Question

Let matrix M =

-8	-24	-12
0	4	0
6	12	10

(a) Find the eigenvalues of M

(b) For each eigenvalue λ of M, find a basis for the eigenspace of λ.

(c) Is the matrix M diagonalizable? If so, find matrices D and P such that D is a diagonal matrix and M=PDP^−1. If not, explain carefully why not.

Answer 1

A be A matrix AFIR A - I some basic fact a aquare Eigen value and eigen vector : suppose non zero vector X is said is said to eigen vector if A X = 1x for some (A-15) x = 0 o's is singular matrin. then 1A-d51:0 a is called eigen value of a A is said to be diagonalizable et there exist a mateix invertible matrix o such that D (diagonal matrix) Now from question Here PAPE 8 -24 -12 Ma 6 12 10 for eigen value IM M-AI | 30 8-d -24 - 12 0 4-d 0 0 12 10-2 0 Expand the determinant along and row (4-4) ((-3-1)(10-1) + 72 ) (4-1)(-80 +84-101 +12+72) = 0 (4-1) 812-21 -8 ) = (4-1) (1? + 21-41-8) = 0 (4-1 (AC A +2) - 4 (A +2)) (4-4) (4+2)(1-4) = 0 = 4,4 Eigen values of M= 14 Eigen vector corresponding to 134 (M-41 ) x = 0 - 2 2} 201 22 23 consider xD -12 -24 -12 0 21 22 ooo 12 6 - 122,-2422 - 12H3=0 2, +222 +13=0 and 1222 +62320 x3 = -2x2 Sneanned with CamScanner

suppose x2=t 23 = -at Xit 2x2 + 23-0 X, + 2+ +(-2+) = 0 *; X3= -26, X234) x=0 2130, 22-4 23 -2f Hence xi [...] as we change value of t we ger different eigen X3 vectors A3 2 one of the eigen vector of Eigen vectos corresponding to (M-150x = 0 Goret consider (M +21) X 30 22 X- T-6 - 24 -12 o 6 68 r 12 12 x3 - 6x1 - 2422 - 1223 -O 6.X2 and 6X1 + 1282 +12X3 SO 2200 from @ Put X250 in 0 and 111 62 i + 12x3=0 - 6x - 12X330 both forms same equation 6X1 + 12x350 - 283 suppose 23: then x = = -25 and 7230 As we change di e operent value of S will get different eigen vectors of M xi S 21 x2 23. ms fog SOI one of the eigen vector of M: ( . Number of linearly independent eigen vectors = 2 7 ta] [13] is said to be diagonalizable Theorem: A Square marrix of order na eft there exist n linearly independent eigen vectors. In fact a linearly i independent vector fooms matrix p. j.e pa [u, va', i. on) Here matrix M has only two linearly independent eigen M is not diagonalizable. es canned with CaniScanner say po,, , ... 0} rectors so