Lecture : Diagonalization A square matrix D is called diagonal if all but diagonal entries are zero: a a D a n 5 n n. () Diagonal matrices are the simplest matrices that are basically equivalent to vectors in R n. Obviously, D has eigenvalues a, a,..., a n, and for each i,,..., n, a......... D e i a i a. i a i e i....... 5. 5. 5 a n We thus conclude that for any diagonal matrix, eigenvalues are all diagonal entries, and e i is an eigenvector associated with a i, the ith diagonal entry, i.e., e i is an eigenvector associated with eigenvalue a i, for i,,..., n. Due to the simplicity of diagonal matrices, one likes to know whether any matrix can be similar to a diagonal matrix. Diagonalization is a process of nding a diagonal matrix that is similar to a given non-diagonal matrix. De nition.. An n n matrix A is called diagonalizable if A is similar to a diagonal matrix D. Example.. Consider A, D 5, P. (a) Verify A P DP (b) Find D k and A k (c) Find eigenvalues and eigenvectors for A. Solution: (a) It su ces to show that AP P D and that P is invertible. Direct calculations lead to det P ) P is invertible Again, direct computation leads to 5 AP, P D 5 5. 5
Therefore AP P D. (b) D 5 5 5 5k, D k k. A P DP P DP P DP P DP P D P A k P D k P (c) Eigenvalues of A Eigenvalues of D, which are λ 5, λ. For D, we know that e is an eigenvectors for λ 5 : D e 5 e () e is an eigenvectors for λ : D e e. () Since AP P D, from () and (), respectively, we see Therefore, AP e P D e P (5 e ) 5P e, AP e P D e P ( e ) P e. P e is an eigenvector associated with eigenvalue λ 5 for matrix A, and P e is an eigenvector of A associated with eigenvalue λ. From this example, we observation that if A is diagonalizable and A is similar to a diagonal matrix D (as in ()) through an invertible matrix P, Then AP P D. P e i is an eigenvector associated with a i, for i,,..., n. This generalization can be easily veri ed in the manner analogous to Example.. Moreover, these n eigenvectors P e, P e,..., P e n are linear independent. To see independence, we consider the linear system nx λ i P e i. i
Suppose the linear system admits a solution (λ,..., λ n ). Then à nx P λ i e i. i Since P is invertible, the above equation leads to nx λ i e i. i Since has e, e,..., e n are linear independent, it follows that the last linear system has only the trivial solution λ i for all i,,..., n. Therefore, P e, P e,..., P e n are linear independent. This observation leads to Theorem.. (Diagonalization).A n n matrix A is diagonalizable i A has n linearly independent eigenvectors. Furthermore, suppose that A has n linearly independent eigenvectors f v, v,..., v n g. Set P [ v, v,..., v n ], D a a a n where a i is the eigenvalue associated with v i, i.e., A v i a i v i.then, P is invertible and P AP D. Proof. We have demonstrated that if A is diagonalizable, then A has n linearly independent eigenvectors. We next show that the reverse is also true: if A has n linearly independent eigenvectors, then A must be diagonalizable. To this end, we only need to verify that AP P D with P and D described above, since P is obviously invertible due to independence of eigenvectors. The proof of AP P D is straightforward by calculation as follows: AP A [ v, v,..., v n ] [A v, A v,..., A v n ] [a v, a v,..., a n v n ], 5 P D P [a e, a e,..., a n e n ] [a P e, a P e,..., a n P e n ]. Now, P e [ v, v,..., v n ] 5 v, P e [ v, v,..., v n ] 5 v,...
