MATH2130-F17 Week 13 Week 14 Week 15, Inner Farid Aliniaeifard - PowerPoint PPT Presentation

Example 1.18 MATH2130 Verify that the polynomials 1+2 t 2 , 4+ t +5 t 2 , and 3+2 t are Farid Aliniaeifard linearly independent. MATH2130 Solution. Let B = { 1 , t, t 2 , t 3 } be the standard basis for P 3 . Week 12 → R 4 where We have by Theorem 1.11 T : P 3 − Week 13 Week 14 p �→ [ p ] B Week 15, Inner Product Therefore by theorem above 1 + 2 t 2 , is an isomorphism. Space 4 + t + 5 t 2 and 3 + 2 t are linearly independent if and only � 1 + 2 t 2 � � 4 + t + 5 t 2 � if B , B , and [3 + 2 t ] B are linearly inde- pendent. So       1 4 3       � 1 + 2 t 2 � � 4 + t + 5 t 2 � 0 1 2       B =  , B =  , [3 + 2 t ] B =     2 5 0 0 0 0

MATH2130 Farid Aliniaeifard MATH2130 Therefore, we only need to show that Week 12         Week 13 1 4 3       Week 14       0 1 2       Week 15,  ,  ,     2 5 0 Inner       Product 0 0 0 Space are linearly dependent. (Do it as an Exercise).

MATH2130 Farid Aliniaeifard MATH2130 Week 12 Week 13 Week 14 Week 9, Lecture 3, Oct.25, the dimension of vector Week 15, space Inner Product Space

MATH2130 Farid Aliniaeifard MATH2130 Theorem 1.19 Week 12 Let T : V − → W be an isomorphism. Week 13 1 v 1 , . . . , v n are linearly independent (dependent) in V if Week 14 Week 15, and only if T ( v 1 ) , . . . , T ( v n ) are linearly independent (de- Inner Product pendent) in W . Space 2 A vector x is in span { v 1 , . . . , v n } if and only if T ( x ) is in span { T ( v 1 ) , . . . , T ( v n ) } .

MATH2130 Example 1.20 Farid Aliniaeifard 1 Verify that the polynomials 1+2 t 2 , 4+ t +5 t 2 , and 3+2 t are linearly independent. MATH2130 2 Is g ( t ) = t − 3 t 2 in span { 1 + 2 t 2 , 4 + t + 5 t 2 , 3 + 2 t } ? Week 12 Week 13 Proof. (1) Let B = { 1 , t, t 2 , t 3 } be the standard basis for P 3 . Week 14 → R 4 where Week 15, We have by Theorem 1.11 T : P 3 − Inner Product Space p �→ [ p ] B is an isomorphism. Therefore by theorem above 1 + 2 t 2 , 4 + t + 5 t 2 and 3 + 2 t are linearly independent if and only if � 1 + 2 t 2 � � 4 + t + 5 t 2 � B , B , [3 + 2 t ] B are linearly independent.

MATH2130 We have Farid Aliniaeifard       1 4 3 MATH2130 � 1 + 2 t 2 �   � 4 + t + 5 t 2 �     0 1 2       Week 12 B =  , B =  , [3 + 2 t ] B =     2 5 0 Week 13 0 0 0 Week 14 Week 15, Inner Therefore, we only need to show that Product Space         1 4 3             0 1 2        ,  ,      2 5 0      0 0 0 are linearly independent. (Do it as an Exercise).

MATH2130 Farid (2) By the above theorem we only need to show that Aliniaeifard         MATH2130 1 4 3       Week 12       0 1 2       [ g ( t )] B ∈ span  ,  , , Week 13     2 5 0       Week 14 0 0 0 Week 15, Inner Product i.e.,   Space         0 1 4 3               1 0 1 2          ∈ span  ,  ,      − 3  2 5 0      0 0 0 0 �

MATH2130 Farid Aliniaeifard • The dimension of a vector space MATH2130 Week 12 Week 13 Theorem 1.21 Week 14 If a vector space V has a basis B = { b 1 , . . . , b n } then any set Week 15, Inner containing more than n vectors must be linearly dependent. Product Space Theorem 1.22 If V is a vector space and V has a basis of n vectors, then every basis of V must consist of exactly n vectors.

MATH2130 Farid Aliniaeifard MATH2130 Definition 1.23 Week 12 1 A vector space is said to be finite-dimensional if it is Week 13 spanned by a finite set of vectors in V Week 14 Week 15, 2 Dimension of V , dim V , is the number of vectors in a Inner Product basis of V . Also dimension of zero space { 0 } is 0 . Space 3 If V is not spanned by a finite set, then V is said to be infinite-dimensional .

