# Thread: Finding eigenvectors

1. ## Finding eigenvectors

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

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 $(A-\lambda I)v=0$.

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

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

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

Send $R_3=R_3-R_2$

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

Send $R_1=R_1-2R_2$

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

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

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

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

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

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

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 $(A-\lambda I)v=0$.

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

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

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

Send $R_3=R_3-R_2$

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

Send $R_1=R_1-2R_2$

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

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

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

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

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.

3. Let $v_1=a+bi$, $v_2=c+di$ and $v_3=e+fi$.

If $a=1, b=0$ then $-1+i+(3-2i)(c+di)=0 \Rightarrow -1+i+3c+3di-2ci+2d=0$ $\Rightarrow -1+3c+2d+i(1+3d-2c)=0$.

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

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

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

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

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

Similarly,

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

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

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

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

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

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

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

$e=0$

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

Does this look about right?

4. Originally Posted by Showcase_22
Let $v_1=a+bi$, $v_2=c+di$ and $v_3=e+fi$.

If $a=1, b=0$ then $-1+i+(3-2i)(c+di)=0 \Rightarrow -1+i+3c+3di-2ci+2d=0$ $\Rightarrow -1+3c+2d+i(1+3d-2c)=0$.

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

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

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

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

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

Similarly,

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

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

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

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

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

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

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

$e=0$

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

Does this look about right?
Seems right to me. But I am like not too proficient with the subject. Maybe you would want to back substitute and check once.

5. 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.

6. 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!