for all v V i 1. Suppose that v V i. Then v = v + cv i for some v V i 1 and c F. Let z = (L a ii I)v. z V i 1 because L(V i 1 ) V i 1 and a ii v V i 1. By induction, (L a 11 I) (L a i 1,i 1 I)(L a ii I)v = (L a 11 I) (L a i 1,i 1 I)z = 0. We have (L a ii I)cv i V i 1, since L(v i ) = a i1 v 1 + a i2 v a ii v i. Thus (L a 11 I) (L a i 1,i 1 I)(L a ii I)cv i = 0, 20

21 and thus (L a 11 I) (L a i,i I)v = 0. Corollary Let F be an algebraically closed field and suppose that A M n,n (F ). Let p A (t) be its characteristic polynomial. Then p A (A) = 0. Here p A (A) = 0 means that p A (A) is the zero matrix. Proof. Let L = L A : F n F n. Then 0 = p L (L). Let β be the standard basis of F n. Since A = M β β (L), we have that p L(t) = p A (t) F [t]. Now p A (A) = p L (M β β (L)) = M β β (p L(L)) = M β β (0) = 0 where the right most zero in the above equation denotes the n n zero matrix. Corollary Suppose that F is a field, and A M n,n (F ). Then p A (A) = 0. Proof. Let K be an algebraically closed field containing F. From the natural inclusion M n,n (F ) M n,n (K) we may regard A as an element of M n,n (K). The computation p A (t) = Det(tI n A) does not depend on the field F or K. By Corollary 13.2, p A (A) = 0. Theorem (Hamilton-Cayley) Suppose that V is a non zero finite dimensional vector space over a field F and L : V V is a linear map. Then p L (L) = 0. Proof. Let β be a basis of V, and let A = M β β (L) M n,n(f ). We have p L (t) = p A (t). By Corollary 13.3, p A (A) = 0. Now Thus p L (L) = 0. M β β (p L(L)) = p L (M β β (L)) = p A(A) = 0. Corollary Suppose that V is a non zero finite dimensional vector space over a field F and L : V V is a linear map. Then the minimal polynomial m L (t) divides the characteristic polynomial p L (t) in F [t]. Proof. By the Hamilton-Cayley Theorem, p L (t) is in the kernel of the homomorphism F [t] F [L] which takes t to L and is the identity on F. Since m L (t) is a generator for this ideal, m L (t) divides p L (t). 14. Invariant Subspaces Suppose that V is a finite dimensional vector space over a field F, and L : V V is a linear map. Suppose that W 1,..., W r are L-invariant subspaces of V such that V = W 1 Wr. Let β i = {v i,1,..., v i,σ(i) } be bases of W i, and let β = {v 1,1,..., v 1,σ(1), v 2,1..., v r,σ(r) } which is a basis of V. Since the W i are L-invariant, the matrix of L with respect to β is a block matrix A M β β (L) = 0 A A r where A i = M β i β i (L W i ) are σ(i) σ(i) matrices. 21