MATH2130 Example 1.24 Farid Aliniaeifard Find dimension of the subspace MATH2130     Week 12 a − 3 b + c         Week 13 2 a + 2 d   H =  : a, b, c, d in R .  Week 14 b − 3 c − d       Week 15, 2 d − b Inner Product Space Solution. We have           a − 3 b + c 1 − 3 1 0           2 a + 2 d 2 0 0 2            = a  + b  + c  + d       b − 3 c − d 0 1 − 3 − 1 2 d − b 0 − 1 0 2

MATH2130 Farid Aliniaeifard MATH2130 Therefore, Week 12           1 − 3 1 0 Week 13               Week 14 2 0 0 2         H = span  ,  ,  ,      Week 15, 0 1 − 3 − 1     Inner   Product 0 − 1 0 2 Space Now, we want to find a basis for H , we had a process for finding the basis.(Do it as an exercise.)

MATH2130 Theorem 1.25 Farid Aliniaeifard Let H be a subspace of a finite dimensional vector space V . MATH2130 Any linearly independent set in H can be expanded to a basis Week 12 for H . Also Week 13 dim H ≤ dim V Week 14 Week 15, Theorem 1.26 Inner Product Space ( The Basis Theorem ) Let V be a p -dimensional vector space p ≥ 1 . 1 Any linearly independent set of exactly p elements in V is automatically a basis for V . 2 Any set of exactly p elements that spans V is automati- cally a basis for V .

MATH2130 Farid Aliniaeifard MATH2130 Week 12 Week 13 Remember: The dimension of Nul A is the number of free Week 14 variables in the equation Ax = 0, and the dimension of Col A Week 15, is the number of pivot columns in A , and the pivot columns Inner Product of A gives a basis for column space of A . Space

MATH2130 Farid Aliniaeifard MATH2130 Week 12 Week 13 Week 14 Week 10, Lecture 1, Oct.30, change of basis Week 15, Inner Product Space

MATH2130 Example 1.27 Farid � 2 � � − 1 � � 0 � � 2 � Aliniaeifard Let b 1 = , b 2 = , c 1 = , c 2 = . Then 0 1 1 1 MATH2130 B = { b 1 , b 2 } and C = { c 1 , c 2 } are two basis for R 2 . Let Week 12 � 0 � Week 13 x = . Then Week 14 2 Week 15, � 0 � � 2 � � − 1 � Inner Product x = = + 2 = b 1 + 2 b 2 Space 2 0 1 � 1 � Therefore, [ x ] B = . Also 2 � 0 � � 0 � � 2 � � 2 � x = = 2 + 0 = 2 c 1 + 0 c 2 so [ x ] C = . 2 1 1 0

Then there is a matrix C←B such that P MATH2130 Farid [ x ] C = P C←B [ x ] B = [[ b 1 ] C [ b 2 ] C ][ x ] B . Aliniaeifard Since MATH2130 � 2 � � 0 � � 2 � Week 12 b 1 = = ( − 1) + = ( − 1) c 1 + c 2 Week 13 0 1 1 Week 14 we have � − 1 � Week 15, Inner Product [ b 1 ] C = . 1 Space Also � − 1 � � 0 � � 2 � b 2 = = 3 / 2 + ( − 1 / 2) = 3 / 2 c 1 − 1 / 2 c 2 1 1 1 Therefore, � − 1 � � 1 � � 2 � 3 / 2 [ x ] C = = 1 − 1 / 2 2 0

MATH2130 Theorem 1.28 Farid Aliniaeifard Let B = { b 1 , . . . , b n } and C = { c 1 , . . . , c n } be bases of a vector space V . Then there is a unique matrix C←B such that P MATH2130 Week 12 [ x ] C = P C←B [ x ] B Week 13 Week 14 Week 15, The columns of C←B are the C -coordinate vectors of the vec- P Inner Product tors in the basis B . That is, Space C←B = [[ b 1 ] C P [ b 2 ] C . . . [ b n ] C ] . Definition 1.29 The matrix C←B in the above theorem is called change-of- P coordinates matrix from B to C .

MATH2130 Farid Aliniaeifard MATH2130 Week 12 Week 13 Week 14 Week 15, Inner Product Space

MATH2130 Farid Aliniaeifard Remark. We have MATH2130 Week 12 [ x ] C = P C←B [ x ] B Week 13 Week 14 so Week 15, Inner − 1 [ x ] C = [ x ] B P Product C←B Space Therefore, C←B ) − 1 B←C = ( P P

MATH2130 Farid • Change of Basis in R n Aliniaeifard MATH2130 Remark. Week 12 1 Let B = { b 1 , . . . , b n } a basis for R n . Let E = { e 1 , . . . , e n } Week 13 be the standard basis for R n . Then P B = [ b 1 | . . . | b n ] is Week 14 the same as E←B . P Week 15, Inner Product 2 Let B = { b 1 , . . . , b n } and C = { c 1 , . . . , c n } be bases for Space R n . Then by row operation we can reduce the matrix [ c 1 . . . c n | b 1 . . . b n ] to [ I | P C←B ] .

