Basis and Dimension, Eigenvalues, and Eigenvectors and Diagonalizability

Related BrainMass Solutions

• Diagonalizability

  The eigenvectors corresponding to the eigenvalue 2 are thus all of the form . The dimension of the nullspace is thus 1, and so B is not diagonalizable. C= The eigenvalues are 0, 2 (with multiplicity 1) and 3 with multiplicity 2.

• theory regarding diagonalization

  So, if there are n distinct eigenvalues, there is a set of n linearly independent eigenvectors and this forms a basis of V.

• Matrix and eigenvalues

  This provides examples of working with a matrix regarding eigenvalues, diagonalizability, and multiplicities.

• Eigenvalues and Dimension of Corresponding Eigenspace (4 Problems)

  50760 Eigenvalues and Dimension of Corresponding Eigenspace Determine the eigenvalues and eigenvectors of each of the following matrices. For each eigenvalue, determine the dimension of the corresponding eigenspace.

• Consider the linear transformation T : complex numbers^n -> complex numbers^n given by T( z1, z2, ... , zn ) = ( a1z1, a2z2, ... , anzn). What is the dimension of the subspace spanned by the eigenvectors of T? Exhibit a basis for this space, and give the eigenvalues.

  What is the dimension of the subspace spanned by the eigenvectors of T? Exhibit a basis for this space, and give the eigenvalues. Please see the attachment.

• Orthogonal Matrix : Diagonalizability

  -r) = 0 (5-r)[(5-r)^2 - 4] = 0 Thus r = 5 or (5-r)^2 = 4 (5-r) = +- 2 r = 7,3 Therefore the eigenvalues are r = 3,5,7 and hence the matrix is diagonalizable.

• Proving A is diagonizable

  It is not hard to find all (k+1) eigenvalues are equal to zero, since its characteristic polynomial is . But we can see that T has rank k, which means the eigenvector space E(0) has dimension 1. So, T can not be diagonalizable.

• Eigenvalues

  Check that T is a linear transformation and compute its eigenvalues and eigenvectors. Please see the attachment.

• find eigenvalues

  In our case: (3.5) It is easy to see that (3.6) And since the dimension of A is , the degeneracy is: (3.7) If we write the vector as a linear combination of the standard basis vectors: (3.8) We see that: (3.9) The geometric multiplicity