Eigenvalue proofs

Related BrainMass Solutions

• Eigenvalues : For an n x n matrix A, show that if one or more of the eigenvalues is zero, A has no inverse. Also show that if, A does have an inverse, the eigenvalues of A^-1 are the reciprocals...

  This means that 1/r is an eigenvalue of A^(-1). So the eigenvluaes of A^(-1) are the reciprocals of the eigenvalue of A. Eigenvalue proofs are provided. The solution is detailed and well presented.

• Idempotent and nilpotent matrix proofs

  218166 Idempotent and nilpotent matrix proofs Please show work. A matrix pxp matrix A is idempotent if A^2=A... Please see the attachment Problem #3 If is idempotent, then .

• Matrix : Convergence, Pseudoinverse and Single Value Decomposition

  Proof: Let  be the largest eigenvalue of A (in absolute terms) and let v be the eigenvector associated with this eigenvalue with . Then: Theorem 2: Let A be an arbitrary square matrix of order n.

• Eigenvectors and eigenvalues

  (b) Suppose that w is an eigenvector of A with eigenvalue β. Find an eigenvector for B which has the same eigenvalue β. Please see the attachment.

• Multiplicities and eigenvalues

  Suppose is an eigenvalue of with geometric multiplicity . First, I claim that is also an eigenvalue of . Since is an eigenvalue of , then we have . Then we have Thus is also an eigenvalue of .

• Minimax Principle - Intermediate Eigenvalue

  25215 Minimax Principle - Intermediate Eigenvalue Use the minimax principle to show that the intermediate eigenvalue {see attachment} is not positive. *Please see attachment for eigenvalue and hint on how to complete the question.

• eigenvalue and eigenvector of matrix

  So A only changes the length of v and thus it proves that equation (*) holds true for eigenvalue To prove that λ is an eigenvalue of matrix, we then use the definition Therefore, λ = - 2 is an eigenvalue of matrix A.

• Solving for Eigenvalues

  So we have We get the characteristic polynomial of is Note: If A has an inverse matrix and has an non-zero eigenvalue a, then 1/a is an eigenvalue of its inverse matrix. You can prove it easily.

• Isomorphisms and eigenvalues

  has only one eigenvalue . for any . So is the only eigenvalue of . So , , . None of them is equal to . For (a), if has an eigenvalue , then there is a nonzero vector , such that . Thus we have . So is a zero vector.