MATH2130 Example 1.30 � − 9 � � − 5 � � � Farid 1 Aliniaeifard Let b 1 = , b 2 = , c 1 = , and c 2 = 1 − 1 − 4 � � MATH2130 3 , and consider the bases for R 2 given by B = { b 1 , b 2 } Week 12 − 5 Week 13 and C = { c 1 , c 2 } . Find the change-of-coordinate matrix from Week 14 B to C . Week 15, Inner Product Solution. We can reduce the matrix [ c 1 c 2 | b 1 b 2 ] to [ I | P C←B ], Space and so we can find C←B . Therefore, we have P � � 1 3 − 9 − 5 Replace R2 by R2+4R1 ← → − 4 − 5 1 − 1 � 1 � 3 − 9 − 5 Scaling R2 by 1 / 7 ← → 0 7 − 35 − 21

MATH2130 Farid Aliniaeifard MATH2130 Week 12 � 1 � � 1 � Week 13 3 − 9 − 5 0 6 4 Replace R1 by R1 − 3R2 ← → Week 14 0 1 − 5 − 3 0 1 − 5 − 3 Week 15, Inner Therefore, Product � � Space 6 4 C←B = P . − 5 − 3

MATH2130 Example 1.31 � � � − 2 � � − 7 � � − 5 � Farid 1 Aliniaeifard Let b 1 = , b 2 = , c 1 = , c 2 = , − 3 4 9 7 MATH2130 and consider the bases for R 2 given by B = { b 1 , b 2 } and C = Week 12 { c 1 , c 2 } . Week 13 1 Find the change-of-coordinates matrix from C to B . Week 14 2 Find the change-of-coordinates matrix from B to C . Week 15, Inner Product Space Solution. (1) Note that we need to find B←C , so compute P � � � 1 � 1 − 2 − 7 − 5 0 5 3 [ b 1 b 2 | c 1 c 2 ] = ↔ . − 3 4 9 7 0 1 6 4 Therefore, � 5 � 3 B←C = P . 6 4

MATH2130 Farid Aliniaeifard MATH2130 Week 12 (2) We now want to compute C←B . Note that P Week 13 Week 14 � 5 � − 1 � � Week 15, 3 2 − 3 / 2 B←C ) − 1 = Inner C←B = ( P P = . 6 4 − 3 5 / 2 Product Space

Remark. Let B = { b 1 , b 2 , . . . , b n } and { c 1 , . . . , c n } be bases MATH2130 for R n . We have (see week 9, lecture 2) Farid Aliniaeifard P B = [ b 1 | b 2 | . . . | b n ] P C = [ c 1 | c 2 | . . . | c n ] . MATH2130 It was shown that Week 12 Week 13 x = P B [ x ] B x = P C [ x ] C . Week 14 Week 15, So we have Inner Product P C [ x ] C = P B [ x ] B . Space Therefore, [ x ] C = P − 1 C P B [ x ] B . We also have [ x ] C = P C←B [ x ] B . So, P − 1 C P B = P C←B

MATH2130 • Change of basis for polynomials Farid Aliniaeifard Example 1.32 MATH2130 Let B = { 1 + t, 1 + t 2 , 1 + t + t 2 } and C = { 2 − t, − t 2 , 1 + t 2 } Week 12 Week 13 be bases for P 2 . Find C←B . P Week 14 Week 15, Solution. Solution. Let E = { 1 , t, t 2 } be the standard basis Inner Product for P 2 . Then Space R 3 T : → P 2 f �→ [ f ] E is an isomorphism.We have       1 1 1   , [1 + t 2 ] E =   , [1 + t + t 2 ] E =   [1 + t ] E = 1 0 1 0 1 1

MATH2130       Farid 2 0 1 Aliniaeifard  , [ − t 2 ] E =  , [1 + t 2 ] E =  .    [2 − t ] E = − 1 0 0 MATH2130 0 − 1 1 Week 12 Week 13 Now we have Week 14         Week 15, 1 1 1   Inner   ,   ,   Product B = 1 0 1 Space   0 1 1 and         2 0 1     ,   ,   C = − 1 0 0   0 − 1 1 be bases for R 3 . We are looking for the matrix C←B . P

MATH2130 Farid Aliniaeifard MATH2130 Week 12 Week 13 Week 14 Week 10, Lecture 2, Nov. 1, Eigenvalues and Week 15, eigenvectors Inner Product Space

Example 1.33 MATH2130 � 3 � � − 1 � � 2 � Farid − 2 Aliniaeifard Let A = , u = , v = . Then 1 0 1 1 MATH2130 � 3 � � − 1 � � − 5 � Week 12 − 2 Au = = Week 13 1 0 1 − 1 Week 14 � 3 � � 2 � � 4 � � 2 � Week 15, − 2 Inner Av = = = 2 Product 1 0 1 2 1 Space Precisely we have Av = 2 v .

