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Change of Basis Consider a Linear Transform, PB , and Its Inverse, P B Change of Basis Consider a linear transform, PB, and its −1 inverse, PB , which map a vector back and forth between its representation in the standard basis and its representation in the basis, B: PB −→ u ← [u]B . −1 PB Change of Basis (contd). Let B consist of N basis vectors, b1 ...bN. Since [u]B is the representation of u in B, it follows that ... u = ([u]B)1 b1 +([u]B)2 b2 + ([u]B)N bN But this is just the matrix vector prod- uct u = B[u]B where B = b1 b2 ... bN . −1 −1 We see that PB = B and PB = B. Similarity Transforms Now consider a linear transformation rep- resented in the standard basis by the ma- trix A. We seek [A]B, i.e., the represen- tation of the corresponding linear trans- formation in the basis B: u −→A Au ↑ B ↓ B−1 [A]B [u]B −→ [Au]B The matrix we seek maps [u]B into [Au]B. From the above diagram, we see that this matrix is the composition of B, A, and B−1: −1 [A]B = B AB. We say that A and [A]B are related by a similarity transform. Diag. of Symmetric Matrices Because of linearity, one might expect that a transformation will have an espe- cially simple representation in the ba- sis of its eigenvectors, X . Let A be its representation in the standard basis and let the columns of X be the eigenvec- tors of A. Then X and XT = X−1 take us back and forth between the standard basis and X : XT −→ u ←− [u]X . X Diag. of Symmetric Matrices (contd.) The matrix we seek maps [u]X into [Au]X : u −→A Au ↑ X ↓ XT [A]X [u]X −→ [Au]X From the above diagram, we see that this matrix is the composition of X, A, and XT: Λ = XTAX. We observe that Λ is diagonal with Λii = λi, the eigenvalue of A associated with eigenvector, xi. Spectral Thm. for Sym. Matrices Any symmetric N × N matrix, A, with N distinct eigenvalues, can be factored as follows: A = XΛXT where Λ is N × N and diagonal, X and XT are N ×N matrices, and the i-th col- umn of X (equal to the i-th row of XT) is an eigenvector of A: λixi = Axi with eigenvalue Λii = λi. Note that xi is orthogonal to x j when i =6 j: 1 if i = j XXT = δ = ij ij 0 otherwise. In other words, XXT = I. Consequently, XT = X−1. Spectral Thm. for Sym. Matrices (contd). Using the definition of matrix product and the fact that Λ is diagonal, we can write A = XΛXT as N Λ T . (A)ij = ∑ (X)ik kk X kj k=1 Since X = x1 x2 ... xN and Λkk = λk N λ (A)ij = ∑ (xk)i k (xk) j k=1 N λ T = ∑ kxkxk ij k=1 N λ T A = ∑ kxkxk k=1 where λkxk = Axk. Spectral Thm. for Sym. Matrices (contd). The spectral factorization of A is: λ T λ T λ T . A = 1x1x1 + 2x2x2 + ··· + NxNxN λ T Note that each nxnxn is a rank one ma- λ T trix. Let Ai = ixixi . Now, because T xi xi = 1: λ λ T ixi = ixixi xi = Aixi i.e., xi is the only eigenvector of Ai and its only eigenvalue is λi. Diag. of Non-symmetric Matrices The situation is more complex when the transformation is represented by a non- symmetric matrix, P. Let the columns of X be P’s right eigenvectors and the rows of YT be its left eigenvectors. Then X and YT = X−1 take us back and forth between the standard basis and X : YT −→ u ←− [u]X . X Diag. of Non-symmetric Matrices (contd.) The matrix we seek maps [u]X into [Pu]X : u −→P Pu ↑ X ↓ YT Λ [u]X −→ [Pu]X From the above diagram, we see that this matrix is the composition of X, P, and YT: Λ = YTPX We observe that Λ is diagonal with Λii = λi, the eigenvalue of P associated with right eigenvector, xi, and left eigenvec- tor, yi. Spectral Theorem Any N × N matrix, P, with N distinct eigenvalues, can be factored as follows: P = XΛYT where Λ is N × N and diagonal, X and YT are N ×N matrices, and the i-th col- umn of X is a right eigenvector of P: λixi = Pxi with eigenvalue Λii = λi and the i-th row of YT is a left eigenvector of P: λ T T iyi = yi P with the same eigenvalue. Spectral Theorem (contd.) Note that xi is orthogonal to y j when i =6 j: 1 if i = j XYT = δ = ij ij 0 otherwise. In other words, XYT = I. Consequently, YT = X−1. Spectral Theorem (contd). The spectral factorization of P is: λ T λ T λ T . P = 1x1y1 + 2x2y2 + ··· + NxNyN λ T Note that each nxnyn is a rank one ma- λ T trix. Let Pi = ixiyi . Now, because T yi xi = 1: λ λ T ixi = ixiyi xi = Pixi and λ T T λ T iyi = yi ixiyi T = yi Pi i.e., xi and yi are the sole right and left eigenvectors of Pi. The only eigenvalue is λi. Stochastic Matrices If P is stochastic, then 1 = ∑Pij. i Let yT = 1 1 ... 1 . It follows that T T . y j = ∑ y i Pij i Consequently, yT is a left eigenvector with unit eigenvalue of every stochastic matrix: yT = yTP. Stochastic Matrices (contd.) What is the representation of an arbi- trary distribution, z, in the basis of eigen- vectors of an arbitrary stochastic ma- trix, P? z = c1x1 + c2x1 + ··· + cNxN = Xc. Solving for c: c = X−1z = YTz. T ... Since y1 = 1 1 1 , it follows that: T c1 = ∑Y1 j z j = ∑z j = 1 j j which is independent of the specific z and P!.
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