# Eigenvector help!!

Discussion in 'Math' started by u-will-neva-no, Apr 6, 2012.

1. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Hey everyone, I have found eigenvalues to be -1,2 and 3. Working out the eigenvectors is problematic for me...

My final form is:

$A=\left{ \begin{array}{lml}

-1 & \, & 2\, & 0\\
2 & \, & -1\, & 0\\
0 & \, & 0\, & 0\\

\end{array} \right\}
$

(multiplied by vector (x,y,z))

okay so my equations are (from the matrix above):
$-x+2y = 0$
$2x-y=0$

Now I could show you all my failed attempts here but I will just post the solution. Please let me know what to do...basically I always mess up on the last part and don't understand what to do after!

Solution is attached. If someone could explain how I get the solution for when
λ = -1 and then I will post my solution for the other two eigenvalues!

Thanks!

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8.6 KB
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2. ### vvkannan Active Member

Aug 9, 2008
138
11
I think the final form you have given is for eigen value 2.
Is your matrix 1 2 0
2 1 0
0 0 2
For λ=-1 equation would look like

[A +I ] [X] = 0

which would be 2 2 0 x
2 2 0 * y = 0
0 0 3 z

The first 2 equations are the same .2x+2y =0 ---> x=-y.So you can take any value of x and y would be -(x).From the 3rd equation 3z=0,z=0.

P.S:Since you didn't specify the original matrix I calculated it using the eigenvalues and the eigenvectors you gave as solution !

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3. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Thanks for the reply, yes that was for α = 2, sorry..

Here is the matrix to solve my eigenvectors:

$A=\left{ \begin{array}{lml}
1 & \, & 2, & 0\\
2 & \, & 1, & 0\\
0 & \, & 0, & 2\\

\end{array} \right\}-\lambda \left{ \begin{array}{lml}
1 & \, & 0, & 0\\
0 & \, & 1, & 0\\
0 & \, & 0, & 1\\

\end{array} \right\}\left{ \begin{array}{lml}
x \\
y \\
z \\

\end{array} \right\}
$

so: for $\lambda = 2$

$A=\left{ \begin{array}{lml}
1 & \, & 2, & 0\\
2 & \, & 1, & 0\\
0 & \, & 0, & 2\\

\end{array} \right\}-2 \left{ \begin{array}{lml}
1 & \, & 0, & 0\\
0 & \, & 1, & 0\\
0 & \, & 0, & 1\\

\end{array} \right\}\left{ \begin{array}{lml}
x \\
y \\
z \\

\end{array} \right\}=0
$

$A=\left{ \begin{array}{lml}
-1 & \, & 2, & 0\\
2 & \, & -1, & 0\\
0 & \, & 0, & 0\\

\end{array} \right\}\left{ \begin{array}{lml}
x \\
y \\
z \\

\end{array} \right\}=0
$

giving
$-x + 2y = 0
2x - y = 0
$

How do I get it into the form attached? I was letting x = t but then got two different values for y..

4. ### vvkannan Active Member

Aug 9, 2008
138
11
-x + 2y = 0
2x - y =0

are your equations .Just two linear equations with two unknowns.Multiply equation one by 2 and add with second one.Only 0 will satisfy the equations.And z can take any value becasue it is not going to make any difference.

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5. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
The question says to find the normalised eigenvectors which is why it has been written as $\frac{1}{\sqrt{2}}$

Would it be possible to explain why it is that??

6. ### vvkannan Active Member

Aug 9, 2008
138
11
Yes.For a normalised eigenvector the sum of squares of its elements must be equal to 1.

So we can convert an eigenvector into normalised one by multiplying the reciprocal of square root of sum of squares of the elements.
For λ = -1 ,our eigenvector is 1
-1
0

Multiply it by 1/√(1² + -1²) so that when we square the individual terms and add we will get 1 .

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7. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
vvkannan, thank you so much for all your help!! I really appreciate it!

8. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Hey again, I have been looking for a situation where all three equations look as though they can not be minimised. I was hoping you, i.e.vvkannan (or anyone else) could look over my workings:

The matrix that I have to solve the eigenvector is:

$A=\left{ \begin{array}{lml}
1 & \, & 2, & 0\\
2 & \, & 0, & 2\\
0 & \, & 2, & -1\\

\end{array} \right\}
$

I noticed that the equations needed to be minimised so I used the reduced row echelon form (please check my workings as I only learnt this yesterday!)

$A=\left{ \begin{array}{lml}
1 & \, & 2, & 0\\
2 & \, & 0, & 2\\
0 & \, & 2, & -1\\

\end{array} \right\}
$

R1 will be row1, R2 = row 2 and R3 = row 3:

$A=\left{ \begin{array}{lml}
1 & \, & 2, & 0\\
0 & \, & -4, & 2\\
0 & \, & 2, & -1\\

\end{array} \right\}
$

I left R1 how it is, then did R2 -> 2R1 and left R3

$A=\left{ \begin{array}{lml}
1 & \, & 2, & 0\\
0 & \, & 1, & -\frac{1}{2}\\
0 & \, & 2, & -1\\

\end{array} \right\}
$

Then divided R2 by -4,left R1 and R2 and did R3 ->R3 -2R2

$A=\left{ \begin{array}{lml}
1 & \, & 2, & 0\\
0 & \, & 1, & -\frac{1}{2}\\
0 & \, & 0, & 0\\

\end{array} \right\}
$

Then R1 ->R1-2R2 and left R2 and R3:

$A=\left{ \begin{array}{lml}
1 & \, & 0, & 1\\
0 & \, & 1, & -\frac{1}{2}\\
0 & \, & 0, & 0\\

\end{array} \right\}
$

So my equations came to:
$
x+z = 0
y - \frac{z}{2} = 0
$

So:

$
x = -z
y = \frac{z}{2}
$

(I may be wrong here so check also )

$A=\left{ \begin{array}{lml}
1 \\
\frac{-1}{2} \\
-1 \\
\end{array} \right\}
$

Then I normalised the eigenvectors:
modulus came to:
$
r = \sqrt{1^2 + \frac{-1}{2}^2 + (-1)^2} = \frac{3}{2}
$

Dividing above matrix by r gave:
$A=\left{ \begin{array}{lml}
\frac{2}{3} \\
\frac{-1}{3} \\
\frac{-2}{3}\\
\end{array} \right\}
$

Can anyone spot any mistakes? Thanks in advance!

9. ### vvkannan Active Member

Aug 9, 2008
138
11
Hi,
I can't find any mistake.Looks right to me

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10. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Thanks for checking!