Math 2331 - Linear Algebra Diagonalization

5.3 Diagonalization Math 2331 – Linear Algebra 5.3 Diagonalization Math 2331, Linear Algebra 1 / 18 5.3 Diagonalization Diagonalization Theorem Examples 5.3 Diagonalization Diagonalization Matrix Powers: Example Diagonalizable Diagonalization Theorem Diagonalization: Examples Math 2331, Linear Algebra 2 / 18 5.3 Diagonalization Diagonalization Theorem Examples Diagonalization The goal here is to develop a useful factorization A = PDP −1 , when A is n × n. We can use this to compute Ak quickly for large k. The matrix D is a diagonal matrix (i.e. entries off the main diagonal are all zeros). Powers of Diagonal Matrix D k is trivial to compute as the following example illustrates. Example   5 0 Let D = . Compute D 2 and D 3 . In general, what is D k , 0 4 where k is a positive integer? Math 2331, Linear Algebra 3 / 18 5.3 Diagonalization Diagonalization Theorem Examples Diagonalization (cont.) Solution: D2 D3 =  = D 2D 5 0 0 4  =   5 0 0 4 52 0 0 42   = 5 0 0 4 0  0   = 0  0 and in general, Dk  = 5k 0 0 4k  Math 2331, Linear Algebra 4 / 18 5.3 Diagonalization Diagonalization Theorem Examples Matrix Powers: Example Example  6 −1 . Find a formula for Ak given that Let A = 2 3     1 1 5 0 −1 A = PDP where P = ,D= and P −1 = 1 2 0 4   2 −1 . −1 1  Solution: A2 = PDP −1


