By consideration of the matter it is necessary to remember that all used objects are vectors, and n-dimensional. At their record no distinctive signs corresponding to classical vectors are used.
Instruction
1. Number k is called own value (number) of a matrix And if there is a vector x such that Ax=kx. (1) At the same time the vector x is called own vector of a matrix And, corresponding to number k. In R^n space (see fig. 1) the matrix And has an appearance as in the drawing.
2. It is necessary to set the task of finding ownof numbers and vectors of a matrix of A. Pust own vector x it is set by coordinates. In a matrix form he will register a matrix column which for convenience should be presented in the transposed line. X = (x1, x2, …, xn) ^T.Исходя from (1), Akh-kkh =0 or Akh-kekh =0 where E is a single matrix (units are located on the diagonal main thing, all the rest elements – zero). Then (A-kE) x =0. (2)
3. Expression (2) is the system of the linear uniform algebraic equations which has the nonzero decision (own vector). Therefore the main determinant of a system (2) is equal to zero, that is |A-kE| =0. (3) The last equality of rather own value k is called the characteristic equation of a matrix And yes in expanded form has an appearance (see fig. 2).
4. This algebraic equation of n-y of degree. The valid roots of the characteristic equation are own numbers (values) of a matrix And.
5. Substituting root k of the characteristic equation in a system (2), receive the uniform system of the linear equations with a degenerate matrix (its determinant is equal to zero). Each nonzero solution of this system represents own vector of a matrix And, corresponding to this own number k (that is to a root of the characteristic equation).
6. Example. To find own values and vectors of a matrix And (see rice 3). Decision. The characteristic equation is presented in fig. 3. Open determinant and find own numbers of a matrix which are roots of this equation (3-k) (-1-k)-5=0, (k-3) (k+1)-5=0, k^2-2k-8=0. Its roots k1=4, k2=-2
7. a) Own vectors corresponding to k1=4 are, through the solution of a system (A-4kE) x =0. At the same time only one its equation as the determinant of a system is obviously equal to zero is required. If to put x = (x1, x2) ^T, then the first equation of x1+x2=0 system (1-4),-3x1+x2=0. If to assume that h1=1 (only not zero), then h2=3. As as is wished there is a lot of nonzero decisions at a uniform system with a degenerate matrix, all set of own vectors corresponding to the first own number x = C1(1, 3), C1=const.
8. b) Find own vectors corresponding to k2=-2. At the solution of a system (A+2kE) x =0, its first equation (3+2) h1+ h2=0, 5х1+ h2=0. If to put h1=1, then h2=-5. Corresponding own vectors x = C2(1, 3), C2=const. The general set of all own vectors of the set matrix: x = C1(1, 3) + C2(1, 3).