13.1 An eigenvector for a given matrix A is any non zero vector v such that A(v) =!v where! is a scalar.


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1 5 CHAPTER THIRTEEN. Eigenvectors. An eigenvector for a given matrix A is any non zero vector v such that Av) v where is a scalar. i.e. an eigenvector is one which A maps into a multiple of itself. The possible values for are called eigenvalues or characteristic values). Example 4 Note that maps,) to 8,6). This means that,) is an eigenvector for with eigenvalue 8. Furthermore since 6 5 is a l.t. it preserves scalar 6 5 multiplication and hence it follows that all scalar multiples of,) are mapped to 8 multiple of themselves. 4 e.g. maps,4) to 6,) 6 5 This means that the set {m,) mr}, i.e. y x is left invariant by the matrix y y x v v u x x
2 6 To find eigenvalues in general We will consider the x case but the argument readily generalises to the nxn case. a b Let A. c d a b x x We wish to investigate c d y y i.e ax + by x cx + dy y i.e. a x) + by cx + d )y i.e. the equations may be written: a b x c d y If a b c d is then Kernel {} only and there are no eigenvectors for A. is not an eigenvector for reasons to be found later). If a b c d is not, then there will be non zero vectors in the Kernel which is what we are looking for. i.e. If the determinant of a b a b c d is zero, then will have eigenvectors with c d corresponding eigenvalues. Note that notationally a b a b c d may be written c d i.e A I. i.e. A has eigenvalue if deta I).
3 7 Example To find eigenvalues and eigenvectors for. We need to look at det i.e. ) ) 6 4 4) +) 4 or i.e. the eigenvalues are 4 and. The set of all eigenvalues is called the SPECTRUM. Spectrum for above matrix is {4,}. To find the eigenvectors corresponding to 4. We wish to solve i.e. x + y 4x x + y 4y x 4x y 4y i.e. y x This means that all position vectors for points on y x are eigenvectors for with eigenvalue 4. We call this S 4. i.e. S 4 {x,y) y x}. In general the set of eigenvectors corresponding to an eigenvalue is called S. To find S. i.e x  x y  y i.e x + y x y x x + y y i.e. any position vector on y x is an eigenvector with eigenvalue. i.e. S {x,y) y x }.
4 8 Example To find S. To find spectrum and S for 4  det 4 i.e i.e. or. i.e. Spectrum is {,} To find S. 4 x x y y i.e. x y x x + 4y y i.e. S {x,y) y x} 4 x x y y i.e. x y x x + 4y y i.e. S {x,y) y x} y x y x For the purposes of calculating eigenvalues, the zero vector is not considered as an eigenvector because A) for all matrices A. Hence every matrix would have as an eigenvector. Furthermore, since for all, every real number would be an eigenvalue for every matrix, clearly an undesirable situation. Hence is not usually an eigenvector. The special case where, i.e, where vectors remain unchanged by the matrix, is of interest in the study of Stochastic Processes, Markov Chains and Probability Theory in general.
5 9 Exercise Find the spectrum and S for i) ii) iii) 4 iv) v) vi) vii) 9 6. If m is a double root of det A I), does it follow that S m is a plane as opposed 4 to a line). Hint consider ).. If v is an eigenvector for A show that v is an eigenvector for A n, for all n. 4. If is an eigenvalue for A, show that is an eigenvalue for A. Is n an eigenvalue for A n? 5. i) Show that if is an eigenvalue for A then is an eigenvalue for A. Assume A exists, of course) ii) Does S for A) S for A )? 6. Find so that Av) A v) v. 7. Show that if A is an idempotent matrix then the spectrum of is a subset of {,}. 8. Show that if is a nilpotent matrix, then the spectrum of is {}. 9. Explain why a rotation of any angle other than 8 in R cannot have eigenvalues.. What are the possible eigenvalues for: i) A reflection? ii) An orthogonal matrix?. Find a matrix having eigenvalues and +. Assume matrix is x). If A is 4 4 find the eigenvalues.
