Exercises: Similariy Transformaion Problem. Diagonalize he following marix: A [ 2 4 Soluion. Marix A has wo eigenvalues λ 3 and λ 2 2. Since (i) A is a 2 2 marix and (ii) i has 2 disinc eigenvalues, we can apply he diagonalizaion [ mehod we discussed in class. Specifically, we obain an arbirary eigenvecor v of λ, say v and, and an arbirary 2 eigenvecor v 2 of λ 2, say v 2 [. Then, we form: Q [ 2 by using v and v 2 as he firs and second columns, respecively. Q has he inverse: [ Q 2 We hus obain he following diagonalizaion of A: A Q diag[3, 2 Q. Problem 2. Consider again he marix A in Problem 5. Calculae A for any ineger. Soluion. We already know ha A: A Q diag[3, 2 Q. Hence: A Q diag[3, 2 Q [ [ [ 3 2 2 2 [ 3 + 2 + 3 + 2 2 3 2 + 2 3 2 Problem 3. Diagonalize he marix A Soluion. Recall ha all symmeric marices are diagonalizable. A is a 3 3 marix. The key is o find hree linearly independen eigenvecors..
From he soluion of Problem, we know ha A has eigenvalues λ and λ 2. x EigenSpace(λ ) includes all x 2 x 3 saisfying x u x 2 v x 3 u for any u, v R. The vecor space EigenSpace(λ ) has dimension 2 wih a basis {v, v 2 } where v (given by u, v ) and v 2 (given by u, v ). Similarly, EigenSpace(λ 2 ) includes all x x 2 x 3 x u x 2 saisfying x 3 u for any u R. The vecor space EigenSpace(λ 2 ) has dimension wih a basis {v 3 } where v 3 - (given by u ). So far, we have obained hree linearly independen eigenvecors v, v 2, v 3 of A. We can hen apply he diagonalizaion mehod exemplified in Problem 5 o diagonalize A. Specifically, we form: Q Q has he inverse: Q We hus obain he following diagonalizaion of A: /2 /2 /2 /2 A Q diag[,, Q. Problem 4. Suppose ha marices A and B are similar o each oher, namely, here exiss P such ha A P BP. Prove: if x is an eigenvecor of A under eigenvalue λ, hen P x is an eigenvecor of B under eigenvalue λ. Soluion. By definiion of similariy, we know A P BP. We proved in he lecure ha λ mus also be an eigenvalue of B. Since x is an eigenvecor of A under λ, we know: Ax λx P BP x λx B(P x) λ(p x) 2
which complees he proof. Problem 5. Suppose ha an n n marix A has n linearly independen eigenvecors v, v 2,..., v n. Prove: for any n vecor x, Ax is a linear combinaion of v, v 2,..., v n. Soluion. Assume ha v i (i [, k) is an eigenvecor of A under eigenvalue λ i. We have Av i λ i v i. Since v, v 2,..., v n are linearly independen, we know ha x mus be a linear combinaion v, v 2,..., v n. Namely, here exis c,..., c n such ha which complees he proof. x c v + c 2 v 2 +... + c n v n Ax c Av + c 2 Av 2 +... + c n Av n Ax c λ v + c 2 λ 2 v 2 +... + c n λ n v n. Problem 6. Prove or disprove: if an n n marix A has rank n, hen i mus have n independen eigenvecors. Soluion. False. Consider n 2 and A v of A mus saisfy: [ Thus, any eigenvecor of A mus have he form dimension of. Problem 7. Prove ha A. I has only one disinc eigenvalue. Thus, any eigenvecor (A I)x [ x 2 {[ is no diagonalizable. R, }. This se of vecors has a Soluion. A has wo eigenvalues λ and λ 2 2. Le v be an eigenvecor of λ. v mus saisfy: Hence, he se of eigenvecors of λ is: This se has dimension. (A λ I)v v R, 3
Le v 2 be an eigenvecor of λ 2. v 2 mus saisfy: (A λ I)v 2 v 2 Hence, he se of eigenvecors of λ 2 is: This se also has dimension. R, I hus follows ha he larges number of linearly independen eigenvecors of A is + 2. Therefore, A is no diagonalizable. Problem 8. Le A, B, and C be hree n n marices for some ineger n. Prove ha if A is similar o B and B is similar o C, hen A is similar o C. Soluion. From he fac ha A is similar o B and B is similar o C, we know: and Hence: which complees he proof. A P BP B Q CQ. A P Q BQP (QP ) B(QP ) Problem 9. Decide wheher A [ 2 4 is similar o B [ 3 2. Soluion. From Problem, we know ha A has disinc eigenvalues 3 and 2. Hence, A is similar o he diagonal marix diag[3, 2. On he oher hand, B clearly also has eigenvalues 3 and 2, and hus, is also similar o diag[3, 2. From he resul of Problem 8, we know ha A is similar o B. 4
[ x y Soluion 2. We will ry o find an inverible marix P z w hold. This is equivalen o AP P B. Hence: [ [ [ [ x y x y 3 2 4 z w z w 2 [ [ x z y w 3x x + 2y 2x + 4z 2y + 4w 3z z + 2w This gives he following equaion se: x z 3x y w x + 2y 2x + 4z 3z 2y + 4w z + 2w x You can verify ha he se of soluions y z is w v [ 2 Le us ry u 2, v. This gives P 2 inverible. We can now conclude ha A is similar o B. u/2 u/2 v u ha makes A P BP u R, v R.. Since de(p ), we know ha P is 5