Linear Algebra M1 - FIB. Contents: 5. Matrices, systems of linear equations and determinants 6. Vector space 7. Linear maps 8.

Linear Algebra M1 - FIB Contents: 5 Matrices, systems of linear equations and determinants 6 Vector space 7 Linear maps 8 Diagonalization Anna de Mier Montserrat Maureso Dept Matemàtica Aplicada II Translation: Oriol Caño, Josep Elgueta

5 Matrices, systems of linear equations and determinants 51 Matrices: basic operations and row echelon form Review of matrix algebra 1 2 Scalars Matrices By a field of scalars K we will mean a set of numbers with two operations (addition and multiplication) such that - the usual properties are satisfied (commutative, associative, distributive, identity elements) - they are invertible (we can subtract and divide) Examples: R, Q, Z p, C Let m, n 1 be integers A m n matrix with elements in the field K consists of mn elements of K arranged in a table with m rows and n columns We will denote by a ij the element in the ith row and jth column A generic matrix is represented by: a 11 a 12 a 1n a 21 a 22 a 2n A = a m1 a m2 a mn We will also use the notation A = (a ij ) m n The set of all m n matrices will be denoted by M m n (K) 3 4

Types of matrices A matrix of type 1 n is called a row matrix A matrix of type m 1 is called a column matrix The null matrix O m,n (or simply O) is the matrix of type m n with all elements equal to 0 A matrix of type n n is called a square matrix The set of all square n n matrices with elements in K is denoted by M n (K) A square matrix (a ij ) n n is upper triangular if aij = 0 for all i > j lower triangular if aij = 0 for all i < j diagonal if it is both upper and lower triangular The matrix Diag(λ 1,, λ n ) is the diagonal matrix (d ij ) n n with d ii = λ i for all i The identity matrix I n is the diagonal matrix Diag(1, 1,, 1) Matrix addition Let A, B M m n (K) with A = (a ij ) and B = (b ij ) Their sum is the matrix A + B = (c ij ) M m n (K) defined by c ij = a ij + b ij Properties If A, B, C M m n (K), the following holds: (Associative) (A + B) + C = A + (B + C) (Commutative) A + B = B + A (Identity element) A + O = O + A = A (Additive inverse) There exists a matrix B such that A + B = B + A = O (This matrix B is denoted A) 5 6 Scalar multiplication Let A M m n (K) with A = (a ij ) and let λ K be a scalar The multiplication of A by the scalar λ is the matrix λa = (b ij ) M m n (K) defined by b ij = λa ij Properties If λ, µ K and A, B M m n (K), the following holds: (Pseudo-associative) λ(µa) = (λµ)a (Distributive 1) λ(a + B) = λa + λb (Distributive 2) (λ + µ)a = λa + µa (Identity) 1A = A Transposition Let A = (a ij ) m n M m n (K) Its transpose is the matrix A t = (b ij ) n m M(K) n m defined by b ij = a ji Clearly (A t ) t = A A square matrix A is symmetric if A t = A skew-symmetric if A t = A Note that ( 1)A = A 7 8

Matrix multiplication Let A = (a ij ) m n M m n (K) and B = (b ij ) n p M n p (K) Their product is the matrix AB = (c ij ) m p M m p (K) with c ij = Observations n a ik b kj = a i1 b 1j + a i2 b 2j + + a in b nj k=1 The product of two matrices may not exist BA may not exist even when AB exists If AB and BA are defined, it may be AB BA Multiplication is an internal map in M n (K) Properties of matrix multiplication For matrices A, B, C the following holds (wherever the operations are defined): (Associative) (AB)C = A(BC) (Distributive) A(B + C) = AB + AC and (A + B)C = AC + BC (Identity element) IA = A = AI, where I is the identity matrix of the required type (Relation with the transpose) (AB) t = B t A t If A M n (K), we will denote by A k the product AA A (ie, A 2 = AA, A 3 = AAA, etc) 9 10 Inverse matrix Let A, B M n (K) We say that B is the inverse matrix of A when AB = BA = I n If this holds we say that A is invertible and we denote by A 1 the inverse matrix Observations If the inverse exists, it is unique Not all matrices have an inverse Invertible matrices have no zero rows or columns Properties of the inverse matrix If A and B are invertible matrices of the same type and λ is a nonzero scalar, the following holds: the matrix A 1 is invertible and (A 1 ) 1 = A the matrix A k is invertible and (A k ) 1 = (A 1 ) k the matrix λa is invertible and (λa) 1 = (λ) 1 A 1 the matrix A t is invertible and (A t ) 1 = (A 1 ) t the product AB is invertible and (AB) 1 = B 1 A 1 11 12

