# Finding eigenvectors

• Oct 16th 2009, 08:56 AM
Showcase_22
Finding eigenvectors
Find the eigenvectors of the matrix $\displaystyle \begin{pmatrix} {-3}&{17}&{-4}\\ {-2}&{9}&{-2}\\ {-2}&{8}&{-1}\\ \end{pmatrix}$

I've already found the eigenvalues which come out to be 1, 2+i and 2-i.

To find the eigenvectors I have to find v such that $\displaystyle (A-\lambda I)v=0$.

I already found the eigenvector for $\displaystyle \lambda=1$. I'm having trouble with it for the complex cases.

Here's my working for $\displaystyle \lambda=2-i$:

$\displaystyle A-(2-i)I=\begin{pmatrix} {-5+i}&{17}&{-4}\\ {-2}&{7+i}&{-2}\\ {-2}&{8}&{i-3}\\ \end{pmatrix}$

Send $\displaystyle R_3=R_3-R_2$

$\displaystyle A-(2-i)I=\begin{pmatrix} {-5+i}&{17}&{-4}\\ {-2}&{7+i}&{-2}\\ {0}&{1-i}&{i-1}\\ \end{pmatrix}$

Send $\displaystyle R_1=R_1-2R_2$

$\displaystyle A-(2-i)I=\begin{pmatrix} {-1+i}&{3-2i}&{0}\\ {-2}&{7+i}&{-2}\\ {0}&{1-i}&{i-1}\\ \end{pmatrix}$

$\displaystyle (A-(2-i)I)v=\begin{pmatrix} {-1+i}&{3-2i}&{0}\\ {-2}&{7+i}&{-2}\\ {0}&{1-i}&{i-1}\\ \end{pmatrix} \begin{pmatrix} {v_1}\\ {v_2}\\ {v_3}\\ \end{pmatrix}=0$

$\displaystyle (-1+i)v_1+(3-2i)v_2=0 \Rightarrow -v_1+3v_2+i(v_1-2v_2)=0 \Rightarrow v_1=3v_2 \ , \ v_1=2v_2$

$\displaystyle (1-i)v_2+(i-1)v_3=0 \Rightarrow v_2-v_3+i(-v_2+v_3) \Rightarrow v_2=v_3$

The only vector that satisfies these conditions is the zero matrix which cannot be an eigenvector by definition!
• Oct 16th 2009, 09:39 AM
aman_cc
Quote:

Originally Posted by Showcase_22
Find the eigenvectors of the matrix $\displaystyle \begin{pmatrix} {-3}&{17}&{-4}\\ {-2}&{9}&{-2}\\ {-2}&{8}&{-1}\\ \end{pmatrix}$

I've already found the eigenvalues which come out to be 1, 2+i and 2-i.

To find the eigenvectors I have to find v such that $\displaystyle (A-\lambda I)v=0$.

I already found the eigenvector for $\displaystyle \lambda=1$. I'm having trouble with it for the complex cases.

Here's my working for $\displaystyle \lambda=2-i$:

$\displaystyle A-(2-i)I=\begin{pmatrix} {-5+i}&{17}&{-4}\\ {-2}&{7+i}&{-2}\\ {-2}&{8}&{i-3}\\ \end{pmatrix}$

Send $\displaystyle R_3=R_3-R_2$

$\displaystyle A-(2-i)I=\begin{pmatrix} {-5+i}&{17}&{-4}\\ {-2}&{7+i}&{-2}\\ {0}&{1-i}&{i-1}\\ \end{pmatrix}$

Send $\displaystyle R_1=R_1-2R_2$

$\displaystyle A-(2-i)I=\begin{pmatrix} {-1+i}&{3-2i}&{0}\\ {-2}&{7+i}&{-2}\\ {0}&{1-i}&{i-1}\\ \end{pmatrix}$

$\displaystyle (A-(2-i)I)v=\begin{pmatrix} {-1+i}&{3-2i}&{0}\\ {-2}&{7+i}&{-2}\\ {0}&{1-i}&{i-1}\\ \end{pmatrix} \begin{pmatrix} {v_1}\\ {v_2}\\ {v_3}\\ \end{pmatrix}=0$

$\displaystyle (-1+i)v_1+(3-2i)v_2=0 \Rightarrow -v_1+3v_2+i(v_1-2v_2)=0 \Rightarrow v_1=3v_2 \ , \ v_1=2v_2$

$\displaystyle (1-i)v_2+(i-1)v_3=0 \Rightarrow v_2-v_3+i(-v_2+v_3) \Rightarrow v_2=v_3$

The only vector that satisfies these conditions is the zero matrix which cannot be an eigenvector by definition!

Hi - Are you assuming v1,v2,v3 to be reals. I guess that is the error.
They can be complex.

I am not 100% sure, but somewhat inclined towards the above so shot this post.
• Oct 16th 2009, 10:39 AM
Showcase_22
Let $\displaystyle v_1=a+bi$, $\displaystyle v_2=c+di$ and $\displaystyle v_3=e+fi$.