This shows AP P D. Example.. Diagonalize A 5 5. Solution: Diagonalization means nding a diagonal matrix D and an invertible matrix P such that AP P D. We shall follow Theorem. step by step. Step. Find all eigenvalues. From computations λ det (A λi) det 5 λ 5 λ λ λ + λ + λ λ + λ + λ + λ + λ (λ ) λ (λ ) λ + (λ+ ) (λ ) (λ+ ). we see that A has eigenvalues λ, λ (this is a double root). Step. Find all eigenvalues nd a basis for each eigenspace Null (A λi i ). For λ, A λ I 5 5 5 5 R R 5 R +R R 5 5. Therefore, linear system reduces to (A I) x x x x + x,
where x is the free variable. Choose x, we obtain an eigenvector x 5. For λ, The corresponding linear system is A λ I 5 5 ( ) 5 5 5. x + x + x It follows that x and x are free variables. As we did before, we need to select (x, x ) to be (, ) and (, ). Choose x, x ) x x x ; choose x, x ) x x x. We thus got two independent eigenvectors for λ : v x x 5 5, v x x 5 5. x x Step. Assemble orderly D and P as follows: there are several choices to pair D and P. Choice# : D 5, P [ v, v, v ] 5 Choice# : D 5, P [ v, v, v ] 5 Choice# : D 5, P [ v, v, v ] 5. Remark. Not every matrix is diagonalizable. For instance, A, det (A λi) (λ ). The only eigenvalue is λ ; it has the multiplicity m. From A I, 5
we see that (A I) has only one pivot. Thus r (A I). By Dimension Theorem, we have dim Null (A I) r (A). In other words, the basis consists of Null (A I) consists of only one vector. For instance, one may choose as a basis. According to the discussion above, if A is diagonalizable, i.e., AP P D for a diagonal matrix a D b then Furthermore, since P is invertible, P e is an eigenvector of A associated with a, P e is an eigenvector of A associated with b. P e and P e are linearly independent (why?). Since we have already demonstrated that there is no more than two linearly independent eigenvectors, it is impossible to diagonalize A. In general, if λ is an eigenvalue of multiplicity m (i.e., the characteristic polynomial then det (A λi) (λ λ ) m Q (λ), Q (λ ) ), dim (Null (A λi)) m. Theorem.. Let A be an n n matrix with distinct real eigenvalues λ, λ,..., λ p with multiplicity m, m,..., m p, respectively. Then,. n i dim (Null (A λ i I)) m i and m + m +... + m p n.. A is diagonalizable i n i m i for all i,,..., p, and m + m +... + m p n. In this case, we have P AP D, where P and D can be obtained as follows. Let B i be a basis of Null (A λ i I) for each i. Then P [B,..., B p ], D λ I m............ 5, I mi (m i m i ) identity... λ p I mp i.e., the rst m columns of P form B, the eigenvectors for λ, the next m columns of P are B, then B, etc. The last m p columns of P are from B p ; the rst m diagonal entries of D are λ, the next m diagonal entries of D are λ, and so on.
. In particular, if A has n distinct real eigenvalues, then A is diagonalizable.. Any symmetric matrix is diagonalizable. Note that, as we saw before, there are multiple choices for assembling P. For instance, if A is 5 5, and n A has two eigenvalues λ, λ with a basis f a, a g for Null (A I) and a basis b, b, o b for Null (A I), respectively, then, we have several choices to select pairs of (P, D) : choice# : P h a, a, b, b, i I b, D I 55 5 choice# : P h a, b, b, a, i b, D 5. Example.. Diagonalize A A 5 5 Solution: We need to nd all eigenvalues and eigenvectors. Since A is an upper triangle matrix, we see that eigenvalues are λ 5, m, λ, m. For λ 5, A 5I 8 5 R+RR 8 5 8 8 8 8 5 8 5 5.
Therefore, x and x are free variable, and x 8x x x x + x. Choose (x, x ) (, ) ) x 8, x ; Choose (x, x ) (, ) ) x, x. We obtain two independent eigenvectors 8 5, 5 (for λ 5). For λ, Hence A ( ) I 8 8 R R R R R R 8 5 R +R R 8 5 8 5 R 8R R 5. x, x. Choose (x, x ) (, ) and (x, x ) (, ), respectively, we have eigenvectors 5 and 5 (for λ ). Assemble pairs (P, D) : There are several choice. For instance, 8 5 P 5, D 5 5 or 8 5 P 5, D 5 5. 8
² Homework. ². Suppose that A is similar to D 5. (a) Find the characteristic polynomial and eigenvalues of A. (b) Let P AP D, where 5 P 5. Find a basis for each eigenspace of A.. Diagonalize the following matrices. (a) A 5, λ, 5. (b) B 5. Show that if A is diagonalizable, so is A.. Suppose that n n matrices A and B have exact the same linearly independent eigenvectors that span R n. Show that (a) both A and B can be simultaneously diagonalized (i.e., there are the same P such that P AP and P BP are diagonal matrices), and (b) AB is also diagonalizable. 5. for each statement, determine and explain true or false. (a) A is diagonalizable if A P DP for some diagonal matrix D. (b) If A is diagonalizable, then A is invertible. (c) If AP P D with D diagonal matrix, then nonzero columns of P are eigenvectors of A. (d) Let A be a symmetric matrix and λ be an eigenvalue of A with multiplicity 5. Then the eigenspace Null (A λi) has dimension 5. 9