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Eigenvectors and Linear Transformations

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Eigenvectors and Linear Transformations Eigenvectors and linear transformations Brian Krummel October 25, 2019 1 Matrix of a linear transformation relative to bases Recall that if T : Rn ! Rm is a linear transformation, then there is a m × n matrix A such that T (x) = Ax for all x in Rn. In particular, A = T (e1) T (e2) ··· T (en) : n Here we secretly used the fact that the standard basis fe1; e2;:::; eng is a basis for R (and m similarly on R ). Thus the j-th column of A are precisely the value T (ej) of T applied to the standard basis vector ej. In some problems, it might be more convenient to choose a basis other than the standard bases for Rn and Rm. For instance, we might choose to work with a basis of eigenvectors. Moreover, we can consider linear transformations T : V ! W between finite-dimensional abstract vector spaces V and W . By representing vectors in V and W by their coordinate vectors for a basis, we can turn T into an equivalent linear transformation Te : Rn ! Rm, which is given by matrix multiplication. Let T : V ! W be a linear transformation between any pair of finite-dimensional vector spaces V and W . Let B = fb1; b2;:::; bng be a basis for V and D = fd1; d2;:::; dmg be a basis for W . We can express each vector x in V as a linear combination x = r1b1 + r2b2 + ··· + rnbn (1) for unique scalar weights rj and thus we can represent the vector x in V by its coordinate vector n [x]B in R : 2 3 r1 6 r 7 6 2 7 [x]B = 6 . 7 : (2) 4 . 5 rn m Similarly, we can represent each vector y in W by its coordinate vector [y]D in R . We want to find a m × n matrix M such that the linear transformation T is equivalent to multiplication of the coordinate vectors by M: y = T (x) means [y]D = M[x]B: 1 Some equivalent characterizations of M are [T (x)]D = M[x]B for each x in V (by substituting y = T (x) into [y]D = M[x]B), and 2 3 r1 6 r 7 6 2 7 [T (r1b1 + r2b2 + ··· + r1bn)]D = M 6 . 7 (3) 4 . 5 rn for every choice of coordinates rj (by (1) and (2)). We call M the matrix of T relative to the bases B and D. For each j = 1; 2; : : : ; n, by setting rj = 1 and rk = 0 for k 6= j so that x = bj in (1) and [x]B = [bj]B = ej in (2), we conclude from (3) that the j-th column of M is Mej = [T (bj)]D so that M = [T (b1)]D [T (b2)]D ··· [T (bn)]D : As a diagram: x in V T T(x) in W -Coordinates -Coordinates B D n Multiply by M m [x]B in R [T(x)]D in R 2 Example 1. Recall that P2 is the space of all polynomials a0 + a1x + a2x of degree at most 2. 2 Let T : P2 ! R be the linear transformation defined by p(2) T (p) = : p(3) Recall that a good way to solve linear algebra problems involving polynomials is to associate a polynomial with its coordinate vector relative to the basis B = f1; x; x2g: 2 3 a0 2 p(x) = a0 + a1x + a2x with [p]B = 4 a1 5 : a2 2 2 We will use the standard basis for R . We want to find 2 × 3 matrix M such that T (p) = M[p]B: 2 a 3 ∗ ∗ ∗ ∗ ∗ ∗ 0 M = such that T (a + a x + a x2) = a : ∗ ∗ ∗ 0 1 2 ∗ ∗ ∗ 4 1 5 a2 Notice that for the basis polynomials 1; x; x2 1 2 22 4 T (1) = ;T (x) = ;T (x2) = = : 1 3 32 9 Placing T (1), T (x), and T (x2) as the 1st, 2nd, and 3rd columns of M respectively, 1 2 4 M = T (1) T (x) T (x2) = : 1 3 9 Example 2. Suppose V is an abstract two-dimensional vector space with basis B = fb1; b2g and W is another vector space with basis D = fd1; d2; d3g. Let T : V ! W be a linear transformation defined by T (b1) = 3d1 + 5d2 + 7d3 T (b2) = 6d1 + 8d2 + 9d3: (4) By taking T (bj) and writing the coefficients of di as the columns of a vector, we find that the D-coordinates of T (b1) and T (b2) are 2 3 3 2 6 3 [T (b1)]D = 4 5 5 [T (b2)]D = 4 8 5 : 7 9 Placing the D-coordinates of T (b1) and T (b2) as the 1st and 2nd columns of M respectively, we obtain 2 3 6 3 M = [T (b1)]D [T (b2)]D = 4 5 8 5 : 7 9 Notice that in (4), we wrote T (b1) and T (b2) from left-to-right, but in M we wrote their coordi- nates as columns of M going up-and-down (this looks like we took a transpose). 