Definition 1.34 MATH2130 An eigenvector of an n × n matrix A is a nonzero vector x Farid Aliniaeifard such that Ax = λx for some scalar λ . A scalar λ is called an eigenvalue of A if there is a nonzero vector x such that MATH2130 Week 12 Ax = λx ; such x is called an eigenvector corresponding Week 13 to λ . Week 14 Example 1.35 Week 15, Inner � � � − 4 � � 3 � Product 2 − 4 Space Let A = , v = , u = . − 1 − 1 1 2 � � � − 4 � � − 12 � � − 4 � 2 − 4 Av = = = 3 − 1 − 1 1 3 1 � − 4 � so is an eigenvector and 3 is an eigenvalue. Au = 1 � � � 3 � � − 2 � � 3 � 2 − 4 = � = λ for any λ. − 1 − 1 2 − 5 2

Example 1.36 MATH2130 � 1 � Farid 5 Aliniaeifard Show that 7 is an eigenvalue of A = . 6 2 MATH2130 Week 12 Solution. The number 7 is an eigenvalue. For some vector Week 13 x we have Week 14 Ax = 7 x Week 15, so Inner Product Ax − 7 x = 0 Space we can write the above equation as ( A − 7 I ) x = 0 so if ( A − 7 I ) x = 0 has a nonzero solution say x ′ , then ( A − 7 I ) x ′ = 0 ⇒ Ax ′ − 7 x ′ = 0 ⇒ Ax ′ = 7 x ′ and so 7 is an eigenvalue.

MATH2130 Farid Aliniaeifard Therefore, we only need to solve MATH2130 ( A − 7 I ) x = 0 , i.e., Week 12 � 1 � � 1 � � x 1 � � 0 � Week 13 6 0 Week 14 ( − 7 ) = 5 2 0 1 x 2 0 Week 15, Inner � − 6 � � x 1 � Product 6 Space ⇒ = 0 5 − 5 x 2 when we solve the equation we have at least a nonzero solu- � 1 � tion . Therefore 7 is an eigenvalue. 1

MATH2130 • How to find all eigenvalues of a matrix A . Farid Aliniaeifard λ is an eigenvalue for A if and only if MATH2130 Ax = λx at least for a nonzero vector x. Week 12 So we can say λ is an eigenvalue of a matrix A if and only if Week 13 Week 14 ( A − λI ) x = 0 at least for some nonzero x. Week 15, Inner Product Which means the equation ( A − λI ) x = 0 does not have only Space trivial solution if and only if det ( A − λI ) = 0 . Lemma 1.37 λ is an eigenvalue of A if and only if det ( A − λI ) = 0 .

MATH2130 Definition 1.38 Farid Aliniaeifard The equation det ( A − λI ) = 0 is called the characteristic equation . MATH2130 Week 12 Week 13 Definition 1.39 Week 14 Let λ be an eigenvalue of n × n matrix A . Then the Week 15, Inner eigenspace of A corresponding to λ is the solution set Product Space of ( A − λI ) x = 0 Remark. Note that we already have the solution set of ( A − λI ) x = 0 is a subspace.

MATH2130 Farid Aliniaeifard MATH2130 Example 1.40 Week 12   Week 13 4 − 1 6   . Week 14 let A = 2 1 6 Week 15, 2 − 1 8 Inner Product (a) Find all eigenvalues of A . Space (b) For each eigenvalue λ of A , find a basis for the eigenspace of A corresponding to λ .

(a) To find all eigenvalues of A we must find all λ such that MATH2130 Farid det ( A − λI ) = 0 . Aliniaeifard Note that MATH2130       Week 12 4 − 1 6 λ 0 0    −    = 0 Week 13 2 1 6 0 λ 0 det ( A − λI ) = det Week 14 2 − 1 8 0 0 λ Week 15,     Inner 4 − λ − 1 6 Product Space  = 0    ⇒ det 2 1 − λ 6 2 − 1 8 − λ you already know how to compute the determinant. We have     4 − λ − 1 6     = − ( λ − 9)( λ − 2) 2 det 2 1 − λ 6 2 − 1 8 − λ so λ = 9 and λ = 2, are the eigenvalues of A .

MATH2130 (b) We first find the basis for eigenspace of A corresponding Farid to λ = 2, which is the same as the finding the basis of the Aliniaeifard solution set of ( A − 2 I ) x = 0 which means we should find the MATH2130 basis for null space of A − 2 I (you know how to do it). The Week 12   x 1 Week 13   such that null space of A − 2 I contains all vectors x 2 Week 14 x 3 Week 15,   Inner x 1 Product Space   = 0. i.e., ( A − 2 I ) x 2 x 3     2 − 1 6 x 1     = 0 2 − 1 6 x 2 2 − 1 6 x 3

The augmented matrix is MATH2130   Farid 2 − 1 6 0 Aliniaeifard   2 − 1 6 0 MATH2130 2 − 1 6 0 Week 12 and the reduced echelon form is Week 13   Week 14 1 − 1 / 2 3 0 Week 15,   Inner 0 0 0 0 Product Space 0 0 0 0 So x 1 is basic and x 2 and x 3 are free. We have x 1 − 1 / 2 x 2 + 3 x 3 = 0 ⇒ x 1 = 1 / 2 x 2 − 3 x 3 Let x 2 = t and x 3 = s . Then x 1 = 1 / 2 t − 3 s.

