Question

Can we have "different" eigenvectors for the same eigenvalue of a matrix A?

Answer 1

\[A=\left[\begin{matrix}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 2 & 1 & 1& 1 \\ -5 & 2 & 5 & -1\end{matrix}\right]\]
\[\lambda=3, -3, 1,-1\]

Answer 2

Consider \(\lambda = 1\)
\[\left[\begin{matrix}-1 & 0 & 1 & 0 \\ 0 & -1 & 0 & 1 \\ 2 & 1 & 0& 1 \\ -5 & 2 & 5 & -2\end{matrix}\right]\]\[->\left[\begin{matrix}1 & 0 & -1 & 0 \\ 0 & -1 & 0 & 1 \\ 0 & 1 & 2 & 1 \\ 0 & 2 & 0 & -2\end{matrix}\right]\]\[->\left[\begin{matrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0\end{matrix}\right]\]
So, \[\left[\begin{matrix}x_1 \\ x_2 \\x_3 \\ x_4\end{matrix}\right]=\left[\begin{matrix} -1 \\ 1 \\ -1 \\ 1\end{matrix}\right]x_4\]Eigenvector corresponding to λ=1 is \((-1, 1, -1, 1)^T\)
Is that right?

Answer 3

This isn't real, is it?

Answer 4

What do you mean?

Answer 5

Earlier I ignored this
http://www.wolframalpha.com/input/?i=Eigen+vectors%7B%7B1%2C1%2C1%7D%2C%7B2%2C2%2C2%7D%2C%7B3%2C3%2C3%7D%7D

Answer 6

-1 1 -1 1 is a eugene vector

http://www.wolframalpha.com/input/?i=%7B%7B0%2C0%2C1%2C0%7D%2C%7B0%2C0%2C0%2C1%7D%2C%7B2%2C1%2C1%2C1%7D%2C%7B-5%2C2%2C5%2C-1%7D%7D

Answer 7

How about the eigenvectors of the identity matrix ?

Answer 8

The answer given to this question is \((1,-1,1,-1)^T\). I don't know why I always get the signs "wrong" (I don't think it is really "wrong" though...)

Answer 9

You can check your answer

M *vector= lambda* vector

maybe the book mixed up the vectors for lambda = 1 and lambda= -1

Answer 10

Eigenvector of lambda = -1 is (1, 5, -1, -5)^T, while I got (-1, -5, 1, 5)^T

Answer 11

Wait.. With reference to the WolframAlpha link given by amistre64, I got the same answer as wolf, but different from the ones given by my teacher. :|

Answer 12

ok, you can always scale the eigenvector, and it still works

M v = \(\lambda\) v
M (av) = λ (av)

so you can choose a= -1 to change your answer.

Answer 13

I would "reduce" v by taking out common factors, but -1 is ambiguous.

Answer 14

I would choose the sign that gives the most positive terms in the e-vector but that is only because I don't like minus signs.

Answer 15

So, we can have multiply the eigenvector by a scalar but it is still the eigenvector corresponding to that eigenvalue lambda. Is that right?

Answer 16

Yes. The eigenvector really represents "direction" and its specific length is not important.

Answer 17

Thanks!!

Answer 18

@Directrix