Elementary row operations Let A M m n (K) Elementary row operations and row echelon form An elementary row operation of A consists of one of the following three operations: (I) switching two rows of A (II) multiplying a row of A by a nonzero scalar (III) adding to one row of A the result of multiplying another row by a nonzero scalar A matrix is (row) elementary if it can be obtained from an identity matrix by performing only one elementary row operation 13 14 Matrix equivalence Row echelon form Let T be an elementary row operation and let M M m n (K) The result of applying the operation T to the matrix M is EM, where E is the elementary matrix obtained by applying T to the identity I m A matrix B is (row) equivalent to a matrix A if B can be obtained from A by a finite sequence of elementary row operations So, if B is equivalent to A we can write B = E r E r 1 E 2 E 1 A, where all E i are elementary matrices A matrix is in (row) echelon form if it satisfies the following: - if it has a zero row (all entries equal to zero), all the rows below it are also zero - in each nonzero row, the first nonzero element is equal to 1 (it is called the leading coefficient or pivot of the row) - the pivot of a row is always to the right of the pivot on the previous row Every matrix is equivalent to a matrix in echelon form The rank of a matrix A is the number of nonzero rows of any matrix in echelon form equivalent to A 15 16

Application: computing the inverse (I) Application: computing the inverse (II) Lemma If E is an elementary matrix, then E is invertible and its inverse E 1 is also an elementary matrix Proof: (I) If B is an elementary matrix corresponding to a type (I) row operation (switching rows i and j), we have BB = I (II) If C λ is the elementary matrix corresponding to a type (II) row operation (multiplying a row by λ 0), we have C λ C λ 1 = I = C λ 1C λ (II) If D k is the elementary matrix corresponding to a type (III) row operation (adding to row i row j multiplied by k), we have D k D k = I = D k D k Let A M n (K) and let M be a matrix in row echelon form equivalent to A Then A is invertible if and only if all the elements in the diagonal of M are equal to 1 Corollary A matrix A M n (K) is invertible if and only if the rank of A is n 17 18 Gauss-Jordan method for computing the inverse Let A M n (K) The proof of the previous theorem implies that if I n = E r E 2 E 1 A, then A 1 = E r E 2 E 1 Given A, its inverse A 1, if it exists, can be found as follows: Start with the matrix (A I n ) Apply row operations to (A I n ), until we get a matrix of the form (I n B) If we succeed, A 1 = B Otherwise, A is not invertible 19

Systems of linear equations 5 Matrices, systems of linear equations and determinants 52 Systems of linear equations A linear equation in the variables x 1,, x n is an expression of the form a 1 x 1 + a 2 x 2 + + a n x n = b, where a 1,, a n, b belong to the field of scalars K A solution is (s 1,, s n ) K n such that a 1 s 1 + a 2 s 2 + + a n s n = b (Obs A linear equation can have no solution, one, or more than one, even infinite, solutions) 20 21 Systems of linear equations Solutions of a system A system of linear equations is a set of linear equations (all of them with the same variables x 1,, x n ) The general form of a system of linear equations is: a 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2 a m1 x 1 + a m2 x 2 + + a mn x n = b m A solution to the system is an n-tuple (s 1,, s n ) K n that satisfies all the equations We will say that a system is inconsistent if it has no solution determined if it has only one solution underdetermined if it has more than one solution The general solution of a system is the set of all its solutions Two systems are equivalent if they have the same set of solutions 22 23

