Announcements Wednesday, November 01 Poll today: Open now! Webwork - PowerPoint PPT Presentation

Announcements Wednesday, November 01 Poll today: Open now! ◮ Webwork due today. ◮ This Friday, Quiz type of question: ‘Find the error’ (1) Circle the error and write down a correction By the cofactor expansion along the first row: � 2   1 0 0 � 3  = 1 · det det 0 2 3 = 2( − 1) − 3 · 1 = − 5  1 − 1 0 − 1 1 do not carry on with the rest of the computations.

Section 5.2 The Characteristic Equation

The Characteristic Polynomial Last section we learn that for a square matrix A : λ is an eigenvalue of A ⇐ ⇒ Ax = λ x has a nontrivial solution ⇐ ⇒ ( A − λ I ) x = 0 has a nontrivial solution ⇐ ⇒ A − λ I is not invertible ⇐ ⇒ det( A − λ I ) = 0 . Compute Eigenvalues The eigenvalues of A are the roots of det( A − λ I ) , which is the characteristic polynomial of A . Definition Let A be a square matrix. The characteristic polynomial of A is f ( λ ) = det( A − λ I ) . The characteristic equation of A is the equation f ( λ ) = det( A − λ I ) = 0 .

The Characteristic Polynomial Example Question: What are the eigenvalues of � 5 � 2 A = ? 2 1 Answer: First we find the characteristic polynomial : �� 5 � λ � 5 − λ � �� � 2 0 2 f ( λ ) = det( A − λ I ) = det − = det 2 1 0 λ 2 1 − λ = (5 − λ )(1 − λ ) − 2 · 2 = λ 2 − 6 λ + 1 . The eigenvalues are the roots of the characteristic polynomial, which we can find using the quadratic formula : λ = 6 ± √ 36 − 4 √ = 3 ± 2 2 . 2

The Characteristic Polynomial Example Definition The trace of a square matrix A is Tr( A ) = sum of the diagonal entries of A . � a � b What do you notice about: the characteristic polynomial of A = ? c d Answer: � a − λ � b det( A − λ I ) = det = ( a − λ )( d − λ ) − bc d − λ c = λ 2 − ( a + d ) λ + ( ad − bc ) ◮ The coefficient of λ is the trace of A and the constant term is det( A ). ◮ Recall that A is not invertible if and only if λ = 0 is a root. Shortcut The characteristic polynomial of a 2 × 2 matrix A is f ( λ ) = λ 2 − Tr( A ) λ + det( A ) .

The Characteristic Polynomial Example Question: What are the eigenvalues of the rabbit population matrix   0 6 8 1 A = 0 0  ?  2 1 0 0 2 Answer: First we find the characteristic polynomial:   − λ 6 8 1 f ( λ ) = det( A − λ I ) = det − λ 0   2 1 0 − λ 2 � 1 � � � λ 2 − 6 · 1 = 8 4 − 0 · − λ − λ 2 = − λ 3 + 3 λ + 2 . Already know one eigenvalue is λ = 2, check : f (2) = − 8 + 6 + 2 = 0. Doing polynomial long division, we get: − λ 3 + 3 λ + 2 = − λ 2 − 2 λ − 1 = − ( λ + 1) 2 . λ − 2 Hence f ( λ ) = − ( λ + 1) 2 ( λ − 2) and so λ = − 1 is also an eigenvalue.

Algebraic Multiplicity Definition The algebraic multiplicity of an eigenvalue λ is its multiplicity as a root of the characteristic polynomial. There is a geometric multiplicity notion, but this one is easier to work with. Example In the rabbit population matrix, f ( λ ) = − ( λ − 2)( λ + 1) 2 . The algebraic multiplicities are � 2 multiplicity 1 , λ = − 1 multiplicity 2 Example � 5 √ √ � 2 In the matrix , f ( λ ) = ( λ − (3 + 2 2))( λ − (3 − 2 2)). The 2 1 √ � 3 + 2 2 alg. multiplicity 1 , √ algebraic multiplicities are λ = 3 − 2 2 alg. multiplicity 1

Multiplicities Theorem If A is an n × n matrix, the characteristic polynomial f ( λ ) = det( A − λ I ) is a polynomial of degree n, and its roots are the eigenvalues of A: f ( λ ) = ( − 1) n λ n + a n − 1 λ n − 1 + a n − 2 λ n − 2 + · · · + a 1 λ + a 0 . Complex numbers If you count the eigenvalues of A , with their algebraic multi- plicities, depending on whether you allow complex eigenvalues , you will get : ◮ Do allow complex numbers: Always n . ◮ Only real numbers: Always at most n , but sometimes less . This is because any degree- n polynomial has exactly n complex roots , counted with multiplicity. Stay tuned!

