eigenvalues and eigen vectors!

• Jul 10th 2007, 12:17 PM
charlotte_usa
eigenvalues and eigen vectors!
helloo!

find the eigenvalues and eigen vectors of

A= |1 2|
|5 4|

i dont think its long to work out, but ive just started this topic and cannot work it out :o
• Jul 10th 2007, 12:45 PM
CaptainBlack
Quote:

Originally Posted by charlotte_usa
helloo!

find the eigenvalues and eigen vectors of

A= |1 2|
|5 4|

i dont think its long to work out, but ive just started this topic and cannot work it out :o

The eigen values are the roots of the equation (quadratic equation in this case):

$\det(A-\lambda I)=0$

Why is that?

RonL
• Jul 10th 2007, 12:46 PM
ThePerfectHacker
The Eigenvalues are solutions to:
$\left| \begin{array}{cc}k-1&-2\\-5&k-4 \end{array} \right| = 0$
Thus,
$(k-1)(k-4) - 10 = 0$
Now solve.
• Jul 10th 2007, 03:33 PM
galactus
A non zero vector x is called an eigenvector of A if Ax is a scalar multiple of x.

$Ax=kx$

Where k is the eigenvalue.

They're tied together. Lambda is mostly used, but I will use k.

You know your eigenvalues, -1 and 6. They're the roots of your characteristic polynomial.

$\begin{bmatrix}1&2\\5&4\end{bmatrix}-\overbrace{(-1)}^{\text{eigenvalue}}\begin{bmatrix}1&0\\0&1\end {bmatrix}=\begin{bmatrix}2&2\\5&5\end{bmatrix}$

$\begin{bmatrix}1&2\\5&4\end{bmatrix}-6\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatr ix}-5&2\\5&-2\end{bmatrix}$

$rref\begin{bmatrix}2&2\\5&5\end{bmatrix}=\begin{bm atrix}1&1\\0&0\end{bmatrix}$

$rref\begin{bmatrix}-5&2\\5&-2\end{bmatrix}=\begin{bmatrix}1&\frac{-2}{5}\\0&0\end{bmatrix}$

In the first matrix, we have $x=-y$

Let y=t, then we have x=-t and y=t

In the second one, we have $x=\frac{2}{5}y$

Let y=t, then $x=\frac{2}{5}t$

Therefore, the eigenvectors are

$\begin{bmatrix}-1\\1\end{bmatrix}\cdot{t}$

and

$\begin{bmatrix}\frac{2}{5}\\1\end{bmatrix}\cdot{t}$
• Jul 11th 2007, 05:54 AM
college_clare
Quote:

Originally Posted by galactus
A non zero vector x is called an eigenvector of A if Ax is a scalar multiple of x.

$Ax=kx$

Where k is the eigenvalue.

They're tied together. Lambda is mostly used, but I will use k.

You know your eigenvalues, -1 and 6. They're the roots of your characteristic polynomial.

$\begin{bmatrix}1&2\\5&4\end{bmatrix}-\overbrace{(-1)}^{\text{eigenvalue}}\begin{bmatrix}1&0\\0&1\end {bmatrix}=\begin{bmatrix}2&2\\5&5\end{bmatrix}$

$\begin{bmatrix}1&2\\5&4\end{bmatrix}-6\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatr ix}-5&2\\5&-2\end{bmatrix}$

$rref\begin{bmatrix}2&2\\5&5\end{bmatrix}=\begin{bm atrix}1&1\\0&0\end{bmatrix}$

$rref\begin{bmatrix}-5&2\\5&-2\end{bmatrix}=\begin{bmatrix}1&\frac{-2}{5}\\0&0\end{bmatrix}$

In the first matrix, we have $x=-y$

Let y=t, then we have x=-t and y=t

In the second one, we have $x=\frac{2}{5}y$

Let y=t, then $x=\frac{2}{5}t$

Therefore, the eigenvectors are

$\begin{bmatrix}-1\\1\end{bmatrix}\cdot{t}$

and

$\begin{bmatrix}\frac{2}{5}\\1\end{bmatrix}\cdot{t}$

http://www.mathhelpforum.com/math-he...a221c356-1.gif

http://www.mathhelpforum.com/math-he...7971eda5-1.gif

i understand how to get upto here, but i dont see how you would get the eigenvectors because the teacher didnt explain the steps. i thought that for the 2255 matrix, it would be 2X1+2X2=0 so x1=-X2 so in the eigenvector, the top coordinate would be -1 and the same principle works for the bottom so it would be -1 again so the eigenvector would be (-1/-1) but i think this is wrong...:confused:
• Jul 11th 2007, 06:47 AM
topsquark
Quote:

Originally Posted by college_clare
http://www.mathhelpforum.com/math-he...a221c356-1.gif

http://www.mathhelpforum.com/math-he...7971eda5-1.gif

i understand how to get upto here, but i dont see how you would get the eigenvectors because the teacher didnt explain the steps. i thought that for the 2255 matrix, it would be 2X1+2X2=0 so x1=-X2 so in the eigenvector, the top coordinate would be -1 and the same principle works for the bottom so it would be -1 again so the eigenvector would be (-1/-1) but i think this is wrong...:confused:

Let's try this in a different way. We know the eigenvalues are -1 and 6. So we may write the equation for the -1 eigenvector as (from the eigenvalue equation $Ax = \lambda x$):
$\left [ \begin{array}{cc} 1 & 2 \\ 5 & 4 \end{array} \right ] \cdot \left [ \begin{array}{c} a \\ b \end{array} \right ] = (-1) \left [ \begin{array}{c} a \\ b \end{array} \right ]$

So we have the simultaneous equations
$a + 2b = -a$
and
$5a + 4b = -b$

Solving this for a and b gives a = a and b = -a. So the eigenvector is
$\left [ \begin{array}{c} a \\ -a \end{array} \right ] = a \left [ \begin{array}{c} 1 \\ -1 \end{array} \right ]$
(This is just the negative of galactus' form for the eigenvector. Since the variable a is unspecified we may easily set t = -a and get my form from his.)

We may get an unambiguous eigenvector if we have some condition on the eigenvector, such as orthonormality or something. Since this wasn't mentioned I will leave the eigenvector like this. The 6 eigenvalue is similar:
$\left [ \begin{array}{cc} 1 & 2 \\ 5 & 4 \end{array} \right ] \cdot \left [ \begin{array}{c} a \\ b \end{array} \right ] = (6) \left [ \begin{array}{c} a \\ b \end{array} \right ]$

I'll leave it to you to work out the rest.

-Dan
• Jul 11th 2007, 07:27 AM
charlotte_usa
Quote:

Originally Posted by topsquark
Let's try this in a different way. We know the eigenvalues are -1 and 6. So we may write the equation for the -1 eigenvector as (from the eigenvalue equation $Ax = \lambda x$):
$\left [ \begin{array}{cc} 1 & 2 \\ 5 & 4 \end{array} \right ] \cdot \left [ \begin{array}{c} a \\ b \end{array} \right ] = (-1) \left [ \begin{array}{c} a \\ b \end{array} \right ]$

So we have the simultaneous equations
$a + 2b = -a$
and
$5a + 4b = -b$

Solving this for a and b gives a = a and b = -a. So the eigenvector is
$\left [ \begin{array}{c} a \\ -a \end{array} \right ] = a \left [ \begin{array}{c} 1 \\ -1 \end{array} \right ]$
(This is just the negative of galactus' form for the eigenvector. Since the variable a is unspecified we may easily set t = -a and get my form from his.)

We may get an unambiguous eigenvector if we have some condition on the eigenvector, such as orthonormality or something. Since this wasn't mentioned I will leave the eigenvector like this. The 6 eigenvalue is similar:
$\left [ \begin{array}{cc} 1 & 2 \\ 5 & 4 \end{array} \right ] \cdot \left [ \begin{array}{c} a \\ b \end{array} \right ] = (6) \left [ \begin{array}{c} a \\ b \end{array} \right ]$

I'll leave it to you to work out the rest.

-Dan

i work out that 6b=15a so b=5/2a and a=a so then i get a final eigenvector of a (1 and 5/2)

but earlier it was suggested that the answer was 2/5 so have i gone wrong?
• Jul 11th 2007, 08:25 AM
topsquark
Quote:

Originally Posted by charlotte_usa
i work out that 6b=15a so b=5/2a and a=a so then i get a final eigenvector of a (1 and 5/2)

but earlier it was suggested that the answer was 2/5 so have i gone wrong?

You are getting an eigenvector of
$a \left [ \begin{array}{c} 1 \\ \frac{5}{2} \end{array} \right ]$

Again, this is simply a modification of galactus' answer. Let a = (2/5)t. Then
$a \left [ \begin{array}{c} 1 \\ \frac{5}{2} \end{array} \right ] = \frac{2}{5} t \left [ \begin{array}{c} 1 \\ \frac{5}{2} \end{array} \right ] = t \left [ \begin{array}{c} \frac{2}{5} \\ 1 \end{array} \right ]$
which is the same as galactus' eigenvector. So the two are essentially the same.

-Dan