Equivalent systems Two systems with the same equations but sorted differently are equivalent And if in a system we multiply an equation by a (nonzero) scalar, or we add to an equation a multiple of another one the resulting system is equivalent to the first one Matrix form of a system Given the system a 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2 a m1 x 1 + a m2 x 2 + + a mn x n = b m the system matrix and the matrices of variables and of constant terms are a 11 a 12 a 1n x 1 b 1 a 21 a 22 a 2n A = x = x 2 b = b 2 a m1 a m2 a mn x n b m We can write the system as a product of matrices: Ax = b 24 25 Augmented matrix The augmented matrix is the matrix (A b), ie, a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2 (A b) = a m1 a m2 a mn b m Obs If row operations are applied to the augmented matrix of a system, the resulting system is equivalent to the first one So, every system of linear equations is equivalent to another one whose augmented matrix is in echelon form Solving a system in row echelon form The row echelon form of a consistent system looks like x 1 + c 12 x 2 + c 13 x 3 + + c 1r x r + + c 1n x n = d 1 x 2 + c 23 x 3 + + c 2r x r + + c 2n x n = d 2 x r + + c rn x n = d r (rearranging the variables if needed) Variables x 1,, x r are called basic and the others are called free The system is solved upwards as follows The basic variable x r is isolated in terms of the free variables: x r = d r c r,r+1 x r+1 c rn x n 26 Now we can isolate x r 1 in terms of x r and the rest of the free variables, etc 27

General solution of a system in row echelon form In a system in row echelon form we can express all the basic variables in terms of the free ones (and the coefficients): x 1 = f 1 + e 1,r+1 x r+1 + + e 1,n x n x 2 = f 2 + e 2,r+1 x r+1 + + e 2,n x n x r = f r + e r,r+1 x r+1 + + e r,n x n This is the general solution of the system Obs For each assignment of a scalar value to each of the free variables x r+1,, x n we get a particular solution of the system We say that the system has n r degrees of freedom 28 Parametric form of the general solution If the general solution of a system is x 1 = f 1 + e 1,r+1 x r+1 + + e 1,n x n x 2 = f 2 + e 2,r+1 x r+1 + + e 2,n x n x r = f r + e r,r+1 x r+1 + + e r,n x n the parametric form of the solution is the expression x 1 f 1 e 1,r+1 x 2 f 2 e 2,r+1 x r = f r + x r+1 e r,r+1 + + x n x r+1 0 1 x n 0 0 e 1,n e 2,n e r,n 0 1 29 Discussing a system: the Rouché-Frobenius Homogeneous systems Let be a system of linear equations with associated matrix A M m n (K) and augmented matrix (A b) Let r, r be the ranks of A, (A b) respectively Then, if r < r, the system is inconsistent if r = r = n, the system is determined if r = r < n, the system is underdetermined with n r degrees of freedom The rank of the associated matrix of a consistent linear system is called the rank of the system A system of linear equations is homogeneous if all the constant terms are equal to 0 Obs A homogeneous system is always consistent (since we have the trivial solution x 1 = = x n = 0) Corollary Let A be the associated matrix of a homogeneous system in n variables; let r be the rank of A Then if r = n, the system is determined and the only solution is the trivial one if r < n, the system is underdetermined and it has non-trivial solutions 30 31

Solving the system: Gaussian elimination To find the general solution of any system of linear equations we proceed as follows: 1 Find the augmented matrix (A b) 2 Find a matrix in row echelon form M equivalent to (A b) 3 Apply the Rouché-Frobenius theorem to determine whether the system is consistent or not 4 In the case the system is consistent, find the general solution from the equivalent system with augmented matrix M 32

