![Given m= 15 ] det (m) = 25-16 = 9 To(m) = 5+5 = 10 Characteristic polynomial Pmln) = det (XI-M) = 22- TrIM)x+ de+(M) 22. 10x](//img.homeworklib.com/questions/176d3b10-d10c-11ea-8f31-b9352adf730f.png?x-oss-process=image/resize,w_560)
![Nole: Necessam ondition for ax? mahx to be. Idem potent is that either it is diagonal or its trace equais / so Any matrix of](//img.homeworklib.com/questions/17f6bdd0-d10c-11ea-a069-756e41352d09.png?x-oss-process=image/resize,w_560)
![Now - we have aita=1 & 9,+992=5 = a + 9 = 5 - - 9 + 92=1 89 =4 = | 9 = 1/2) 19,= Also b6 + 9b₂=8 51+ b2 so [b2=1 ² 862 = 8 →](//img.homeworklib.com/questions/18877f10-d10c-11ea-865d-4b350919df3d.png?x-oss-process=image/resize,w_560)
![Noы Слин ЕЕ, = Ce] - І, Јв, ) = ( ; - 1- гоо] 1- 4+, -,+ , ) Loo), / we are loo, Treuf Also eigen rauses of m are 1,9 а - €i](//img.homeworklib.com/questions/19176ff0-d10c-11ea-9344-7b7c690404b5.png?x-oss-process=image/resize,w_560)
Given m= 15 ] det (m) = 25-16 = 9 To(m) = 5+5 = 10 Characteristic polynomial Pmln) = det (XI-M) = 22- TrIM)x+ de+(M) 22. 10x + 9 Equate it to zero we get 1x²_100 +9=0l charactens se Equation → x?94-X +9=0 = 217-9)-1(8-9)=0 - (2-1) (3-9) = 0 =) . 221, 7-9 1 = 1 and 11 = 9] Now in can be decomposed into two Idempotent matrizes E, and E Sucis that E,+E, = I, 1, 6, 71, E, EM and E, €2=10]
Nole: Necessam ondition for ax? mahx to be. Idem potent is that either it is diagonal or its trace equais / so Any matrix of the form lab with a²+bc = a is (ci-a) dem polent. ca, bz1 Now let E, = [ a b ] & E,= [a b ) Now E, TE, = I =) (914a2 bitha 7 1 4+2 2-4,-921 - 9179,- bitb,=0 2-9,792 = 1 etc, co - A, E, + 12 E₂ = M = 4(9:6,]+91 97, 7+(35) 9179a₂=5 5,+962 = 8 C1 + 942 = 2 1-a, +9 (1-92) = 5
Now - we have aita=1 & 9,+992=5 = a + 9 = 5 - - 9 + 92=1 89 =4 = | 9 = 1/2) 19,= Also b6 + 9b₂=8 51+ b2 so [b2=1 ² 862 = 8 → 5,= -1 Also 9+96=2 Atų =0 - - - 24 = 2 = {2 = 14 - [ C= -1/4
Noы Слин ЕЕ, = Ce] - І, Јв, ) = ( ; - 1- гоо] 1- 4+, -,+ , ) Loo), / we are loo, Treuf Also eigen rauses of m are 1,9 а - €i реn vаще, м