6 Exercise. Answers. i) {,4} S {x,y) y + x }, S 4 {x,y) y x}. ii) {6, } S 6 {x,y) y x}, S {x,y) 5y x}. iii) {6, } S 6 {x,y) 5x y}, S {x,y) y x}. iv) {7, } S 7 {x,y) y x}, S {x,y) x + y } v) {} S {x,y) y x} vi) {,} S {x,y) x + y }, S {x,y) y x}. vii) φ i.e. no eigenvalues. 4. No. is a double root of but S is a line {x,y) y x}. 4. Yes. 5. ii) Yes. 6. ±. 9. All rotations except I) change directions of vectors.. i) ±. ii) ±. m m m n n + 6 m e.g. 4.,. Eigenvalues for x matrices Consider a d g b e h c f A i a  b c x This maps x,y,z) to x,y,z), as in the R case, when d e  f y g h i  z And as before, we obtain the eigenvalues from the equation deta I).
7 Example has eigenvalue when det. i.e. ) ) ) i.e. or Note is a double root. To find S x x y y occurs when z z x x x + y y y x x y + z z Note that S is a plane {x,y,z} y x} To find S i.e. x x y y z z x x x + y y x ; y z x y + z z i.e. S is a line {m,, ) m R}
8 Example To find eigenvalues and S for det i.e i.e ) + ) ) 4) i.e., or 4. To find S z  y  x  z y x i.e. x y + z x x + y + z y x + y z i.e. S is a line {m,, 4) mr} Similar techniques show that S {m,, ) mr} and S 4 {m,,) mr} This means each of the three lines of vectors S, S and S 4 remains invariant after The previous examples suggest the following theorem 4 z y x
9 Theorem. If is an eigenvalue for matrix A then S is a subspace of the domain A. Proof Let v and v S Then Av ) v and Av ) v i.e. Av + v ) Av ) + Av ) since A is a l.t. v + v v + v ) v + v S i.e. S is closed under vector addition. Similarly Acv ) cav ) since A is a l.t. c v ) cv ) cv S i.e. S is closed under scalar multiplication. Therefore S is a subspace of domain of A. Theorem. If A: V W is a l.t. and is an eigenvalue then det A Proof ) If then some vectors in the domain of A are mapped to zero times themselves, i.e. mapped to. This mean that Kernel of A contains something other than, i.e. KerA A is not. det A ) The above argument is completely reversible. Exercise. x. Find x so that has eigenvalue 4.
10 4. Find eigenvalues for 5 5. Hence deduce its inverse.. Find eigenvalues and the corresponding S for: Does it follow that if is a double root of det A I) where A is a x matrix then S is a plane? Investigate by considering. 5. Show that   is idempotent. Find its eigenvalues and describe its geometric significance. 6. Find spectrum of. 7. Investigate for eigenvalues and the resulting subspaces S. 8. By considering the range of matrix A show that i) is not onto ii) A is not iii) Ker A iv) is an eigenvalue Furthermore show that the range is {m,,) m R}, that 6 is an eigenvalue and hence that the only possible subspace for S 6 is {x,y,z) x y z}.
11 5 9. Find examples and which are eigenvalues for matrices A and A respectively but such that + is not an eigenvalue for A + A.. i) If is an eigenvalue for A, is an eigenvalue for ma m )? ii) Is m an eigenvalue for ma?. If and are eigenvalues for A and A respectively, is an eigenvalue for A A?. Find a matrix other than I or the zero matrix which has every vector as an eigenvector. How many eigenvalues does the matrix have? Exercise. Answers. x. {5, } A 5 5. {,, } S {x,y,z) x y z} S {x,y,z) S {x,y,z) x y z } 7 y z x } 4. No. is a double root but S is a line {x,y,z) x y z}. 5. or. It maps everything to y x. 6. {,, } 7. or. is a double root. S {x,y,z) x y} S {x,y,z) x ;y z}.. i) No. ii) Yes.. No. A magnification a a which has a single eigenvalue a.