2 Matrix of a transformation and similarity Notice that we could have done the same process with T : V ! V and B as a basis for V . Then T has a matrix relative to the basis B, which we denote by M = [T ]B so that [T (x)]B = [T ]B[x]B: n Now let B = fb1; b2;:::; bng is any basis for R . (For instance, B might be a basis of eigenvectors for a diagonalizable matrix A.) Recall that to solve problems with a basis B, we often did so by placing the basis vectors bj as the columns of an n × n matrix P : P = b1 b2 ··· bn : 3 Since B is a basis, P is invertible. Let x is a vector in Rn and express x as x = r1b1 + r2b2 + ··· + rnbn for scalar weights ri so that the B-coordinate vector of x is 2 3 r1 6 r 7 6 2 7 [x]B = 6 . 7 : 4 . 5 rn Then 2 3 r1 6 r 7 6 2 7 P [x]B = b1 b2 ··· bn 6 . 7 = r1b1 + r2b2 + ··· + rnbn = x: 4 . 5 rn Multiplying by P −1: −1 [x]B = P x for each x in Rn. Hence multiplication by P −1 transforms the vector x in Rn into its B-coordinate vector [x]B. Let T : Rn ! Rn defined by T (x) = Ax in standard coordinates and let's find the matrix M = [T ]B of T relative to B coordinates. We can express the behavior of M = [T ]B as the diagram: T x in V T(x) in V Multiply by A -1 -1 P P -Coordinates -Coordinates Multiply by Multiply by B B n Multiply by B n [x]B in R [T(x)]B in R For each vector x in Rn, multiplying A times x is the same as converting x to B-coordinates by multiplying by P −1, then multiplying by the matrix M, then converting back to standard coordinates by multiplying by P . This gives us Ax = PMP −1x for each x in Rn. That is, A = PMP −1: Equivalently, −1 M = [T ]B = P AP: Therefore, A is similar to M if after some change of basis A becomes M. 4 Example 3 (Old example revised). Recall from a few lectures back that 5 1 A = 1 5 has 1 eigenvector b = with eigenvalue 6 1 1 −1 eigenvector b2 = with eigenvalue 4: x21 Ae2 Thus 5 1 1 −1 6 0 1 −1 −1 A = = : 1 5 1 1 e2 0 4 1 1 2 Thus after changing the basis for R to the basis ofe1 eigenvectorsAe1 x1 B = fb1; b2g, A becomes the diagonal matrix 6 0 D = : 0 4 Geometrically, this means: x2 Av1 v1 x1 v2 x2 x2 Av2 T Example 4. Suppose V = R2 with the basis x1 x1 1 0 B = b = ; b = : 1 1 2 −1 What this means is that the basis B provides a coordinate grid for R2: x2 x1 We can describe every vector in R2 uniquely in terms of its coordinates on this grid; for instance: 1 1 0 1 x = = + 2 ) [x] = −1 1 −1 B 2 Notice that when we express x as a column vector in the usual way, this assumes that we have the standard basis fe1; e2g with the standard left-to-right, up-and-down coordinate grid. But the choice of coordinate grid, i.e. the choice of basis, is not sacred and we can describe directed line segments in terms of any basis we want. 5 Suppose we are given a linear transformation T : R2 ! R2 defined by −1 1 T (x) = Ax = x −2 1 for each x in R2. Setting 1 0 P = b b = ; 1 2 1 −1 we compute M = [T ]B equals 1 0 −1 −1 1 1 0 1 0 −1 1 1 0 M = P −1AP = = 1 −1 −2 1 1 −1 1 −1 −2 1 1 −1 1 0 0 −1 0 −1 = = : 1 −1 −1 −1 1 0 Therefore the matrix [T ]B of the transformation T relative to the basis B is 0 −1 [T ] = : B 1 0 This means that T (b1) = b2;T (b2) = −b1: By direct computation −1 1 1 0 T (b ) = = = b ; 1 −2 1 1 −1 2 −1 1 0 −1 T (b ) = = = −b : 2 −2 1 −1 −1 1 Geometrically, the basis B determines a coordinate grid on R2.
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