Eigenvalues and Eigenvectors

Transcription

1 5 Eigenvalues and Eigenvectors 5.2 THE CHARACTERISTIC EQUATION

2 DETERMINANATS nn Let A be an matrix, let U be any echelon form obtained from A by row replacements and row interchanges (without scaling), and let r be the number of such row interchanges. Then the determinant of A, written as det A, is times the product of the diagonal entries u 11,, u nn in U. If A is invertible, then u 11,, u nn are all pivots (because A ~ I n and the u ii have not been scaled to 1 s). ( 1) r Slide 5.2-2

3 DETERMINANATS Otherwise, at least u nn is zero, and the product u 11 u nn is zero. Thus det product of r ( 1) A pivots in U 0,, when A is invertible when A is not invertible Slide 5.2-3

4 DETERMINANATS Example 1: Compute det A for. Solution: The following row reduction uses one row interchange: A U ~ ~ ~ A 1 Slide 5.2-4

5 DETERMINANATS 1 So det A equals. The following alternative row reduction avoids the row interchange and produces a different echelon form. The last step adds times row 2 to row 3: A ~ This time det A is as before. ( 1) (1)( 2)( 1) 2 1/ ~ U / 3 0 ( 1) (1)( 6)(1/ 3) 2 2, the same Slide 5.2-5

6 THE INVERTIBLE MATRIX THEOREM (CONTINUED) nn Theorem: Let A be an matrix. Then A is invertible if and only if: s. The number 0 is not an eigenvalue of A. t. The determinant of A is not zero. Theorem 3: Properties of Determinants nn Let A and B be matrices. a. A is invertible if and only if det. det AB (det A)(det B) det A det A b.. T c.. A 0 Slide 5.2-6

7 PROPERTIES OF DETERMINANTS d. If A is triangular, then det A is the product of the entries on the main diagonal of A. e. A row replacement operation on A does not change the determinant. A row interchange changes the sign of the determinant. A row scaling also scales the determinant by the same scalar factor. Slide 5.2-7

8 THE CHARACTERISTIC EQUATION Theorem 3(a) shows how to determine when a matrix of the form is not invertible. A λi det( Aλ I) 0 The scalar equation characteristic equation of A. is called the nn A scalar λ is an eigenvalue of an matrix A if and only if λ satisfies the characteristic equation det( Aλ I) 0 Slide 5.2-8

9 THE CHARACTERISTIC EQUATION Example 3: Find the characteristic equation of A Solution: Form A λi, and use Theorem 3(d): Slide 5.2-9

10 THE CHARACTERISTIC EQUATION The characteristic equation is or 5 λ λ 8 0 det( Aλ I) det λ λ (5 λ)(3 λ)(5 λ)(1 λ) 2 (5 λ) (3 λ)(1 λ) 0 2 (λ 5) (λ 3)(λ 1) 0 Slide

11 THE CHARACTERISTIC EQUATION Expanding the product, we can also write λ 14λ 68λ 130λ 75 0 nn det( A λ) I If A is an matrix, then is a polynomial of degree n called the characteristic polynomial of A. The eigenvalue 5 in Example 3 is said to have multiplicity 2 because (λ 5) occurs two times as a factor of the characteristic polynomial. In general, the (algebraic) multiplicity of an eigenvalue λ is its multiplicity as a root of the characteristic equation. Slide

12 SIMILARITY nn If A and B are matrices, then A is similar to B if 1 there is an invertible matrix P such that P AP B, or, equivalently,. A PBP 1 Q BQ 1 P 1 Writing Q for, we have. A So B is also similar to A, and we say simply that A and B are similar. Changing A into transformation. 1 P AP is called a similarity Slide

13 SIMILARITY Theorem 4: If matrices A and B are similar, then they have the same characteristic polynomial and hence the same eigenvalues (with the same multiplicities). Proof: If B nn 1 P AP then, B λi P AP λ P P P ( AP λ P) P ( A λ I) P Using the multiplicative property (b) in Theorem (3), we compute 1 det( λ ) det ( λ ) B I P A I P 1 det( P ) det( A λ I) det( P) (2) Slide

14 SIMILARITY det( P ) det( P) det( P P) det I 1 det( B λ I) det( A λ I) Since 1 1, we see from equation (1) that. Warnings: 1. The matrices and are not similar even though they have the same eigenvalues. Slide

15 SIMILARITY 2. Similarity is not the same as row equivalence. (If A is row equivalent to B, then B EA for some invertible matrix E ). Row operations on a matrix usually change its eigenvalues. Slide