So MATH2130         x 1 1 / 2 t − 3 s 1 / 2 − 3 Farid Aliniaeifard   =   = t   + s   x 2 t 1 0 x 3 s 0 1 MATH2130 Week 12 so the eigenspace of A corresponding to 2 is Week 13       Week 14 1 / 2 − 3   Week 15,   + s   : s, t ∈ R  t 1 0 Inner  Product 0 1 Space and the basis for the eigenspace of A corresponding to 2 is       1 / 2 − 3     ,   1 0  .  0 1 Now you will find the eigenspace and the basis of it for λ = 9 (Do it as an exercise).

MATH2130 Farid Aliniaeifard MATH2130 Week 12 Week 13 Week 10, Lecture 3, Nov. 3, Characteristic polyno- Week 14 mial and diagonalization Week 15, Inner Product Space

MATH2130 Farid Theorem 1.41 Aliniaeifard The eigenvalues of a triangular matrix are the entries on its MATH2130 main diagonal. Week 12 Week 13 Example 1.42 Week 14   Week 15, a b c Inner   . Product Let A = 0 d e Then eigenvalues of A are Space 0 0 f a, d, and f . Why? because     a b c λ 0 0   −   ) = det ( A − λI ) = det ( 0 d e 0 λ 0 0 0 f 0 0 λ

MATH2130 Farid Aliniaeifard MATH2130 Week 12   Week 13 a − λ b c Week 14   ) = ( a − λ )( d − λ )( f − λ ) det ( 0 d − λ e Week 15, 0 0 f − λ Inner Product Space Therefore, the eigenvalues are a, d and f , the entries on the main diagonal.

MATH2130 Theorem 1.43 Farid Aliniaeifard If v 1 , . . . , v r are eigenvectors that correspond to distinct eigenvalues λ 1 , . . . , λ r of an n × n matrix A , then the set MATH2130 { v 1 , . . . , v r } is linearly independent. Week 12 Week 13 Week 14 Example 1.44 Week 15,   Inner 4 − 1 6 Product   . Then 2 and 9 are eigenvalues of A . Space let A = 2 1 6 2 − 1 8 The eigenspace corresponding to 2 has a basis       1 / 2 − 3     ,   1 0  .  0 1

MATH2130 Farid Aliniaeifard Also, the eigenspace corresponding to 9 has a basis     MATH2130 1   Week 12   1  . Week 13  1 Week 14 Week 15, Inner Then Product             Space 1 / 2 1 − 3 1       ,     ,   1 1 and 0 1     0 1 1 1 are linearly independent.

MATH2130 Farid • When 0 is an eigenvalue of an n × n matrix A : Aliniaeifard MATH2130 If 0 is an eigenvalue, then there is a nonzero vector x such Week 12 that Ax = 0 x Week 13 ⇒ Ax = 0 Week 14 Week 15, which means that Ax = 0 has a nonzero solution, which also Inner Product means A is not invertible and det A = 0. Space Theorem 1.45 Let A be an n × n matrix. Then A is invertible if and only if one of the following holds: 1 The number 0 is not eigenvalue of A . 2 The determinant of A is not zero.

MATH2130 Farid Aliniaeifard • Similarity: MATH2130 Week 12 Week 13 Definition 1.46 Week 14 Two n × n matrices A and B are said to be similar if there Week 15, Inner exists an invertible matrix P such that A = PBP − 1 . Product Space Definition 1.47 The expression det ( A − λI ) is called the characteristic poly- nomial .

Let A and B are similar. Then there exists an invertible MATH2130 matrix P such that Farid A − λI = PBP − 1 − λI Aliniaeifard A = PBP − 1 ⇔ Note that PP − 1 = I , so MATH2130 Week 12 A − λI = PBP − 1 − λPP − 1 = P ( B − λI ) P − 1 Week 13 Week 14 Now Week 15, det ( A − λI ) = det ( P ( B − λI ) P − 1 ) Inner Product = det ( P ) det ( B − λI ) det ( P − 1 ) Space = det ( P ) det ( P − 1 ) det ( B − λI ) = det ( B − λI ) Therefore, A and B have the same characteristic polynomial and so they have the same eigenvalues. Proposition 1.48 Similar matrices have the same characteristic polynomial and so they have the same eigenvalues.

