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Section 8.8 Eigenvalues and Eigenvectors Matrix-vector multiplication can be thought of geometrically as a linear transformation changing one vector into another 4 2 −3 −4 = 5 1 4 −11 4x 5x + + 2y y = = 0 6 4 2 x 0 = 5 1 y 6 4 2 2 0 = 5 1 −4 6 Ax = b What vector x gets transformed into the vector b ? Ax = λx 4 2 2 −2 2 = = −1 5 1 − 5 5 −5 4 2 1 6 1 = = 6 1 1 6 5 1 eigenvector eigenvalue What vector x gets transformed into λx ( a scalar mult. of the orig. x ) ? Section 8.8 Eigenvalues and Eigenvectors Solve: Ax = λ x ( An×n ) Ax − λ x = 0 A−λI )x = 0 ( ↑ ↑ matrix vector vector Need ( A − λ I ) to not be invertible, because if it was we would only have the trivial solution x = 0. ⇓ Set det ( A − λ I ) = 0 This leads to an equation in λ called the characteristic equation. The roots of the characteristic equation are the eigenvalues λ . For each eigenvalue λ , find its eigenvector by solving ( A − λ I ) x = 0 1 −1 1 0 Find the eigenvalues of 1 2 1 . 0 3 −1 det ( A − λ I ) = 0 ⇒ −1 − λ 1 0 1 2−λ 3 0 1 −1 − λ = ( −1 − λ ) 2−λ 1 3 −1 − λ + ( −1)1 1 1 0 −1 − λ = ( −1 − λ ) ( 2 − λ )( −1 − λ ) − 3 − ( −1 − λ ) = 0 = ( −1 − λ ) ( −2 − λ + λ 2 ) − 3 − ( −1 − λ ) = 0 = ( −1 − λ ) λ 2 − λ − 5 − ( −1 − λ ) = 0 ( ) = ( −1 − λ ) ( λ 2 − λ − 5 ) − 1 = 0 = ( −1 − λ ) ( λ − λ − 6 ) = 0 2 = ( −1 − λ )( λ − 3)( λ + 2 ) = 0 λ1 = −1, λ2 = −2, λ3 = 3 There must be an easier way ☺ Shortcut to finding the characteristic equation 2× 2 λ 2 − ( trace ( A ) ) λ + det ( A) = 0 3× 3 sum of the diagonal entries λ 3 − ( trace ( A) ) λ 2 + ( C11 + C22 + C33 ) λ − det ( A) = 0 sum of the diagonal cofactors The only problem now is that you have to factor a cubic Find one root, then use synthetic (or long) division to get the quadratic factor, then factor that (if possible) n × n triangular or diagonal matrix The eigenvalues are on the diagonal! 2 −1 1 0 Find the eigenvalues of 1 2 1 0 3 −1 trace ( A ) = −1 + 2 − 1 =0 2 1 C11 = = −2 − 3 = −5 3 −1 det ( A ) = −1 C22 = −1 0 =1 0 −1 C33 = −1 1 = −2 − 1 = −3 1 2 ⇒ C11 + C22 + C33 = −7 2 1 1 1 = −1( −2 − 3) − 1( −1) = 6 + (1)( −1) 3 −1 0 −1 Possible roots: Characteristic Equation: λ − 0 ⋅ λ − 7 λ − 6 = 0 ⇒ λ − 7λ − 6 = 0 3 using the 3 × 3 shortcut λ 3 − ( trace ( A) ) λ 2 + ( C11 + C22 + C33 ) λ − det ( A) = 0 2 3 ( since the constant term is ±1, ± 2, ± 3, and ± 6 − 6) Plug these in until you find all the roots. λ = 2 ? ⇒ 8 − 14 − 6 ≠ 0 No λ 3 − 7λ − 6 = 0 λ = −2? ⇒ −8 + 14 − 6 = 0 Yes λ = 1? ⇒ 1 − 7 − 6 ≠ 0 No λ = −1? ⇒ −1 + 7 − 6 = 0 Yes λ = 3? ⇒ 27 − 21 − 6 = 0 Yes λ1 = −1, λ2 = −2, λ3 = 3 Long division: λ2 − λ − 6 λ + 1 λ 3 + 0λ 2 − 7 λ − 6 Or you could find one root, then use synthetic (or long) division to find the quadratic factor. Synthetic division: −1 1 0 −7 −6 ↓ −1 1 6 1 −1 − 6 0 −(λ3 + λ2 ) − λ 2 − 7λ − ( −λ 2 − λ ) ⇒ λ − 7λ − 6 = ( λ + 1) ( λ − λ − 6 ) 3 2 − 6λ − 6 λ 3 − 7λ − 6 = 0 ⇒ ( λ + 1)( λ + 2 )( λ − 3) = 0 − ( −6λ − 6 ) 0 −1 1 0 Find the eigenvectors of 1 2 1 0 3 −1 Eigenvalues: λ1 = −1, λ2 = −2, λ3 = 3 For λ1 = −1 z is free, let z = 1 ( A − λ I ) x = 0 ⇒ ( A + I ) v1 = 0 0 1 0 1 3 1 0 3 0 0 R1 ↔ R2 0 0 | | | 1 3 1 0 1 0 0 3 0 | | | 0 0 0 −3R2 + R3 1 3 1 0 1 0 0 0 0 | | | 0 x + 3y + z = 0 ⇒ x = −z 0 ⇒ y = 0 0 −1 v1 = 0 1 For λ2 = −2 z is free, let z = 3 ( A − λ I ) x = 0 ⇒ ( A + 2 I ) v2 = 0 1 1 0 1 4 1 0 3 1 | | | 0 0 − R1 + R2 0 1 1 0 0 3 1 0 3 1 | | | 0 0 0 − R2 + R3 1 1 0 0 3 1 0 0 0 | | | 1 0 x + y = 0 ⇒ x = −y ⇒ x = 3 z 0 3 y + z = 0 ⇒ y = −31 z 0 1 v2 = −1 3 For λ3 = 3 ( A − λ I ) x = 0 ⇒ ( A − 3I ) v3 = 0 −4 1 0 | 0 R1 ↔ R2 1 −1 1 | 0 0 3 −4 | 0 x − y + z = 0 ⇒ x = y − z ⇒ x = 13 z 1 −1 1 | 0 1 −1 1 | 0 1 −1 1 | 0 4R + R 4 2 0 −3 4 | 0 0 −3 4 | 0 −3 y + 4 z = 0 ⇒ y = 3 z −4 1 0 | 0 1 0 3 −4 | 0 0 3 −4 | 0 R + R 0 0 0 | 0 2 3 1 z is free, let z = 3 v3 = 4 3 3 8.8 # 18 1 6 0 Find the eigenvalues and eigenvectors of 0 2 1 . 0 1 2 det ( A − λ I ) = 0 ⇒ 1− λ 6 0 0 2−λ 1 0 1 2−λ = (1 − λ ) 2−λ 1 1 2−λ = (1 − λ ) ( 2 − λ )( 2 − λ ) − 1 = 0 (1 − λ ) 4 − 4λ + λ 2 − 1 = 0 (1 − λ ) λ 2 − 4λ + 3 = 0 ⇒ (1 − λ )( λ − 1)( λ − 3) = 0 λ1 = λ2 = 1, λ3 = 3 For λ1 = λ2 = 1 For λ3 = 3 ( A − I ) v1 = 0 0 6 0 0 1 1 0 1 1 0 1 1 0 6 0 0 0 0 | | | | | | 0 0 0 − R3 + R2 0 6 0 0 1 1 0 0 0 0 0 0 −6 R1 + R2 0 0 0 x is free, let x = 1 1 0 0 | | | 1 −6 0 ( A − 3I ) v3 = 0 0 R1 ↔ R2 0 0 | | | −2 6 0 0 −1 1 0 1 −1 0 y = −z ⇒ y = 0 0 ⇒ z =0 0 1 0 0 | | | −3 −1 0 1 | | 0 0 | 0 0 0 0 0 0 −1 2 R1 1 0 0 −3 −1 0 1 | | 1 −1 | 0 0 0 R2 + R3 ⇒ x = 3 y ⇒ x = 3z ⇒y=z z is free, let z = 1 1 v1 = 0 0 3 v3 = 1 1 8.8 # 14 7 0 Find the eigenvalues and eigenvectors of . 0 13 7 0 0 13 is a diagonal matrix (its eigenvalues are on the diagonal) λ1 = 7 and λ2 = 13 For λ1 = 7 For λ2 = 13 ( A − 7 I ) v1 = 0 ( A − 13I ) v2 = 0 0 0 0 6 1 v1 = 0 | | 0 x is free,let x = 1 0 ⇒ y = 0 −6 0 0 | 0 | 0 ⇒x=0 0 y is free,let y = 1 0 v2 = 1 The eigenvectors of a diagonal matrix will be the columns of the identity matrix. 4