5 Matrices, systems of linear equations and determinants 53 Determinants Definition of determinant Let A = (a ij ) M n (K) A minor of A is any matrix obtained by deleting a certain number of rows and the same number of columns of A The minor associated to the element a ij is the matrix A ij obtained by deleting row i and column j from A The minor A ij is an (n 1) (n 1) square matrix The determinant of A is recursively defined by - if n = 1, then det(a) = a 11 - if n 2, then det(a) = a 11 det(a 11 ) a 12 det(a 12 ) + + ( 1) 1+j a 1j det(a 1j ) + + ( 1) n+1 a 1n det(a 1n ) The cofactor of the element a ij is ( 1) i+j det(a ij ) 33 34 Computing the determinant (Instead of det(a), sometimes we write A ) Matrices 2 2 and 3 3: a b c d = a det((d)) b det((c)) = ad bc a b c d e f = a e f g h i h i b d f g i + c d e g h = a(ei fh) b(di fg) + c(dh eg) = aei + cdh + bfg ceg afh bdi One can prove by induction: If A has a zero row or column then det(a) = 0 If A = Diag(a 1, a 2,, a n ), then det(a) = a 1 a 2 a n Let A M n (K) and i, j [n] Then det(a) = n a ik ( 1) i+k det(a ik ) k=1 (Expansion of the determinant along row i) det(a) = n a kj ( 1) k+j det(a kj ) k=1 (Expansion of the determinant along column j) 35 36

Determinants and row operations Let A, B M n (K) If B is the matrix obtained from A switching two rows, then det(b) = det(a) (row operation of type (I)) multiplying the i-th row of A by λ, then det(b) = λ det(a) (row operation of type (II)) adding to a row a multiple of another one, then det(b) = det(a) (row operation of type (III)) Corollary If M is obtained from A by performing row operations, det(m) = K det(a), where K 0 Characterization of invertible matrices A matrix A M n (K) is invertible if and only if det(a) 0 Corollary A matrix A M n (K) has rank n if and only if det(a) 0 Let A M m n (K) The rank of A is r if and only if the largest minor of A with nonzero determinant has type r r So, if A and M are equivalent matrices, then det(a) 0 det(m) 0 37 38 Determinants and operations with matrices If A, B M n (K), then det(ab) = det(a) det(b) det(a t ) = det(a) if A is invertible, det(a 1 ) = det(a) 1 However, in general det(a + B) det(a) + det(b) 39

6 Vector spaces Properties Addition in R n satisfies the following properties, for all u, v, w R n : s1) (associative) u + (v + w) = (u + v) + w s2) (commutative) u + v = v + u s3) (identity element) u + 0 = u where 0 = (0 0 0) t s4) (additive inverse) for all u there exists u such that u + u = 0 The scalar multiplication in R n satisfies the following properties, for all u, v R n and λ, µ R: p1) λ(µu) = (λµ)u p2) λ(u + v) = λu + λv p3) (λ + µ)u = λu + µu p4) 1u = u (All properties hold because they hold in R and the operations are defined componentwise) 40 42 R n and its operations x 1 R n x 2 = { : x i R, 1 i n} x n x 1 y 1 x 2 Let x = and y = y 2 be elements of Rn and λ R x n y n Addition in R n : Scalar Multiplication in R n : x 1 + y 1 λx 1 x 2 + y 2 x + y = λx 2 λx = x n + y n λx n (ie, both operations are defined componentwise) 62 Vector spaces A vector space over a field K consists of 1 a non empty set E 2 a map E E E (addition +) and 3 a map K E E (scalar multiplication ) such that the following holds for all u, v, w E and all λ, µ K: e1) (associative) u + (v + w) = (u + v) + w e2) (commutative) u + v = v + u e3) (identity element) there exists a unique element 0 E E such that u + 0 E = u e4) (additive inverse) for each u E there exists a unique u E such that u + u = 0 E e5) λ(µu) = (λµ)u e6) λ(u + v) = λu + λv e7) (λ + µ)u = λu + µu e8) 1u = u, where 1 is the identity element of the product in K 41 43