PDP −1 = PD P −1 P DP −1 = PDDP −1 = PD 2 P −1 Again A3 = A2 A = PD 2 P −1


PDP −1 = PD 2 P −1 P DP −1 = PD 3 P −1 Math 2331, Linear Algebra 5 / 18 5.3 Diagonalization Diagonalization Theorem Examples Matrix Powers: Example (cont.) In general, Ak = PD k P −1  =  = 1 1 1 2  2 · 5k − 4k 2 · 5k − 2 · 4k 5k 0 0 4k  −5k + 4k −5k + 2 · 4k 2 −1 −1 1   . Diagonalizable A square matrix A is said to be diagonalizable if A is similar to a diagonal matrix, i.e. if A = PDP −1 where P is invertible and D is a diagonal matrix. Math 2331, Linear Algebra 6 / 18 5.3 Diagonalization Diagonalization Theorem Examples Diagonalizable When is A diagonalizable? (The answer lies in examining the eigenvalues and eigenvectors of A.) Note that      6 −1 1 1 =5 , 2 3 1 1  6 −1 2 3  1 2  6 −1 2 3   = 1 1 1 2  5 0 0 4  Math 2331, Linear Algebra 1 1 1 2 1 2 0  =4 Altogether        6 −1 1 1 5 4 1 1 = = 0 2 3 1 2 5 8 1 2 Equivalently,    −1 7 / 18 5.3 Diagonalization Diagonalization Theorem Examples Diagonalizable (cont.) In general, A  and if  v1 v2 v1 v2 · · · v1 v2 · · · v1 v2 · · · vn  λ1 0    0 λ2 · · · vn  . ..  .. . 0 0 vn      vn   λ1 0 .. . 0   = ··· ··· .. . ···  0 0     λn is invertible, A equals  0 ··· 0 λ2 · · · 0    v1 v2 · · · .. . .  . . 0 ··· vn −1 λn Math 2331, Linear Algebra 8 / 18 5.3 Diagonalization Diagonalization Theorem Examples Diagonalization Theorem Theorem (Diagonalization) An n × n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. In fact, A = PDP −1 , with D a diagonal matrix, if and only if the columns of P are n linearly independent eigenvectors of A. In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors in P. Math 2331, Linear Algebra 9 / 18 5.3 Diagonalization Diagonalization Theorem Examples Diagonalization: Example Example Diagonalize the following matrix, if  2  1 A= −1 possible.  0 0 2 1  0 1 Step 1. Find the eigenvalues of A.   2−λ 0 0 2−λ 1  det (A − λI ) = det  1 −1 0 1−λ = (2 − λ)2 (1 − λ) = 0. Eigenvalues of A: λ = 1 and λ = 2. Math 2331, Linear Algebra 10 / 18 5.3 Diagonalization Diagonalization Theorem Examples Diagonalization: Example (cont.) Step 2. Find three linearly independent eigenvectors of A. By solving (A − λI ) x = 0, for each value of λ, we obtain the following:  0 v1 =  −1  1  Basis for λ = 1:  Basis for λ = 2:  0 v2 =  1  , 0 Math 2331, Linear Algebra   −1 v3 =  0  1 11 / 18 5.3 Diagonalization Diagonalization Theorem Examples Diagonalization: Example (cont.) Step 3: Construct P from the vectors in step 2.   0 0 −1 0  P =  −1 1 1 0 1 Step 4: Construct D from the corresponding eigenvalues.   1 0 0 D= 0 2 0  0 0 2 Math 2331, Linear Algebra 12 / 18 5.3 Diagonalization Diagonalization Theorem Examples Diagonalization: Example (cont.) Step 5: Check your work by verifying that AP = PD      2 0 0 0 0 −1 0 0 −2 0  =  −1 2 0  AP =  1 2 1   −1 1 −1 0 1 1 0 1 1 0 2      0 0 −1 1 0 0 0 0 −2 0   0 2 0  =  −1 2 0  PD =  −1 1 1 0 1 0 0 2 1 0 2 Math 2331, Linear Algebra 13 / 18 5.3 Diagonalization Diagonalization Theorem Examples Diagonalization: Example Example Diagonalize the following matrix,  2  A= 0 0 if possible.  4 6 2 2 . 0 4 Since this matrix is triangular, the eigenvalues are λ1 = 2 and λ2 = 4. By solving (A − λI ) x = 0 for each eigenvalue, we would find the following:     1 5    λ1 = 2 : v1 = 0 , λ2 = 4 : v2 = 1  0 1 Every eigenvector of A is a multiple of v1 or v2 which means there are not three linearly independent eigenvectors of A and by Theorem 5, A is not diagonalizable. Math 2331, Linear Algebra 14 / 18 5.3 Diagonalization Diagonalization Theorem Examples Diagonalization: Example Example   2 0 0 Why is A =  2 6 0  diagonalizable? 3 2 1 Solution: Since A has three eigenvalues: λ1 = , λ2 = , λ3 = and since eigenvectors corresponding to distinct eigenvalues are linearly independent, A has three linearly independent eigenvectors and it is therefore diagonalizable. Theorem (6) An n × n matrix with n distinct eigenvalues is diagonalizable. Math 2331, Linear Algebra 15 / 18 5.3 Diagonalization Diagonalization Theorem Examples Diagonalization: Example Example Diagonalize the following matrix, if possible.  −2 0 0 0  0 −2 0 0 A=  24 −12 2 0 0 0 0 2     Solution: Eigenvalues: −2 and 2 (each with multiplicity 2). Solving (A − λI ) x = 0 yields the following eigenspace basis sets.  Basis for λ = −2 :  1  0   v1 =   −6  0 Math 2331, Linear Algebra   0  1   v2 =   3  0 16 / 18 5.3 Diagonalization Diagonalization Theorem Examples Diagonalization: Example (cont.)  Basis for λ = 2 :  0  0   v3 =   1  0   0  0   v4 =   0  1 {v1 , v2 , v3 , v4 } is linearly independent ⇒ P = [v1 v2 v3 v4 ] is invertible ⇒ A = PDP −1 , where  1 0 0 0  0 1 0 0 P=  −6 3 1 0 0 0 0 1      and  −2 0 0 0  0 −2 0 0  . D=  0 0 2 0  0 0 0 2 Math 2331, Linear Algebra 17 / 18 5.3 Diagonalization Diagonalization Theorem Examples Diagonalization: Theorem Theorem (7) Let A be an n × n matrix whose distinct eigenvalues are λ1 , . . . , λ p . a. For 1 ≤ k ≤ p, the dimension of the eigenspace for λk is less than or equal to the multiplicity of the eigenvalue λk . b. The matrix A is diagonalizable if and only if the sum of the dimensions of the distinct eigenspaces equals n, and this happens if and only if the dimension of the eigenspace for each λk equals the multiplicity of λk . c. If A is diagonalizable and βk is a basis for the eigenspace corresponding to λk for each k, then the total collection of vectors in the sets β1 , . . . , βp forms an eigenvector basis for Rn . Math 2331, Linear Algebra 18 / 18