12 6. If A is a matrix then deta I) is called the characteristic polynomial of A. e.g. If A is then det 4 4. i.e. 6 + is the characteristic polynomial Hamilton Cayley Theorem If A is a matrix with characteristic polynomial F ), then FA) [] where [] is the zero matrix. We will first illustrate the meaning of this theorem. For A as above, F ) Then FA) means A 6A + I. The theorem allows us to deduce that A 6A + I. [] Check A A 8 4 I Then A 6A + I does equal as required by the theorem. A general proof of the Hamilton Cayley Theorem is beyond the scope of this text but we will show the truth of the theorem in the x case by direct substitution. Proof a b Let A be then characteristic polynomial is given by det a b c d c d i.e. a+d) + ad bc We now need to show that A a+d)a + ad bc)i []
13 7 A a + bc ab + bd ac + dc bc  d a ad  ab bd a+d)a  ac  dc  ad  d ad bc ab dc)i ad  bc Hence A a+d)a + ad bc)i as required. Example Consider The characteristic polynomial is det i.e. + i.e. is the only eigenvalue. we do not include complex eigenvalues) It therefore follows from the Hamilton Cayley Theorem that A + I [] i.e. A I i.e. A A Hence A. The Hamilton Cayley Theorem can be used occasionally to find high powers of some matrices as follows. Question 5 If A ±. find A. The characteristic polynomial is 4 i.e.
14 8 i.e. A 4I i.e. A 4I A 4I) Previously in the chapter we examined. We found that its eigenvalues were 4 and and that S 4 {x,y) m,) R} and S {m, ) m R}. We will now look at in greater detail. Call the matrix A. To diagonalise A This means to multiple A by matrices so that A becomes a diagonal matrix. This can be effected as follows. Let P be the matrix mapping i to any vector in S 4, say),) and mapping j to any vector in S, say), ). i.e. P is Now we know A,) 8,) and A, ),) since they belong to S 4 and S respectively) 8 AP 8 4 But P 4 i.e. AP P But det P, P exists. P 4 AP i.e. A can be diagonalised to a diagonal matrix whose non zero elements are its eigenvalues.
15 9 4 We say A is SIMILAR to i.e. if A P BP then A is similar to B. Note that detp AP) det P det A det P det A det P det P det A Since the determinant of P AP is clearly the product of the eigenvalues, see above, it follows that det A product of its eigenvalues This is true for nxn matrices but it is not true that every square matrix is similar to a diagonal matrix. Exercise. Question 4 is an example illustrating this.) Previously, we found that had eigenvalues,, and 4 and S {m,, 4) mr}, S {m,, ) mr}and S 4 {m,,) mr}. We may deduce from this that P is a matrix such that 4 P P 4 In fact every symmetric matrix can be made similar to a diagonal matrix. More advanced texts treat similar matrices more fully where they have considerable importance.
16 Exercise.. Find the eigenvalues for 7 and hence find its determinant.. Repeat for.. A is the matrix Find P such that P APB. where B is a diagonal matrix. 4. Repeat question where A 5. Use the Hamilton Cayley Theorem to find the inverse of. Note that this is the matrix in Question. 6. Investigate the conjecture that if A is a x matrix with eigenvalues,, then trace A + + [trace A is the sum of the diagonal elements of A] 7. Show that cos sin sin cos is orthogonal and that it represents a rotation of a.c. about the z axis. 8. Show that if A is a x orthogonal matrix then {Ai), Aj), Ak)} is a set of mutually perpendicular unit vectors. Is the set an orthonormal basis for the range of A? 9. Show that A 4 5 has only one real eigenvalue. Find the corresponding S. Explain why it is not possible to find a matrix P such that P APB where B is a diagonal matrix.
17 . Is diagonalisable?. Find the characteristic polynomial of and consequently name a matrix a b c whose characteristic polynomial is x 5x 6x 7.. Show that if A and B are similar matrices then they have the same eigenvalues. Do they have the same eigenvectors?. Find the spectrum of. Find the corresponding eigenspaces and describe the geometric significance of the matrix. 4. True or False? a) If A A then A [] or I b) If A + B I then A commutes with B c) If A I) [] then A does not have an inverse d) If all the eigenvalues of A are zero then A k [] for some integer k. e) If A is a diagonal matrix then AB BA for every matrix B. Exercise. Answers. Spectrum {,5, 4}. Determinant is zero.. Spectrum {}. Determinant. A is similar to. P is a a a b b b c for any a,b,c. c
18 4. P. a bc where d e d b e c a c a i.e. Pi), Pj) are l.i. An example is Since A I by Hamilton Cayley Theorem A A. 6. True 8. Yes 9. z y 5 x. No No. Spectrum {,} S {x,y) y x} S {x,y) y x}. It is a projection on to y x. 4. a) False b) True c) False e.g. d) True Hamilton Cayley Theorem) e) False
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