Question

eigenvectors

Answer 1

after I plug an eigenvalue into A-lamba I, I get the matrix
1 1 0
0 0 1
0 0 0

How do you determine the e vector in this case?

Answer 2

can you give the original matrix that you are trying to find the eigenvector for

Answer 3

2 4 3
-4 -6 -3
3 3 1

Answer 4

I think I have a solution. I am double checking.

Answer 5

I ended up doig x+y =0, z = 0
x = -y
y = y
0 = 0

vector
-1
1
0

Answer 6

The eigenvector is found by taking the null space of the A-lambda I matrix. In this case, it appears to be [1 -1 0].

Answer 7

Yep, your answer is equivalent.

Answer 8

Does it satisfy the eigenvalue when multiplied by A?

Answer 9

That is does A[-1 1 0] = lambda[-1 1 0]

Answer 10

er typo there somewhere

Answer 11

I guess the first question is, what was the eigenvalue you got?

Answer 12

1 and -2, but using -2 in this case

Answer 13

Excellent! Then -2 works as planned A[-1 1 0] = [2 -2 0] = -2[-1 1 0]

Answer 14

All of this makes sense when considering that we wish to satisfy
\[Ax = \lambda x\] so that we need \[Ax - \lambda x = 0\] that is \[(A - \lambda I) x = 0\] which is just the null space.

Answer 15

right

Answer 16

Just a side note, for strict checking purposes wolframalpha is very good.

Answer 17

Its just confusing me cause the matrices usually looks like
1 0 -1
0 1 1
0 0 0 or similar

and I work it out to
x1=x3
x2=-x3
x3 free

vector
1
-1
1

Answer 18

1 1 0 x + y + 0z =0
0 0 1 y free
0 0 0 z = 0

x=-y
y = y
z = 0

-1
1
0
makes sense i guess

Answer 19

That's because there's an infinite number of eigenvectors you can pick that are all scalar multiples of each other. So if we plug in what you got to our eigenvector \[\Large \bar e = \left(\begin{matrix}x_1 \\ x_2 \\ x_3 \end{matrix}\right)\] \[\Large \left(\begin{matrix}x_1 \\ x_2 \\ x_3 \end{matrix}\right) = \left(\begin{matrix}x_3 \\ -x_3 \\ x_3 \end{matrix}\right) = x_3\left(\begin{matrix}1 \\ -1 \\ 1 \end{matrix}\right) \]