Some examples of vector spaces R n Z n 2 : strings of n bits Addition is bit by bit: eg, 0 1 1 1 1 1 1 0 0 0 0 0 Scalar multiplication: 0u = 0 Z n 2 and 1u = u M m n (K) (the m n matrices with entries in the field K) Upper triangular matrices in M n (R) P(R): the set of polynomials with coefficients in R P d (R): the polynomials with degree at most d with coefficients in R The trivial vector space with only the zero vector: {0 E } The solutions of a homogeneous system of linear equations Properties If v E and λ is a scalar, the following holds: 0v = 0 E λ 0 E = 0 E If λ v = 0 E, then λ = 0 or v = 0 E The additive inverse of v is ( 1)v; usually we will write v 44 45 63 Vector subspaces and linear combinations A subset S E is a vector subspace (VS) if (s1) S (s2) for all u, v S, u + v S (s3) for all u S and all λ K, λu S The vector 0 E belongs to all vector subspaces Some examples of vector subspaces P d (R) is a vector subspace of the vector space P(R) Upper triangular matrices of M n (R) are a VS of M n (R) The solutions of a homogeneous system of linear equations in n variables and coefficients in K is a VS of K n Intersection of subspaces Lemma If S and S are vector subspaces of E, then S S is also a subspace of E The union of vector subspaces is not a vector subspace in general, as in the case ( ) ( ) x S = { : x R}, and S x = { : x R}; x x ( ) ( ) 1 2 We have + S S 1 2 Linear combination Given vectors u 1,, u k in E, a linear combination of u 1,, u k is any vector of the form λ 1 u 1 + + λ k u k, where λ 1,, λ k are scalars 46 47

Subspaces spanned by some vectors Let u 1,, u k be vectors of E The subspace spanned by u 1,, u k is the set u 1,, u k = {λ 1 u 1 + λ 2 u 2 + + λ k u k : λ 1,, λ k K}, ie, the set of all linear combinations of u 1,, u k Proposition The subspace u 1,, u k is a vector subspace Moreover, it is the smallest subspace that contains u 1,, u k If a space S can be written as S = u 1,, u l, we will say that {u 1,, u l } is a spanning set of S The spanning set of a space is not unique Observe that v is a linear combination of u 1,, u k if and only if v u 1,, u k Examples of u 1,, u l 1 0 0 R n 0 =, 1,, 0 0 0 1 The matrix space M m n (R) is spanned by the matrices M ij that have all their entries equal to 0, except the (i, j) entry, which is equal to 1, 1 i n and 1 j m For example, M 2 (R) = M 11, M 12, M 21, M 22, where M 11 = ( ) 1 0, M 0 0 12 = P d (R) = 1, x,, x d ( ) 0 1, M 0 0 21 = (These examples can be generalized to any field K) ( ) 0 0, M 1 0 22 = ( ) 0 0 0 1 48 49 64 Linear independence Let u 1,, u k E The equation To span the upper triangular matrices, it is enough to use the previous matrices M ij with i j A subspace whose vectors are given in terms of parameters {a + (b a)x + (c b)x 2 + (a c)x 3 : a, b, c R} ={a(1 x + x 3 ) + b(x x 2 ) + c(x 2 x 3 ) : a, b, c R} = 1 x + x 3, x x 2, x 2 x 3 λ 1 u 1 + λ 2 u 2 + + λ k u k = 0 E always has the solution λ 1 = = λ k = 0 If this is the only solution we will say that the vectors u 1,, u k are linearly independent (LI) If there is a solution with some λ i 0, we will say that the vectors are linearly dependent (LD) (We will also say that the set {u 1,, u k } is LI or LD, resp) Examples: The vector 0 E is linearly dependent Given a vector u 0 E, the vector u is linearly independent If u is any given vector and λ is a scalar, {u, λu} is LD 50 51

To determine if a set of vectors u 1, u 2,, u k of R n is linearly independent we proceed as follows: (1) we form a matrix A with the given vectors, putting them in columns (2) we compute the rank r of A (3) if r = k, then the k vectors are LI if r < k, then they are LD; if we have computed the rank putting the matrix A in row echelon form, then the vectors that correspond to the columns with a pivot are a maximal LI subset; if we have computed the rank by minors, the vectors that correspond to the columns of the largest minor of A with nonzero determinant are a maximal LI subset In general, to determine if a set of vectors u 1, u 2,, u k of a vector K-space E is linearly independent we proceed as follows: (1) from the vector equation λ 1 u 1 + λ 2 u 2 + + λ k u k = 0 E we obtain a homogeneous system with unknowns λ 1, λ 2,, λ k (2) we discuss the system, if it is determined, the vectors u1, u 2,, u k are LI underdetermined, the vectors u1, u 2,, u k are LD 52 53 Properties Let S = {u 1,, u k } be a set of vectors of a K-vector space E If 0 E is in S, then u 1,, u k are LD If u 1,, u k are LI, then 0 E is not in S If u 1,, u k are LI, every subset of S is LI If u 1,, u k are LD, every set that contains S is LD If u 1,, u k are LD and u 1 is a linear combination of the other vectors in S, then Characterizations A set of vectors S is LD if, and only if, there exists a vector v in S that is a linear combination of the other vectors in S Corollary Let v E If u 1,, u k are LI, then v, u 1,, u k are LI if, and only if, v u 1,, u k u 1, u 2,, u k = u 2,, u k 54 55

