Eigenvalues and Eigenvectors
Let A ∈ R n × n be a matrix. If λ ∈ R and v ∈ R n , v � = 0, with Av = λ v , then we call 1. λ an eigenvalue of A , 2. v an eigenvector of A , 3. and ( λ, v ) an eigenpair of A Eigen, adjective: “own”, “intrinsic”. First use in Linear Algebra in 1904 by David Hilbert. 1
Let λ ∈ R and v ∈ R n with v � = 0. The following are equivalent: 1. ( λ, v ) is an eigenpair of A 2. Av = λ v 3. ( A − λ Id) v = 0 4. v ∈ ker ( A − λ Id) Conclusion: λ is an eigenvalue of A if A − λ Id is a singular matrix. This is the case exactly then if det ( A − λ Id) = 0. 2
Let λ ∈ R and v ∈ R n with v � = 0. The following are equivalent: 1. ( λ, v ) is an eigenpair of A 2. Av = λ v 3. ( A − λ Id) v = 0 4. v ∈ ker ( A − λ Id) Conclusion: λ is an eigenvalue of A if A − λ Id is a singular matrix. This is the case exactly then if det ( A − λ Id) = 0. 2
Let λ ∈ R and v ∈ R n with v � = 0. The following are equivalent: 1. ( λ, v ) is an eigenpair of A 2. Av = λ v 3. ( A − λ Id) v = 0 4. v ∈ ker ( A − λ Id) Conclusion: λ is an eigenvalue of A if A − λ Id is a singular matrix. This is the case exactly then if det ( A − λ Id) = 0. 2
Let λ ∈ R and v ∈ R n with v � = 0. The following are equivalent: 1. ( λ, v ) is an eigenpair of A 2. Av = λ v 3. ( A − λ Id) v = 0 4. v ∈ ker ( A − λ Id) Conclusion: λ is an eigenvalue of A if A − λ Id is a singular matrix. This is the case exactly then if det ( A − λ Id) = 0. 2
Let λ ∈ R and v ∈ R n with v � = 0. The following are equivalent: 1. ( λ, v ) is an eigenpair of A 2. Av = λ v 3. ( A − λ Id) v = 0 4. v ∈ ker ( A − λ Id) Conclusion: λ is an eigenvalue of A if A − λ Id is a singular matrix. This is the case exactly then if det ( A − λ Id) = 0. 2
Let A ∈ R n × n be a matrix. The characteristic polynomial of A is a 11 − λ a 12 a 1 n . . . a 21 a 22 − λ . . . a 2 n p A ( λ ) := det ( A − λ Id) = det . . . ... . . . . . . a n 1 a n 2 . . . a nn − λ For λ ∈ R we have p A ( λ ) = 0 ⇐ ⇒ det( A − λ Id) = 0 . The matrix A − λ Id is singular if and only if λ is a root of the characteristic polynomial of A . 3
Let A ∈ R n × n and let p A be the characteristic polynomial. By the Fundamental Theorem of Algebra, we can write p A ( λ ) = ( λ − λ 1 ) · · · · · ( λ − λ n ) where the λ 1 , . . . , λ n ∈ C are the roots of the polynomial. (The leading term λ n has coefficient ( − 1) n .) The λ 1 , . . . , λ n are not necessarily distinct. The algebraic multiplicity µ a ( A , λ ) is the number how often an eigenvalue appears as a root of the characteristic polynomial. 4
Let A ∈ R n × n and let p A be the characteristic polynomial. By the Fundamental Theorem of Algebra, we can write p A ( λ ) = ( λ − λ 1 ) · · · · · ( λ − λ n ) where the λ 1 , . . . , λ n ∈ C are the roots of the polynomial. (The leading term λ n has coefficient ( − 1) n .) The λ 1 , . . . , λ n are not necessarily distinct. The algebraic multiplicity µ a ( A , λ ) is the number how often an eigenvalue appears as a root of the characteristic polynomial. Generally, the roots of a characteristic polynomial may be complex numbers. (Fundamental Theorem of Algebra) 4
Let A ∈ R n × n , λ ∈ C , and v ∈ C n , v � = 0. We call λ an eigenvalue of A , we call v an eigenvector of A , and ( λ, v ) an eigenpair of A if Av = λ v , The following are equivalent: 1. ( λ, v ) is an eigenpair of A 2. Av = λ v 3. ( A − λ Id) v = 0 4. v ∈ ker ( A − λ Id) 5
Let A ∈ C n × n , λ ∈ C , and v ∈ C n , v � = 0. We call λ an eigenvalue of A , we call v an eigenvector of A , and ( λ, v ) an eigenpair of A if Av = λ v , The following are equivalent: 1. ( λ, v ) is an eigenpair of A 2. Av = λ v 3. ( A − λ Id) v = 0 4. v ∈ ker ( A − λ Id) 6
Example 2 − 3 1 2 − λ − 3 1 A = 1 − 2 1 , p A ( λ ) = det 1 − 2 − λ 1 , 1 − 3 2 1 − 3 2 − λ We compute p A ( λ ) = − λ 3 + 2 λ 2 − λ = − λ ( λ − 1) 2 The roots of the polynomial p A are precisely 0 and 1. The eigenvalue 0 has algebraic multiplicity 1 and the eigenvalue 1 has algebraic multiplicity 2: µ a ( A , 0) = 1 , µ a ( A , 1) = 2 , 7