MATH2130 • Diagonalization (Heads up) Farid Aliniaeifard Example 1.49 MATH2130 � 2 � Week 12 0 If D = , Then Week 13 0 3 Week 14 � 2 � � 2 � � 2 2 � Week 15, 0 0 0 Inner D 2 = = Product 3 2 0 3 0 3 0 Space � 2 3 � 0 D 3 = 3 3 0 and for k we have � 2 k � 0 D k = 3 k 0

MATH2130 Definition 1.50 Farid Aliniaeifard A matrix D is a diagonal matrix if it is of the form MATH2130   d 1 0 0 . . . 0 Week 12   0 d 2 0 . . . 0 Week 13    .   . . . . . . . . . . Week 14  . . . . . Week 15, 0 0 0 . . . d n Inner Product Space Definition 1.51 A matrix is called diagonalizable if A is similar to a diago- nal matrix, i.e., there is an invertible matrix P and a diagonal matrix D such that A = PDP − 1 .

MATH2130 Farid Theorem 1.52 Aliniaeifard An n × n matrix A is diagonalizable if and only if it has n MATH2130 linearly independent eigenvectors. Week 12 Week 13 Example 1.53 Week 14 Week 15, • How to diagonalize a matrix: Inner Product 1 First check that if the matrix has n linearly dependent Space eigenvectors, if so, the matrix is diagonalizable. 2 Find a basis for the set of all eigenvectors, say { v 1 , . . . , v n } . 3 Let P = [ v 1 | . . . | v n ] , then D = P − 1 AP is an diagonal matrix with eigenvalues on its diagonal.

MATH2130 Example 1.54 Farid � 1 � Aliniaeifard 2 Find if A = is diagonalizable, if so find an in- MATH2130 0 − 3 Week 12 vertible matrix P and a diagonal matrix D such that D = Week 13 P − 1 AP . Week 14 Solution. First we should find basis for eigenspaces. Note Week 15, Inner that det ( A − λI ) = (1 − λ )( − 3 − λ ). So, A has two eigenvalues Product Space 1 and − 3. The eigenspace corresponding to 1 has the basis �� 1 �� and the eigenspace corresponding to − 3 has the 0 �� − 1 / 2 �� � 1 � − 1 / 2 basis . Then we have P = , and 1 0 1 � 1 � 0 . Check that D = P − 1 AP . D = 0 − 3

MATH2130 Farid Aliniaeifard MATH2130 Week 12 Week 13 Week 14 Week 11, Lecture 1, Nov. 6, Diagonalization Week 15, Inner Product Space

MATH2130 Farid Example 1.55 Aliniaeifard � 2 � 0 If D = , Then MATH2130 0 3 Week 12 � 2 � � 2 � � 2 2 � Week 13 0 0 0 D 2 = Week 14 = 3 2 0 3 0 3 0 Week 15, Inner � 2 3 � Product Space 0 D 3 = 3 3 0 and for k we have � 2 k � 0 D k = 3 k 0

MATH2130 Definition 1.56 Farid Aliniaeifard A matrix D is a diagonal matrix if it is of the form MATH2130   d 1 0 0 . . . 0 Week 12   0 d 2 0 . . . 0 Week 13    .   . . . . . . . . . . Week 14  . . . . . Week 15, 0 0 0 . . . d n Inner Product Space Definition 1.57 A matrix is called diagonalizable if A is similar to a diago- nal matrix, i.e., there is an invertible matrix P and a diagonal matrix D such that A = PDP − 1 .

MATH2130 Example 1.58 Farid Aliniaeifard � � 7 2 . Find a formula for A k , given that A = Let A = MATH2130 − 4 1 � � � 5 � Week 12 1 1 0 PDP − 1 . Where P = Week 13 and D = . − 1 − 2 0 3 Week 14 Week 15, Inner Solution. We can find the inverse of P which is Product Space � � 2 1 P − 1 = − 1 − 1 Then A 2 = ( PDP − 1 )( PDP − 1 ) = PD ( P − 1 P ) DP − 1 =

MATH2130 � � � 5 � 2 � � 1 1 0 2 1 PD 2 P − 1 = = Farid − 1 − 2 0 3 − 1 − 2 Aliniaeifard � � � 5 2 � � � MATH2130 1 1 0 2 1 Week 12 3 2 − 1 − 2 0 − 1 − 2 Week 13 Again, Week 14 A 3 = AA 2 = ( PDP − 1 )( PD 2 P − 1 ) = Week 15, Inner PD ( P − 1 P ) D 2 P − 1 = PD 3 P − 1 . Product Space In general, for k > = 1, � � � 5 k � � � 1 1 0 2 1 A k = PD k P − 1 = 3 k − 1 − 2 0 − 1 − 2 � � 2 . 5 k − 3 k 5 k − 3 k = . 2 . 3 k − 2 . 5 k 2 . 3 k − 5 k

MATH2130 Farid Aliniaeifard Theorem 1.59 MATH2130 (The diagonal theorem) An n × n matrix A is diagonalizable Week 12 if and only if A has n linearly independent eigenvectors. Week 13 Week 14 Definition 1.60 Week 15, An eigenvector basis of R n corresponding to A is a basis Inner Product { v 1 , . . . , v n } of R n such that v 1 , . . . , v n are eigenvectors of A . Space • An n × n matrix A is diagonalizable if and only if there are eigenvectors v 1 , . . . , v n such that { v 1 , . . . , v n } are a basis for R n , i.e., { v 1 , . . . , v n } is an eigenvector basis for R n corre- sponding to A .

