Question

Find the eigenvalues and eigenvectors of the matrix shown below

Answer 1

\[\left[\begin{matrix}-1 & 4 & -3\\ 1 & -2 & -1\\ 2 & 2 & -5\end{matrix}\right]\]

Answer 2

What I've done so far:
eigenvalues are -2 and -3 (repeated eigenvalue at -3)
eigenvector for -2 is <2,1,2>
and one of the eigenvectors for -3 is <1,1,2>.
My problem is my homeworks asking me for the second eigenvector for the eigenvalue -3, and I cant say I know how to find it...

Answer 3

plugging in -3 for the eigenvalue gives you:
\[\left[\begin{matrix}2 & 4 & -3 \\ 1 & 1 & -2\\ 2 & 2 & -4\end{matrix}\right]\]
which row reduces to:
\[\left[\begin{matrix}2 & 0 & -1 \\ 0 & 2 & -1\\ 0 & 0 & 0\end{matrix}\right]\]

Answer 4

and as far as I can tell there's only one solution to that matrix being <1,1,2>...

Answer 5

besides of course, the trivial solution

Answer 6

Yeah I think that was a mistake when I typed it in. It's -2 in my work

Answer 7

Just noticed another typo... the entry above that should be -1, not -2. so either way it should still row reduce to the final matrix I typed up

Answer 8

Is it at all possible that there's an imaginary eigenvector?

Answer 9

Well, from what I understand you can have imaginary eigenvalues(or rather, complex eigenvalues). Not sure about imaginary eigenvectors, haven't actually gotten to that part in my linear algebra course I'm in now, so this is just from what I've read ahead.

Answer 10

Unfortunately it doesn't look like imaginary eigenvalues. It's an online homework assignment so I know for certain (the assignment checked that much) that both eigenvalues are -2 and -3. Neither are complex. I also know that the eigenvector for -2 is <2,1,2> and for -3 one of the vectors is <1,1,2>. I've tried inputting a zero vector to no avail. I've tried doing it by hand, with wolfram, with matlab... Nothings giving me a second eigenvector for -3... :P

Answer 11

Plugging it into wolfram I got only real eigenvalues/vectors for this matrix.

Answer 12

I'm beginning to think that this question is broken...

Answer 13

It looks like that while it should have 3 eigenvalues, two of them are identical.

Answer 14

I've tried that as well...

Answer 15

I really don't know how to derive the answer, but the repeated eigenvalue also corresponds to <0,0,0> according to wolfram.

Answer 16

It's a sad day indeed when even wolfram can't help you...

Answer 17

At this point I think I'll just email my professor.. I think I'm either missing something pretty big, or the online homework is broken... But thanks for the help everyone

Answer 18

do you try (0,0,0)

Answer 19

yep! I tried <0,0,0> , <1,1,2> and <-1,-1,-2>

Answer 20

linear algebra

Answer 21

Tell me about it :P It's my second time through and I can't say its any more fun now than it was last year

Answer 22

No I took it in highschool but I wasn't able to receive any transfer credit for it

Answer 23

Yeah, when I applied the eigenvalue of -3 to the matrix, I ended up with something like(using c, because that's what I usually use as the third variable in a three coordinate vector):
\[c\left[\begin{matrix}\frac{1}{2} \\ \frac{1}{2} \\ 1\end{matrix}\right] =c \left[\begin{matrix}1 \\ 1 \\ 2\end{matrix}\right]\]
This isn't linearly independent, and from what I'm reading on wolfram mathworld, eigenvectors that are not linearly independent are returned as zero vectors.

Answer 24

There are varying "intensities" of linear algebra. I'm taking it at a community college. It transfers to the university I'm going to in the fall, but only for the lower intensity one(second year level), if I needed the third year version that's also at that university, it wouldn't apply. Only majors at that university that I know require the higher one are electrical engineering majors. I don't actually need it at all though...

Answer 25

Honestly, after Calculus III last semester, the linear algebra course I'm taking is the easiest math course I've ever taken. I was deciding between a couple different majors at the beginning of the semester, two being Aerospace Engineering and Chemical Engineering(which both would have required it) I decided the other way I was leaning, but with the possibility of graduate school and another professor saying he wish he hadn't dropped it once he got to grad school I decided to stick with it.