Similarity Definition Two n × n matrices A and B are similar if there is an invertible n × n matrix C such that A = CBC − 1 . The intuition C keeps record of a basis C = { v 1 , . . . , v n } of R n . B transforms the C -coordinates of x : B [ x ] C = [ Ax ] C in the same way that A transforms the standard coordinates of x Why does it work? ◮ First, C = { v 1 , v 2 , . . . , v n } is a basis for R n ( C is invertible), so w = c 1 v 1 + c 2 v 2 + c n v n = C [ w ] C ◮ Using C -coordinates for any vector w , is [ w ] C = C − 1 w . ◮ Then A = CBC − 1 implies C − 1 A = BC − 1 . Using C -coordinates: [ Ax ] C = C − 1 ( Ax ) = B ( C − 1 x ) = B ([ x ] C ) .

Similarity Example � 1 � 2 � 2 � � � 2 0 1 A = CBC − 1 . A = B = C = = ⇒ − 1 4 0 3 1 1 What does B do geometrically? Scaling: x -direction by 2 and y -direction by 3. B acting on the usual coordinates Bx B 3-eigenspace x Be 2 e 2 e 1 Be 1 y 2-eigenspace By Now A will do to the standard coordinates what �� 2 � � 1 �� B does to the C -coordinates, where C = , . 1 1

From C -coordinates to standard coordinates � 2 � vectors in C v 1 = 1 3-eigenspace � 1 � v 2 = 1 � 1 � 2-eigenspace [ x ] C = x 1 v 2 � 3 � v 1 x = v 1 + v 2 = 2 � − 2 � y [ y ] C = − 1 y = − 2 v 1 − v 2 � − 5 � = − 3

A does to the usual coordinates what B does to the C -coordinates � 4 � Av 1 = 2 v 1 = 2 3-eigenspace � 3 � Av 2 = 3 v 2 = 3 Ax � 2 � 2-eigenspace B [ x ] C = = [ Ax ] C 3 Av 2 � 7 � Av 1 Ax = 2 v 1 + 3 v 2 = 5 � − 4 � B [ y ] C = = [ Ay ] C − 3 Ay = − 4 v 1 − 3 v 2 Ay � − 11 � = − 7 � 1 � 1 � � 3 � � 7 � � � − 5 � � − 11 � 2 2 ✧ Check: Ax = = Ay = = − 1 4 2 5 − 1 4 − 3 − 7

Similar Matrices Have the Same Characteristic Polynomial Fact If A and B are similar , then they have the same characteristic polynomial . Consequence : similar matrices have the same eigenvalues ! Though dif- ferent eigenvectors in general. Why? Suppose A = CBC − 1 . We can show that det( A − λ I ) = det( B − λ I ). A − λ I = CBC − 1 − λ I = CBC − 1 − C ( λ I ) C − 1 = C ( B − λ I ) C − 1 . Therefore, C ( B − λ I ) C − 1 � � det( A − λ I ) = det = det( C ) det( B − λ I ) det( C − 1 ) = det( B − λ I ) , because det( C − 1 ) = det( C ) − 1 .

Similarity Caveats Warning 1. Matrices with the same eigenvalues need not be similar. For instance, � 2 � 2 � � 1 0 and 0 2 0 2 both only have the eigenvalue 2, but they are not similar . 2. Similarity is lost in row equivalence . For instance, � 2 � 1 � � 1 0 and 0 2 0 1 are row equivalent, but they have different eigenvalues .

Once more: The Invertible Matrix Theorem Addenda We have a couple of new ways of saying “ A is invertible” now: The Invertible Matrix Theorem Let A be a square n × n matrix, and let T : R n → R n be the linear transformation T ( x ) = Ax . The following statements are equivalent . 1. A is invertible . 2. T is invertible. 11. A has a left inverse (there exists B such that BA = I n ). 3. A is row equivalent to I n . 12. A has a right inverse (there exists B such that AB = I n ). 13. A T is invertible. 4. A has n pivots. 14. The columns of A form a basis for R n . 5. Ax = 0 has only the trivial solution. 15. Col A = R n . 6. The columns of A are linearly independent. 7. T is one-to-one. 16. dim Col A = n . 8. Ax = b is consistent for all b in R n . 17. rank A = n . 9. The columns of A span R n . 18. Nul A = { 0 } . 10. T is onto. 19. dim Nul A = 0. 19. The determinant of A is not equal to zero . 20. The number 0 is not an eigenvalue of A .