Example What are the eigenvectors? 2 − 3 1 1 0 = 1 − 2 1 1 0 , 1 − 3 2 1 0 2 − 3 1 − 1 − 1 = 1 − 2 1 0 0 , 1 − 3 2 1 1 2 − 3 1 3 3 = 1 − 2 1 1 1 . 1 − 3 2 0 0 8
Example � � � � cos( θ ) − sin( θ ) cos( θ ) − λ − sin( θ ) A = , p A ( λ ) = det , sin( θ ) cos( θ ) sin( θ ) cos( θ ) − λ We compute p A ( λ ) = sin( θ ) 2 + (cos( θ ) − λ ) 2 = sin( θ ) 2 + cos( θ ) 2 − 2 λ cos( θ ) + λ 2 = λ 2 − 2 λ cos( θ ) + 1 Any root of this polynomial must satisfy cos( θ ) 2 − 1 = ( λ − cos( θ )) 2 The left-hand side is negative unless θ is an integer multiple of π , so the eigenvalues are complex unless θ is an integer multiple of π . 9
Example � � � � cos( θ ) sin( θ ) cos( θ ) − λ sin( θ ) A = , p A ( λ ) = det , − sin( θ ) cos( θ ) − sin( θ ) cos( θ ) − λ We compute p A ( λ ) = sin( θ ) 2 + (cos( θ ) − λ ) 2 = sin( θ ) 2 + cos( θ ) 2 − 2 λ cos( θ ) + λ 2 = λ 2 − 2 λ cos( θ ) + 1 Any root of this polynomial must satisfy − sin( θ ) 2 = ( λ − cos( θ )) 2 The left-hand side is negative unless θ is an integer multiple of π , so the eigenvalues are complex unless θ is an integer multiple of π . 10
Example The eigenvalues are λ 1 = cos( θ ) + sin( θ ) i , λ 2 = cos( θ ) − sin( θ ) i . We check that � � � � � � cos( θ ) sin( θ ) 1 1 = λ 1 , − sin( θ ) cos( θ ) i i � � � � � � cos( θ ) sin( θ ) 1 1 = λ 2 . − sin( θ ) cos( θ ) − i − i 11
The characteristic polynomial p A of A ∈ C n × n is defined as a 11 − λ a 12 a 1 n . . . a 21 a 22 − λ a 2 n . . . p A ( λ ) := det ( A − λ Id) = det . . . ... . . . . . . a n 1 a n 2 a nn − λ . . . The scalar λ ∈ C is an eigenvalue of A if and only if it is a root of the characteristic polynomial. Can we use special structures of the matrix to find the eigenvalues? 12
Example Let A ∈ C n × n be a triangular matrix. a 11 − λ a 12 . . . a 1 n 0 a 22 − λ . . . a 2 n p A ( λ ) = det ( A − λ Id) = det . . . ... . . . . . . 0 0 . . . a nn − λ = ( a 11 − λ ) · ( a 22 − λ ) · · · · · ( a nn − λ ) The eigenvalues of a triangular matrix are the diagonal elements: � p A ( λ ) = ( a ii − λ ) . 1 ≤ i ≤ n 13
How to find the eigenvectors? If λ ∈ C is an eigenvalue of A ∈ R n , then the eigenvectors for that eigenvalue are the solutions of the homogeneous linear system of equations ( A − λ Id) · v = 0 . Possible strategy: Bring A − λ Id into reduced row echelon form and determine the nullspace from there. 14
Example 1 2 0 0 0 0 0 0 1 3 − 1 0 � x 1 + 2 x 2 = 0 � 0 0 0 0 0 1 � x ∈ R 6 ker = x 3 + 3 x 4 − x 5 = 0 � 0 0 0 0 0 0 � � x 6 = 0 � 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 − 1 0 0 0 3 0 = span , , 0 − 1 − 1 0 0 − 3 0 0 0 15
Theorem A matrix A ∈ C n × n is nonsingular if and only if 0 is not an eigenvalue of A. 16
Theorem A matrix A ∈ C n × n is nonsingular if and only if 0 is not an eigenvalue of A. Proof. The following are equivalent: 1. 0 is an eigenvalue of A 2. The matrix A − 0 Id has a non-trivial kernel. 3. The matrix A has a non-trivial kernel. 4. There exists v ∈ C n , v � = 0, with Av = 0. 5. The matrix A is singular. 16
Theorem A matrix A ∈ C n × n is nonsingular if and only if 0 is not an eigenvalue of A. Proof. The following are equivalent: 1. 0 is an eigenvalue of A 2. The matrix A − 0 Id has a non-trivial kernel. 3. The matrix A has a non-trivial kernel. 4. There exists v ∈ C n , v � = 0, with Av = 0. 5. The matrix A is singular. 16
Theorem A matrix A ∈ C n × n is nonsingular if and only if 0 is not an eigenvalue of A. Proof. The following are equivalent: 1. 0 is an eigenvalue of A 2. The matrix A − 0 Id has a non-trivial kernel. 3. The matrix A has a non-trivial kernel. 4. There exists v ∈ C n , v � = 0, with Av = 0. 5. The matrix A is singular. 16
Recommend
More recommend
