Please see attached file for full description.
Show that v is an eigenvector of A and find the corresponding eigenvalue.
Show that lamda is an eigenvalue of A and find one eigenvector corresponding to this eigenvalue.
8. A = [ 2 2, 2 -1], lamda = -2
10. A = [0 4, -1 5]; lamda = 4
Use the method of Example 4.5 to find all of the eigenvalues of the matrix A. Give bases for each of the corresponding eigenspaces. Illustrate the eigenspaces and the effect of multiplying eigenvectors by A as in figure 4.8.
24. A = [ 2 4; 6 0];
26. A = [ 1 2; -2 3].
Please see the attached file for detailed solution.
To prove that a vector is the eigenvector of the matrix, we can use the definition
(*), that is, A only change the length of v, not its direction.
It is shown that A only scales the vector v (i.e., changes the length) and thus it proves that equation (*) hold true for the following eigenvalue .
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.
To find the eigenvector corresponding to eigenvalue λ = -2, we compute the mull space A- (-2)I. we find
The solution shows detailed steps of finding the eigenvalues and corresponding eigenvectors of a matrix. It also illustrates the effect of multiplying eigenvectors by the matrix graphically.