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Download Applying transformations in succession Suppose that A and B are 2
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Applying transformations in succession Suppose that A and B are 2 × 2 matrices representing the maps TA and TB . Then, TB (TA(x)) = B (TA(x)) = B (A x) = (B A) x, so the combined transformation is represented by the matrix B A. That is, TB ◦ TA = TBA. Note that the above is “first TA, then TB ”, and not the other way around! Inverses of transformations The matrix for a rotation or a reflection is invertible: their determinants are non-zero. Suppose that A is the 2 × 2 matrix representing TA, and it has inverse A−1. Then TA−1 (TA (x)) = A−1(Ax) = (A−1A)x = I2x = x, and similarly, TA(TA−1 (x)) = x. So to get the matrix for the inverse, we just take the inverse of the matrix! Eigenvalues Definition A number λ is called an eigenvalue for A if there is a non-zero n × 1 vector v such that A v = λ v. The vector v is called an eigenvector for λ. Example Let A = 2 1 −4 7 ! . Then A 1 1 ! = 3 3 ! =3 1 1 ! , so λ = 3 is an eigenvalue of A with (1, 1)T a corresponding eigenvector. How can we find all of the eigenvectors of a given square matrix? Theorem Let A be a 2 × 2 matrix. Then λ is an eigenvalue of A if and only if (A − λ I2) is singular if and only if det (A − λ I2) = 0. Example Find all eigenvalues of the 2 × 2 matrix A = det (A − λ I2) = det = det 5 −6 2 −2 ! 1 0 0 1 −λ − : λ 0 0 λ ! !! !! 5−λ −6 2 −2 − λ (5 − λ) (−2 − λ) − (−6)2 −10 − 5 λ + 2 λ + λ2 + 12 λ2 − 3 λ + 2 (λ − 1) (λ − 2). = det = = = = 5 −6 2 −2 ! 5 −6 2 −2 ! Now, this determinant is zero exactly when λ = 1 or 2, so these are the eigenvalues of A. General method for the eigenvalue problem Definition Let A be an n × n matrix. Then the characteristic polynomial p for A is p(λ) = |A − λ In|. Corollary. The eigenvalues of A are the roots of the characteristic polynomial p. That is, the numbers λ such that p(λ) = 0. Solving the eigenvalue problem 1. Use determinants to calculate p(λ) = |A − λ In |. 2. Find the values λ such that p(λ) = 0. 3. For each λ, use linear algebra to solve the equation (A − λ In) v = 0. Example Find the eigenvectors in the previous example. First let λ = 1. Then A − λ In = 5 −6 2 −2 ! − 1 0 0 1 ! = 4 −6 2 −3 ! . We need to solve 4 −6 2 −3 ! 0 0 u= ! . The augmented matrix is 4 −6 0 2 −3 0 ! 1 R , R → 1 R give the RREF and the row operations R2 → R2 − 2 1 1 4 1 3 1 −2 0 0 0 0 ! . The general solution to this system is u2 = t, u1 = 3 2 t. Taking t = 2 (any non-zero t would do), we obtain an eigenvector u= 3 2 ! . It is easy to check that Au = 5 −6 2 −2 ! 3 2 ! = 3 2 ! = 1 u, so u is indeed an eigenvector for the eigenvalue 1. A similar calculation shows that v = (2, 1)T is an eigenvector for λ = 2.