If $\displaystyle a=1, b=0$ then $\displaystyle -1+i+(3-2i)(c+di)=0 \Rightarrow -1+i+3c+3di-2ci+2d=0$$\displaystyle \Rightarrow -1+3c+2d+i(1+3d-2c)=0. \displaystyle \Rightarrow 3c+2d=1 and \displaystyle 3d-2c=-1. \displaystyle \Rightarrow 2d-\frac{4}{3}c=-\frac{2}{3} \displaystyle 3c+\frac{4}{3}c-\frac{2}{3}=1 \Rightarrow \frac{13}{3}c=\frac{5}{3} \Rightarrow 13c=5 \Rightarrow c=\frac{5}{13} This gives \displaystyle d=\frac{2c-1}{3}=\frac{\frac{10}{13}-1}{3}=-\frac{1}{13} So \displaystyle v_2=\frac{1}{13}(5-i) Similarly, \displaystyle (1-i)\left( \frac{1}{13}(5-i) \right)+(i-1)(e+fi)=0 \displaystyle \frac{1}{13}(1-i)(5-i)+ei-e-f-fi=0 \displaystyle \frac{1}{13}(5-5i-i-1)+ei-e-f-fi=0 \displaystyle \frac{5}{13}-\frac{5}{13}i-\frac{1}{13}i-\frac{1}{13}+ei-e-f-fi=0 \displaystyle \frac{4}{13}-e-f+ i \left(\frac{4}{13}+e-f \right)=0 \displaystyle \Rightarrow e+f=\frac{4}{13} and \displaystyle f-e=\frac{4}{13} \displaystyle 2f=\frac{8}{13} \Rightarrow f=\frac{4}{13} \displaystyle e=0 So the eigenvector corresponding to \displaystyle \lambda=2-i is \displaystyle u_1=\begin{pmatrix} {1}\\ {\frac{1}{13}(5-i)}\\ {\frac{4}{13}i}\\ \end{pmatrix} Does this look about right? • Oct 16th 2009, 11:01 AM aman_cc Quote: Originally Posted by Showcase_22 Let \displaystyle v_1=a+bi, \displaystyle v_2=c+di and \displaystyle v_3=e+fi. If \displaystyle a=1, b=0 then \displaystyle -1+i+(3-2i)(c+di)=0 \Rightarrow -1+i+3c+3di-2ci+2d=0$$\displaystyle \Rightarrow -1+3c+2d+i(1+3d-2c)=0$.

$\displaystyle \Rightarrow 3c+2d=1$ and $\displaystyle 3d-2c=-1$.

$\displaystyle \Rightarrow 2d-\frac{4}{3}c=-\frac{2}{3}$

$\displaystyle 3c+\frac{4}{3}c-\frac{2}{3}=1 \Rightarrow \frac{13}{3}c=\frac{5}{3} \Rightarrow 13c=5 \Rightarrow c=\frac{5}{13}$

This gives $\displaystyle d=\frac{2c-1}{3}=\frac{\frac{10}{13}-1}{3}=-\frac{1}{13}$

So $\displaystyle v_2=\frac{1}{13}(5-i)$

Similarly,

$\displaystyle (1-i)\left( \frac{1}{13}(5-i) \right)+(i-1)(e+fi)=0$

$\displaystyle \frac{1}{13}(1-i)(5-i)+ei-e-f-fi=0$

$\displaystyle \frac{1}{13}(5-5i-i-1)+ei-e-f-fi=0$

$\displaystyle \frac{5}{13}-\frac{5}{13}i-\frac{1}{13}i-\frac{1}{13}+ei-e-f-fi=0$

$\displaystyle \frac{4}{13}-e-f+ i \left(\frac{4}{13}+e-f \right)=0$

$\displaystyle \Rightarrow e+f=\frac{4}{13}$ and $\displaystyle f-e=\frac{4}{13}$

$\displaystyle 2f=\frac{8}{13} \Rightarrow f=\frac{4}{13}$

$\displaystyle e=0$

So the eigenvector corresponding to $\displaystyle \lambda=2-i$ is $\displaystyle u_1=\begin{pmatrix} {1}\\ {\frac{1}{13}(5-i)}\\ {\frac{4}{13}i}\\ \end{pmatrix}$

Seems right to me. But I am like not too proficient with the subject. Maybe you would want to back substitute and check once.
• Oct 16th 2009, 06:57 PM
HallsofIvy
By the way, are you working with a vector space over the real numbers or the complex numbers. Since the matrix you give has only real entries, it could be either one. But if you are working with a vector space over the real numbers, then only "1" is an eigenvalue.
• Oct 17th 2009, 03:50 AM
Showcase_22
The question itself doesn't say, but i'm guessing it's over the complex numbers.

Also, I think there's something wrong with my above working. Checking on octave gives:

A =

-3 17 -4
-2 9 -2
-2 8 -1

P =

1.00000 - 0.00000i
0.38462 + 0.07692i
0.00000 - 0.30769i

octave-3.0.1:4> B=A*P
B =

3.5385 + 2.5385i
1.4615 + 1.3077i
1.0769 + 0.9231i

octave-3.0.1:6> B./P
ans =

3.5385 + 2.5385i
4.3077 + 2.5385i
-3.0000 + 3.5000i

But B./P should be an eigenvector, either 2-i or 2+i.

Furthermore, actually computing the answer on matlab gives:

A =

-3 17 -4
-2 9 -2
-2 8 -1

octave-3.0.1:2> [V,D] = eig(A)
V =

0.87447 + 0.00000i 0.87447 - 0.00000i -0.70711 + 0.00000i
0.33634 + 0.06727i 0.33634 - 0.06727i -0.00000 + 0.00000i
0.33634 + 0.06727i 0.33634 - 0.06727i 0.70711 + 0.00000i

D =

2.00000 + 1.00000i 0.00000 + 0.00000i 0.00000 + 0.00000i
0.00000 + 0.00000i 2.00000 - 1.00000i 0.00000 + 0.00000i
0.00000 + 0.00000i 0.00000 + 0.00000i 1.00000 + 0.00000i

...and this is clearly different to what I have.

From what I can see, my initial method to find the eigenvectors should have worked!