The number k is called a private value (a number) of the matrix A, if there exists a vector x such that Ax=kx. (1)the vector x is called proper vector of the matrix A corresponding to the number k.In the space R^n (see Fig.1) the matrix A has the form as shown.
You need to put the task of finding the eigenvalues and eigenvectors of the matrix A. Let the eigenvector x is given coordinate. In matrix form it can be written by the matrix-column, which for convenience, should provide a transposed line. X=(x1,x2,...,xn)^T. on the Basis of (1), Ah-Kh=0 or Ah-kekh=0, where E is the identity matrix (the unit is located on the main diagonal, all other elements are zero). Then (A-kE)x=0. (2)
The expression (2) is a system of linear homogeneous algebraic equations which has a nontrivial solution (eigenvector). Therefore, the main determinant of system (2) is equal to zero, i.e. |A-kE|=0. (3) the Last equality relative to the eigenvalue k is called the characteristic equation of the matrix A and the expanded view has the form (see Fig.2).
It is the algebraic equation of n-th degree. The actual roots of the characteristic equation are the numbers (values) of the matrix A.
Substituting the root k of the characteristic equation of the system (2), a homogeneous system of linear equations with a singular matrix (its determinant is zero). Each nonzero solution of this system is an eigenvector of matrix A corresponding to the private number k (i.e. the root of characteristic equation).
Example. Find eigenvalues and vectors of matrix A (see figure 3).Solution. The characteristic equation is shown in Fig. 3. Expand the determinant and find the eigenvalues of the matrix that are roots of the equation (3-k)(-1-k)-5=0, (k-3)(k+1)-5=0, k^2-2k-8=0.Its roots are k1=4, k2=-2
a) the eigenvectors corresponding to k1=4, are, through the solution of the system (A-4kE)x=0. This requires only one equation, because the determinant of the system is obviously equal to zero. If we put x=(x1, x2)^T, the first equation of the system (1-4)x1+x2=0, -3x1+x2=0. If we assume that x1=1 (not zero), then x2=3. Since nonzero solutions of a homogeneous system with a degenerate matrix of any number, then the set of all eigenvectors corresponding to the first private number x =C1(1, 3), C1=const.
b) Find the eigenvectors corresponding to k2=-2. When solving the system (A+2kE)x=0, the equation (3+2)x1+x2=0, 5x1+x2=0.If we put x1=1, x2=-5. The corresponding eigenvectors x =C2(1, 3), C2=const. The total set of all eigenvectors of a given matrix: x =C1(1, 3)+ C2(1, 3).