A. MillerM340May 97Similarity, diagonalization, and eigenvaluesRemark. From now on all matrices are square, i.e., n × n for some n andour field of scalars is the complex numbers, C.Definition 1 A is similar to B (written A ∼ B) iff there exists an invertiblematrix P such that A = P BP−1.Theorem 2 1. A ∼ A2. A ∼ B implies B ∼ A, and3. A ∼ B and B ∼ C implies A ∼ C.proof:For (1) A = IAI−1.For (2) if A = P BP−1, then B = QAQ−1where Q = P−1.For (3) suppose A = P BP−1and B = QCQ−1. ThenA = (P Q)C(P Q)−1.QEDDefinition 3 A is diagonalizable iff A is similar to a diagonal matrix.Definition 4 λ ∈ C is an eigenvalue of A iff for some X 6= Z we haveAX = λX. Such an X is called an eigenvector of A.Theorem 5 An n×n matrix A is diagonalizable iff Cnhas a basis consistingof eigenvectors of A.proof:Suppose first that Cnhas a basis X1, . . . , Xnconsisting of eigenvectorsof A, so AXi= λiXi. Let P be the n × n matrix with colj(P ) = Xjfor1each j = 1, . . . , n and let D be the diagonal matrix with λ1, λ2, . . . , λnon itsdiagonal, i.e.,D =λ10 0 · · · 00 λ20 · · · 0... 0...0......... 0 λn−100 0 0 0 λn.Since the Xj’s are a basis P is invertible.Claim AP = P D.proof:For any jcolj(AP ) = Acolj(P )= λjcolj(P )since colj(P ) = Xjan eigenvector associated to λj. Alsocolj(P D) = P colj(D)= P colj(λjI)= λjP colj(I)= λjcolj(P I)= λjcolj(P )Hence AP and P D have the same columns, and so they are equal. Thisproves the Claim and so A = P DP−1.Now we prove the converse. Suppose that A = P DP−1. Where D is adiagonal matrix with λ1, . . . , λnon its diagonal. We have thatAP = P Dand by the same argument as the Claim, we have for each j = 1, . . . , n thatAcolj(P ) = λjcolj(P )Then since P is invertible the columns P are a basis of Cnand the formulaimplies that each colj(P ) is an eigenvector of A.QED2Theorem 6 If λ1, λ2, . . . , λmare distinct eigenvalues of A and AXi= λiXifor i = 1, . . . , m are (nontrivial) eigenvectors of A, then X1, X2, . . . , Xmarelinearly independent.proof:Nontrivial means that Xi6= Z all i.Suppose they are linearly dependent. Then (by an exercise) either X1= Zor there exists k such that Xk∈ span({X1, . . . , Xk−1}). If we choose k mini-mal such that Xk∈ span({X1, . . . , Xk−1}), then it follows that X1, . . . , Xk−1are linearly independent.Now let a1, . . . , ak−1be such thatXk= a1X1+ · · · + ak−1Xk−1.Multiplying by A gives usAXk= a1AX1+ · · · + ak−1AXk−1and using AXi= λiXigives usλkXk= a1λ1X1+ · · · + ak−1λk−1Xk−1.Case 1. λk= 0.In this case we haveZ = a1λ1X1+ · · · + ak−1λk−1Xk−1.and since X1, . . . , Xk−1are linearly independent we have that λiai= 0 forall i = 1, . . . , k − 1. But the λ’s are all distinct, so λi6= λk= 0 for alli = 1, . . . , k − 1. Hence ai= 0 and so Xk= a1X1+ · · · + ak−1Xk−1= Z,which contradicts the fact that Xkis nontrivial.Case 2. λk6= 0.In this case we can divide by λkand getXk= a1λ1λkX1+ · · · + ak−1λk−1λkXk−1.subtracting this fromXk= a1X1+ · · · + ak−1Xk−13givesZ = (a1− a1λ1λk)X1+ · · · + (ak−1− ak−1λk−1λk)Xk−1.Since X1, . . . , Xk−1are linearly independent we have that ai− aiλ1λk= 0 forall i = 1, . . . , k − 1. Since Xkis nontrivial at least one i ≤ k − 1 exists suchthat ai6= 0, but thenai− aiλiλk= 0implies that λi= λkcontradicting the fact that the λ’s are distinct.In either case we get a contradiction and so our “suppose not” is impos-sible and therefore X1, . . . , Xmare linearly independent.QEDCorollary 7 If an n × n matrix A has n distinct eigenvalues, then it isdiagonalizable.proof:Since any set of n independent vectors in Cnmust be a basis, it must bethat Cnhas a basis consisting of eigenvectors of A and so A is