Determinants and Eigenvalues Facts About Determinants Amazing

Determinants and Eigenvalues Facts About Determinants Amazing

Determinants and eigenvalues Math 40, Introduction to Linear Algebra Wednesday, February 15, 2012 Amazing facts about determinants det A can be found by “expanding” along any row or any column Theorem. The determinant of a triangular Consequence: matrix is the product of its diagonal entries. 123 4 056 7 A = det(A)=1 5 8 10 = 400 008 9 · · · 0 0 0 10 Amazing facts about determinants EROs barely change the determinant, and they do so in a predictable way. EROs effect on det A Strategy to compute swap two rows changes sign det A more quickly for general matrices A = multiply row multiply det ⇒ by scalar c by scalar c Perform EROs to get REF of A and add c ຘ row i no change at all! compute det A based to row j on det of REF Amazing facts about determinants Theorem. A square matrix A is invertible if and only if det A =0. det(A) = det(AT ) determinant is multiplicative det(AB) = det(A) det(B) Consequence: 1 1 det(A) det(A− ) = det(AA− ) = det(I)=1 det(A 1)= 1 ⇒ − det A 1 1 If A is invertible, det(A− )= det A . Example using properties of determinant Example If det A = -3 for a 5 x 5 matrix A, find the determinant of the matrix 4A3. We have det(4A3) =45 det(A3) =45[det(A)]3 = 45( 3)3 − Another property of the determinant? Question True or false: det(A + B) = det A + det B ? False! Consider 11 1 1 A = ,B= − 11 11 − Then 20 A + B = det(A + B)=4= 0 = det A + det B 02 Eigenvalues and eigenvectors Introduction to eigenvalues Let A be an n x n matrix. If Ax = λx for some scalar λ and some nonzero vector x, then we say λ is an eigenvalue of A and x is an eigenvector associated with λ. n n Viewed as a linear transformation from R to R A sends vector x to a scalar multiple of itself ( λ x ) . Eigenvalues, eigenvectors for a 2x2 matrix 12 A = 54 12 2 12 2 Notice that = =6 54 5 30 5 12 6 6 =6 54 15 15 eigenvectors eigenvalues 6 2 =3 12 1 1 1 15 5 − = =( 1) − 54 1 1 − 1 − Any (nonzero) scalar multiple of an eigenvector is itself an eigenvector (associated w/same eigenvalue). A(cx)=c(Ax) = c(λx)=λ(cx) Graphic demonstration of eigenvalues and eigenvectors: eigshow eigshow demonstrates how the image A x changes as we rotate a unit vector x in R 2 around a circle in particular, we are interested in knowing when A x is parallel to x Finding eigenvalues of A We want nontrivial solutions to Ax = λx Ax λx = 0 Ax λIx = 0 ⇐⇒ − ⇐⇒ − (A λI)x = 0 When does this ⇐⇒ − homogeneous system have a solution other than x = 0 ? Must have that A λ I is not invertible, − which means that det(A λI)=0 − eigenvalues of A given eigenvalue λ, ⇒ associated eigenvectors are find values of λ nonzero vectors in such that det(A λI)=0 null(A λI) − − Example of finding eigenvalues and eigenvectors Find eigenvalues and 10 1 Example − A = 2 15 corresponding eigenvectors of A. − 00 2 1 λ 0 1 − − 0 = det(A λI) = 2 1 λ 5 − − − 002λ − 1 λ 0 characteristic = (2 λ) − − 2 1 λ polynomial − − λ3 +2λ2 + λ 2 − − = (2 λ)(1 λ)( 1 λ) − − − − λ =2, 1, or 1 − Example of finding eigenvalues and eigenvectors Example Find eigenvalues and 10 1 − A = 2 15 corresponding eigenvectors of A. − 00 2 λ =2, 1, or 1 − λ =2 Solve (A 2I)x = 0. − 10 1 0 10 1 0 − − EROs [ A 2I 0]= 2 350 01 1 0 − | − −−−→ − 0000 00 0 0 x x 1 eigenvectors of A for λ = 2 are 1 − 3 − x = x = x3 = x3 1 1 2 − x3 x3 1 c 1 for c =0 for any x3 R 1 ∈ Example of finding eigenvalues and eigenvectors Find eigenvalues and 10 1 Example − A = 2 15 corresponding eigenvectors of A. − 00 2 λ =2, 1, or 1 − 1 − eigenvectors of A for λ = 2 are c 1 for c =0 λ =2 Solve (A 2I)x = 0. − 1 E = eigenspace of A = set of all eigenvectors 0 2 for λ =2 of A for λ =2 ∪ { } = null( A 2I) − 1 − = span 1 1 Example of finding eigenvalues and eigenvectors Example Find eigenvalues and 10 1 − A = 2 15 corresponding eigenvectors of A. − 00 2 λ =2, 1, or 1 − 1 − λ =2 E2 = span 1 1 1 λ =1 Solve (A I)x = 0. = E1 = span 1 − ⇒ 0 0 λ = 1 Solve (A + I)x = 0. = E 1 = span 1 − − ⇒ 0 Eigenvalues, eigenvalues... where are you? 11 Example Find eigenvalues of A. A = 11 − 1 λ 1 0 = det(A λI) = − = (1 λ)2 +1 − 11λ − − − 2 = λ 2λ +2 − 2 √4 8 = λ = ± − =1 i ⇒ 2 ± Eigenvalues are complex!.

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