65 Bases and dimension Let E be a K-vector space A set of vectors B = {b 1, b 2,, b n } is a basis of E if (b1) B is linearly independent (b2) E = b 1, b 2,, b n, ie, b 1, b 2,, b n span E The canonical basis 1 0 0 of K n 0 is {, 1,, 0 } 0 0 1 of M m n (K) is given by the mn matrices M ij that have all the entries equal to 0 except the (i, j) entry, which is equal to 1 of P d (K) is {1, x, x 2,, x d } ({x d, x d 1,, 1} is also a canonical basis; we will need to specify which one we use) 56 Let B = {b 1,, b n } be a basis of E Proposition Each vector of E can be written in a unique way as a linear combination of the vectors in B Let v E If v = α 1 b 1 + + α n b n, we say that α 1 v B = α n is the vector of coordinates of v in the basis B Proposition Let {u 1,, u k } be a set of vectors of E that are LI Then k n Corollary Every basis of E has n elements 57 Dimension The cardinal of any basis of a vector space E (or of a VS) will be called the dimension of the space, denoted by dim(e) The dimensions of the spaces that we will mostly work with are: dim(k n ) = n, dim(m m n (K)) = nm, and dim(p d (K)) = d + 1 The dimension of the subspace {0 E } is 0 Let E be of dimension n and let W = {w 1,, w n } be a subset of E if W is a LI set, then W is a basis of E if W spans E, then W is a basis of E If S is a subspace of E then dim(s) dim(e) if dim(s) = dim(e), S = E Change of basis Let B = {b 1,, b n } and B = {b 1,, b n} be two bases of a K-vector space E Let u be a vector of E We look for the relationship between the coordinate vectors u B and u B We call change-of-basis matrix from B to B the matrix whose columns are the vectors of coordinates (b 1 ) B,, (b n ) B It is denoted by P B B PB B = (b 1 ) B (b 2 ) B (b n ) B Then u B = P B B u B PB B = ( PB B ) 1 59 58

7 Linear Maps Properties Let f : E F be a linear map Then f (0 E ) = 0 F f ( λ i v i ) = λ i f (v i ), for all u E if S is a subspace of E, f (S) is a subspace of F if S is a subspace of F, f 1 (S ) is a subspace of E Proposition Let B = {b 1,, b n } be a basis of E Then f is uniquely determined by f (b 1 ),, f (b n ) ie, the image of any vector u E can be obtained from the image of a basis: if u = α 1 b 1 + + α n b n, then f (u) = α 1 f (b 1 ) + + α n f (b n ) Corollary If S = v 1,, v k is a subspace of E, then f (S) = f (v 1 ),, f (v k ) 60 62 71 Definitions, examples and properties Let E and F be two K-vector spaces A map f : E F is linear if: (a1) for all u, v E, f (u + v) = f (u) + f (v) (a2) for all u E and all λ K, f (λu) = λf (u) If E = F, we say that f is an endomorphism Examples Trivial map f : E F given by f (u) = 0 F, u E, is linear Identity map I E : E E given by I E (u) = u, u E, is linear Let A M m n (K) The map f : K n K m given by f (v) = Av, v K n, is linear The following is a non-linear map (( )) a b f : M 2 2 (R) P 2 (R), f = x 2 (a+d)x+(2c b) The map f : R 2 R 2, f ( x y c d ) ( x = 2 y 2 ) is not linear x + y Let B = {b 1,, b n } be a basis of E, W a basis of F and m the dimension of F The matrix associated to f in the bases B and W is the matrix in M m n (K) whose columns are the coordinates in the basis W of the images of the vectors in the basis B It is denoted by M B W (f) MW B (f ) = f (b 1 ) W f (b 2 ) W f (b n ) W M m n(k) To find the vector of coordinates of the image of a vector u E we just need to compute the matrix product f (u) W = M B W (f )u B 61 63

