Linear Algebra MA237 Fall 2009 Dr. ByrneHomework for Section 5.1 Due 11/161. Determine whether the given vectors v1, v2, v3, v4 are eigenvectors of the given matrix.If they are, state the corresponding eigenvalue.4221A; v1 =21; v2 = 12; v3 = 11; v4 = 00v1: 1052142211vAEigenvector: =5v2:001242212vAEigenvector: =0v3:631142213vANot an eigenvector.v4:000042214vANot an eigenvector. (!)For the matrices below find (a) the characteristic polynomial, (b) the real eigenvalues, (c) the eigenvector and (d) eigenspace corresponding to each eigenvalue. (e) Describe the eigenspace geometrically (i.e., as two lines or something else).2. 1485A(a) characteristic polynomial: 0148510011485IA  03215 02762(b) real eigenvalues:   0932762 so =-3,9(c)eigenvectors: -3-eigenvector is  11, 9-eigenvector is 12(d) eigenspaces:  11sand 12s(e) eigenspaces geometrically are 2 lines3. 005261104A (Hint: 5 is an eigenvalue.)(a) characteristic polynomial: 03029223(b) real eigenvalues:  = 5, -6, -1(c) eigenvectors: 5-eigenvector =11131 -6-eigenvector =010-1-eigenvector =125951(Hint: use row reduction to solve the linear systems, the eigenvectors are the spanning vectors of the solution, the next problem is shown in more detail.)(d) eigenspaces: 11131sand 010sand 125951s(e) eigenspaces geometrically are 3 lines.4. 011101110A(a) characteristic polynomial: det(A-I)=det(111111)=0(-)*(2-1) - 1*(--1)+1*(1+) = -3+++1+1+=-3+3+2=3-3-2(b) eigenvalues: To factor polynomials of degree 3, first check possible rational roots: 1, 2try +1: (-1) 3-3-2 … doesn’t worktry -1: (+1) 3-3-2 … works … yields 2--2 = (-2)*(+1)roots are -1,+2 and -1(c) eigenvectors: =-1: Solve (A-I)x=0(A-I)x =111111111x=0  000000111 x+y+z=0  x=-y-z but y and z are free101011stststzyxTwo eigenvectors!=2: Solve (A-I)x=0(A-I)x =211121112x=0  000110101 x-z=0, y-z=0  only z is free111sssszyx(d) two distinct eigenspaces: 111sand 101011st(e) the geometry of the eigenspaces is the union of a line and a