Question

how to find the eigenvector of this

1 0 1
0 1 0
1 0 1

I have found the eigenvalues which are 0, 1, 2

Answer 1

lol, um, so how did you find them?

Answer 2

-L along the diagonals; and take the determinant; set it equal to 0 to find Ls

Answer 3

hard to tell what the vectors are without a vector to go by tho ...

Answer 4

Ax = Lx

Ax-Lx = 0

(A-L)x = 0

and solve for x

Answer 5

subtract the Ls from the diags and rref to find "x"

Answer 6

1-0 0 1
0 1-0 0
1 0 1-0

1-1 0 1
0 1-1 0
1 0 1-1

1-2 0 1
0 1-2 0
1 0 1-2

Answer 7

rref{{1-0, 0, 1},{ 0, 1-0, 0},{ 1, 0, 1-0}}

1 0 1 x1 -1
0 1 0 x2 = x3 0
0 0 0 x3 1

rref{{1-1, 0, 1},{ 0, 1-1, 0},{ 1, 0, 1-1}}

100 0
001 = x3 0
000 1


rref{{1-2, 0, 1},{ 0, 1-2, 0},{ 1, 0, 1-2}}

10-1 1
01 0 = x3 0
00 0 1

Answer 8

im not to sure about the second one tho

Answer 9

I used the rule of sarus and got x^3-3x^2+2x (used x for lambda) and then i got 0, 1, 2 as eigenvalues

Answer 10

ill trust you did that part good then ;)

Answer 11

the rest is just row reducing that eugened matrixes with the 0 vector

Answer 12

ok thanks :)

Answer 13

http://www.wolframalpha.com/input/?i=%7B%7B1%2C+++0%2C++1%7D%2C%7B++0%2C+++1%2C+++0%7D%2C%7B++1%2C+++++0%2C+++1%7D%7D

yeah, i missed v2 :)