MATH2130 Farid Aliniaeifard MATH2130 Week 12 Week 13 Week 14 Week 11, Lecture 2, Nov. 8, diagonalizable Week 15, matrices, eigenvectors and linear transformations Inner Product Space

How to diagonalize an n × n matrix A . MATH2130 Step 1. First find the eigenvalues of A . Farid Step 2. Find a basis for each eigenspace. That is, if Aliniaeifard MATH2130 det ( A − λI ) = ( x − λ 1 ) k 1 ( x − λ 2 ) k 2 . . . ( x − λ p ) k p , Week 12 Week 13 we should find the basis of eigenspace corresponding to each Week 14 λ i . Week 15, Step 3. If the number of all vectors in bases in Step 2 is n , Inner Product then A is diagonalizable, otherwise it is not and we stop. Space Step 4. Let v 1 , v 2 , . . . , v n be all vectors in bases in Step 2, then P = [ v 1 | v 2 | . . . | v n ] . Step 5. Constructing D form eigenvalues. If the multiplic- ity of an eigenvalue λ i is k i , we repeat λ i , k i times, on the diagonal of D .

MATH2130 Farid Example 1.61 Aliniaeifard Diagonalize the following matrix, if possible. MATH2130   Week 12 1 3 3 Week 13   . A = − 3 − 5 − 3 Week 14 3 3 1 Week 15, Inner Product That is, find an invertible matrix P and a diagonal matrix D Space such that A = PDP − 1 . Solution. Step 1. Find eigenvalues of A . 0 = det ( A − λI ) = − λ 3 − 3 λ 2 + 4 = − ( λ − 1)( λ + 2) 2 . Therefore, λ = 1 and λ = − 2 are the eigenvalues.

Step 2. Find a basis for each eigenspace. The eigenspace MATH2130 corresponding to λ = 1 is the solution set of Farid Aliniaeifard ( A − I ) x = 0 . MATH2130 A basis for this space is Week 12     Week 13 1   Week 14   1  .  Week 15, 1 Inner Product Space The eigenspace corresponding to λ = − 2 is the solution set of ( A − ( − 2) I ) x = 0 . A basis for this space is       − 1 − 1     ,   1 0  .  0 1

MATH2130 Farid Aliniaeifard Step 3. Since we find three vectors MATH2130         1 − 1 − 1 Week 12     ,   ,   1 1 0  . Week 13  Week 14 1 0 1 Week 15, Inner So A is diagonalizable. Product Space Step 4.   1 − 1 − 1   P = 1 1 0 1 0 1

MATH2130 Step 5. Farid   Aliniaeifard 1 0 0   D = 0 − 2 0 MATH2130 Week 12 0 0 − 2 Week 13 It is a good idea to check that P and D work, i.e., Week 14 Week 15, A = PDP − 1 or AP = PD. Inner Product Space If we compute we have     1 2 2 1 2 2     . AP = − 1 − 2 0 PD = − 1 − 2 0 1 0 − 2 1 0 − 2

MATH2130 Farid Aliniaeifard Example 1.62 MATH2130 Diagonalize the following matrix, if possible. Week 12   Week 13 2 4 3 Week 14   A = − 4 − 6 − 3 Week 15, 3 3 1 Inner Product Space Solution. First we find the eigenvalues, which are the roots of characteristic polynomial det ( A − λI ) . 0 = det ( A − λI ) = − λ 3 − 3 λ 2 + 4 = − ( λ − 1)( λ + 2) 2

MATH2130 So λ = 1 and λ = − 2 are eigenvalues. Farid Aliniaeifard A basis for eigenspace corresponding to λ = 1 is   MATH2130   1   Week 12   − 1 Week 13   1 Week 14 Week 15, Inner and a basis for eigenspace corresponding to λ = − 2 is Product Space     − 1     1  .  0 Since we can not find 3 eigenvectors that are linearly inde- pendent, so A is not diagonalizable.

MATH2130 Farid Aliniaeifard MATH2130 Week 12 Theorem 1.63 Week 13 An n × n matrix with n distinct eigenvalues i.e., Week 14 Week 15, det ( A − λI ) = ( x − λ 1 )( x − λ 2 ) · · · ( x − λ n ) with distinct λ i ’s, Inner Product Space is diagonalizable.

