Question

HW 17, Problem 3.

Compute the characteristic values and corresponding characteristic vectors of the given matrix. Write the vectors in their general form and give a specific numerical example. Also prove that if A is the given original Matrix and D, is its diagonization matrix then A and D are similar.

4 1 6 6 3 2 -5 -2

Answer 1

Given matrix is 4 6 6 A = 1 3 -5 2. - 2 The characteristic equation of A is IA-d Il 0 lie. lu-d 6 6. 3-d 2 -) - 5 -2-d TT 2-5d²+ 8d -4 = 0 (2-1) (2-2)2 = 0 07 d = 1,2,2 Hence, the eigenvalues of are 1, 2, 2- The eigenvector X, of A Corresponding to the eigenvalue d=1 are given by (A - I) X, = 0 4-1 ) = 0 07 6 3-1 -5 6 2 - 2-1 10) JEJ 3 Iow 6 2 1 6 2 -3 -5 Rit / Ri 1 2. 2 X, 1 2 2. 9 -3 2, 11-5

R2 R2 - Ri 2 2 TE Rs - R3+R, -39-2, = 0, 2, + 24, + 22, = 0 2 =-34, and x + 2y + 2 (-34, ) = 0 2,=44, Hence, 4y, g, Z y (-34, general form eg- taking Y, =) then XI = 3 Again, the eigenvector Xa of A Corresponding to the eigenvalue = 2 given by CA- ZI) X2 = 0 are 2 1 6 6 2 - 5 - 4 X2 Y2 22 RitR 1 3 3. 2 Y2 22 -5 -4 RRo-R, 1 3 3 - 2 JE Y Rg R₂ + Ri o - 2 -1

1 3 3 R3 R3-R2 -2 -1 o 22 0 -242-22 and Ka+ 3y2 + 3 22=0 x2 + 3y2 + 3(-242) = 0 1 22=-242 =) X 2 = 342 Hence, X2 Yz 22 342 Y2 -242 eg.- taking Y2=1 3 then X2 11 ها - لم diagonal matrix # The necessary and sufficient condition for a Similar to a Square matrix to be is that geometric multiplicity of each of its eigenvalues coincides with the algebraic multiplicity, rank 2. Here, the coefficient matrix for these eigenvector X2 is of So these equations have only one linearly independent solution. Thus the geometric multiplicity of the eigenvalue 2 is one while its algebraic multiplicity is 2. Since the geometric multiplicity of this eigenvalue is not equal to its algebraic multiplicity. Therefore, A is not similar to a diagonal matrix.