72 Characterization of linear maps A linear bijective map is called an isomorphism Characterization of the type of linear map Let f : E f be a linear map and let M be the matrix associated to f in given bases of E and of F f is injective rank(m) = dim(e) f is surjective rank(m) = dim(f ) f is an isomorphism rank(m) = dim(e) = dim(f ) If E and F have the same dimension, then f is an isomorphism f is injective f is surjective 73 Composition of linear maps Let E, F and G be vector spaces with bases B, W and V, respectively Proposition If f : E F and g : F G are linear maps, the composition g f : E G is also a linear map and Proposition M B V (g f ) = MW V (g)mb W (f ) If f : E F is an isomorphism, the inverse map f 1 : F E is also linear and MB W (f 1 ) = (MW B (f )) 1 64 65 74 Change of basis We look for the change to the matrix associated to a linear map when we change the basis in the source and/or the target space Let f : E F be a linear map, B and B bases of E, and W and W bases of F E B I E P B B E B f M B W (f ) f F W P W W I F F W M B W (f ) f = I F f I E M B W (f ) = P W W MW B (f ) PB B 66

The problem of diagonalization Let f : E E be an endomorphism Is there any basis B of E in which the matrix M B (f ) is simple? More concretely, diagonal? 8 Diagonalization Def An endomorphism f : E E is diagonalizable if there exists a basis B of E such that M B (f ) is diagonal Obs The matrix M B (f ) may be non-diagonal, but if we know that the endomorphism f diagonalizes in another basis B, then the matrix (PB B ) 1 M B (f )PB B is diagonal So, being diagonalizable is equivalent to the existence of an invertible matrix P such that P 1 M B (f )P is diagonal 67 68 Eigenvalues and eigenvectors Def The scalar λ is an eigenvalue of an endomorphism f if there exists a vector v 0 E such that f (v) = λv All vectors v 0 E such that f (v) = λv are called eigenvectors of eigenvalue λ The endomorphism f : E E diagonalizes if and only if there is a basis of E completely made of eigenvectors of f 69 Computing eigenvalues Let M be the matrix associated to f : E E in a basis B Def The characteristic polynomial of the endomorphism f is p f (x) = det(m xi n ) The eigenvalues of f are the roots of the characteristic polynomial The algebraic multiplicity of an eigenvalue λ is the multiplicity of λ as a root of p f (x) and it is denoted by m λ The equation p f (x) = 0 is called the characteristic equation The characteristic polynomial does not depend on the basis in which we compute the matrix M 70

Eigenspaces Characterization of diagonalizable endomorphisms Let λ be an eigenvalue of the endomorphism f : E E The eigenspace of the eigenvalue λ is the set Properties E λ = {u E : f (u) λu = 0 E } E λ is a vector subspace of E 1 dim(e λ ) m λ The dimension of E λ is called the geometric multiplicity of λ Let f : E E be an endomorphism of a vector space E of dimension n The endomorphism f is diagonalizable if and only if it has n eigenvalues (counting multiplicities) and for each eigenvalue the algebraic and the geometric multiplicity coincide Corollary If f has n different eigenvalues, then it is diagonalizable 71 72 Diagonalization algorithm To decide whether the endomorphism f : E E is diagonalizable or not, we will proceed as follows: (1) We find the matrix associated to f in any basis and we calculate the characteristic polynomial p f (x) (2) We find the eigenvalues and their multiplicities solving p f (x) = 0 (3) If the multiplicities of the eigenvalues add up to less than dim(e), the endomorphism does not diagonalize Otherwise go to (4) (4) For each eigenvalue λ, we find the eigenspace E λ and its dimension dim(e λ ) (5) If for all λ we have m λ = dim(e λ ), the endomorphism diagonalizes Otherwise it does not If the endomorphism diagonalizes, a basis in which it diagonalizes is given by the union of the bases of the spaces E λ 73