MATH2130 Theorem 1.64 Farid Aliniaeifard Let characteristic polynomial of A is MATH2130 ( x − λ 1 ) k 1 ( x − λ 2 ) k 2 . . . ( x − λ p ) k p . Week 12 Week 13 Week 14 1 For each 1 ≤ i ≤ p The dimension of eigenspace corre- Week 15, Inner sponding to λ i is at most k i . Product Space 2 The matrix A is diagonalizable if and only if the sum of the dimensions of the eigenspaces equals n , and this happens if and only if 1 the characteristic polynomial factors completely into lin- ear factors and 2 the dimension of the eigenspace for each λ i equals the multiplicity of λ i .

MATH2130 Farid Aliniaeifard MATH2130 Week 12 Week 13 If A is diagonalizable and B i is a basis for the eigenspace Week 14 corresponding to λ i for each i , then the total collection of Week 15, vectors in the sets B 1 , . . . , B p forms an eigenvector basis for Inner Product R n . Space

MATH2130 Farid Aliniaeifard MATH2130 Week 12 Week 13 Week 14 Week 11, Lecture 3, Nov. 10, Eigenvectors and Week 15, linear transformations Inner Product Space

MATH2130 Farid Aliniaeifard • Eigenvectors and linear transformations MATH2130 Week 12 When A is diagonalizable there exist an invertible matrix P Week 13 and a diagonal matrix D such that A = PDP − 1 . Our goal Week 14 is to show that the following two linear transformations are Week 15, Inner essentially the same. Product Space R n R n R n R n → → x �→ Ax u �→ Du

MATH2130 Farid Aliniaeifard MATH2130 Week 12 Remark. Let B = { b 1 , . . . , b n } be a basis for a vector space Week 13 V . Then the coordinate mapping Week 14 R n T : V → Week 15, Inner x �→ [ x ] B Product Space is a one-to-one linear transformation form V onto R n .

MATH2130 Farid • The matrix of a linear transformation: Let V be an n - Aliniaeifard dimensional vector space and W be an m -dimensional vector MATH2130 space. Week 12 Week 13 Week 14 Week 15, Inner Product Space

Let B and C be bases for V and W , respectively. The con- MATH2130 nection between [ x ] B and [ T ( x )] C is easy to find. Let B = Farid { b 1 , b 2 , . . . , b n } be the basis of V . If x = r 1 b 1 + r 2 b 2 + . . . + r n b n , Aliniaeifard then   MATH2130 r 1 Week 12   r 2   Week 13 x B =    . . .  . Week 14 Week 15, r n Inner Product Note that Space T ( x ) = T ( r 1 b 1 + r 2 b 2 + . . . + r n b n ) = r 1 T ( b 1 )+ r 2 T ( b 2 )+ . . . + r n T ( b n Since the coordinate mapping from W to R m is a linear trans- formation, we have [ T ( x )] C = [ r 1 T ( b 1 ) + r 2 T ( b 2 ) + . . . + r n T ( b n )] C = r 1 [ T ( b 1 )] C + r 2 [ T ( b 2 )] C + . . . + r n [ T ( b n )] C =

MATH2130 Farid   Aliniaeifard r 1   r 2 MATH2130   [ [ T ( b 1 )] C [ T ( b 2 )] C . . . [ T ( b n )] C ]  =   . Week 12 .  . Week 13 r n Week 14 Week 15, [ [ T ( b 1 )] C [ T ( b 2 )] C . . . [ T ( b n )] C ] [ x ] B . Inner Product Space So [ T ( x )] C = M [ x ] B , where M = [ [ T ( b 1 )] C [ T ( b 2 )] C . . . [ T ( b n )] C ] .

MATH2130 Theorem 1.65 Farid Let V be an n -dimensional vector space with basis B = Aliniaeifard { b 1 , b 2 , . . . , b n } , and let W be an m -dimensional vector space MATH2130 with basis C . If T is a linear transformation form V to W , Week 12 then Week 13 [ T ( x )] C = M [ x ] B , Week 14 where M = [ [ T ( b 1 )] C [ T ( b 2 )] C . . . [ T ( b n )] C ] . M is Week 15, Inner called matrix for T relative to the bases B and C . Product Space

Example 1.66 MATH2130 Farid Let B = { b 1 , b 2 } be a basis for V and C = { c 1 , c 2 , c 3 } be a Aliniaeifard basis for W . Let T : V → W be a linear transformation such MATH2130 that Week 12 Week 13 T ( b 1 ) = 3 c 1 − 2 c 2 + 5 c 3 T ( b 2 ) = 4 c 1 + 7 c 2 − c 3 Week 14 Week 15, Find matrix M for T relative to B and C . Inner Product Space Solution. We have that M = [[ T ( b 1 )] C [ T ( b 2 )] C ] . We have     3 4     . [ T ( b 1 )] = − 2 [ T ( b 2 )] = 7 5 − 1