Need Help On Eigenvalue and Eigenvector

[shermaine80]

Hi Guys,

I have got some enquires for eigenvalue and eigenvector.

Consider the 1st matrix:

A = [ 1 2 3]
[ 0 5 6]
[ 0 6 5]

The characteristic polynomial is

det(A-λI) = [ 1-λ 2 3]
[ 0 5-λ 6]
[ 0 6 5-λ]

= (1-λ) [ (5-λ)^2 - 36]

The eigenvaules are D(λ) = 0 ---> λ1 = 1, λ2=-1 and λ3 =11
May i know how do we get the λ1 , λ2 and λ3? Can someone advise me?
Dont seem quadratic is working for this??


Matrix (2)

A = [3 5 3]
[0 4 6]
[0 0 1]

Is the characteristic polynomial: det(A - λI) = [ 3-λ 5 3]
[ 0 4-λ 6]
[ 0 0 1-λ]

= (3-λ)[(4-λ)(1-λ)- 0]?


Does the characteristic polynomial of matrix 2 also the same as Matrix (3)

A = [ 3 0 0]
[ 4 4 0]
[ 5 6 1]

Please advise. Thanks :)

 

[Tomsk]

When you take the det of a 3x3 matrix, you can take the elements of ANY row or column and multiply them by the 2x2 determinants you get from removing the row and column that that element is in, not forgetting the signs, which follow the

(+-+...)
(-+-...)
(+-+...)

pattern, I forget what it's called. So for those matrices, with their handy 0s in the right places, you can use the rows or columns with the two zeros in, which simplifies things a great deal.

 

[shermaine80]

Thanks! How about the eigenvaules are D(λ) = 0 ---> λ1 = 1, λ2=-1 and λ3 =11
May i know how do we get the λ1 , λ2 and λ3? Can someone advise me?
Dont seem quadratic is working for this??

 

[KoGs]

You're on the right track. Now just do what you did for the whole thing, instead of having 3 different matrixes.

 

[shermaine80]

Can you kindly advise how to get the values of
λ1 = 1, λ2=-1 and λ3 =11? As i dun seem to get it.

 

[KoGs]

Do what you did to get the characteristic polynomials, just do it for the entire 3 x 3 matrix. From there you will get an expression with λ^3. Then set the characteristic polynomial = 0. The roots of the characteristic polynomials are nothing more than your λ's.

 

[KoGs]

det(A-λI) = [ 1-λ 2 3]
[ 0 5-λ 6]
[ 0 6 5-λ]

= (1-λ) [ (5-λ)^2 - 36]

This work you have done is exactly right. Now set this = 0. And find what your λ's are.

Edit: This IS right, and this is all you need. You have to figure out why. In order to do this you have to know how to take a determinant of an entire 3*3 matrix.

 

[shermaine80]

Thanks! I understand it now.
I have another question:

[ 2 2 3]a]
[ 0 6 6]b] = 0
[ 0 6 6]c]

How do i find [a]

[c] ?
please advise as i pretty poor in matrix.

 

[KoGs]

2(a) + 2(b) + 3(c) = d11
0(a) + 6(b) + 6(c) = d21
0(a) + 6(b) + 6(c) = d31

You want the whole thing = 0. So you want d11 = d22 = d33 = 0.

 

[shermaine80]

Hi Kogs,

Do u hv msn? can i msn u to ask something awhile?
Thanks!

 

[KoGs]

I'd rather just talk here. What do you need?

I misread the original thing. I guess only d21 = 0. But you will notice d21 = d31.

 

[shermaine80]

Thanks. But the answer for for [a
b
c]
is [1
0
0]

any idea how to get [1
0
0]?
please advise. Thanks!

 

[KoGs]

I must be reading the question wrong then. Because that answer doesn't look right.

 

[shermaine80]

i will scanned the notes and u hv a look ...cheers :)

 

[shermaine80]

Here's the note :)
the one circled, may i know how to get it? Please advise. Thanks!

 

[KoGs]

Can you post on website somewhere?

 

[shermaine80]

does this link wors: https://www.physicsforums.com/attachment.php?attachmentid=8444&d=1164702136

 

[KoGs]

No it does not.

 

[shermaine80]

i re-attached again

 

[shermaine80]

or u can try this :http://www.electro-tech-online.com/...alue-eigenvector-picture-003.jpg?d=1164703364

 

[KoGs]

It doesn't work because I have to be logged in as you to see it.

 

[shermaine80]

hi, can i msn u instead?

 

[shermaine80]

hi,

can i checked with you..how we normlized Eigenvectors?

 

[HallsofIvy]

Why did you change matrices? The matrix in the attachment is the same one you mentioned first:
$$\left[\begin{array}{ccc}1 & 2 & 3 \\0 & 5 & 6 \\ 0 & 6 & 5\end{array}\right]$$
but you asked how to find [1 0 0] as an eigenvalue of
$$\left[\begin{array}{ccc}2 & 2 & 3 \\0 & 6 & 6 \\ 0 & 6 & 6\end{array}\right]$$
and it isn't! Did you miscopy?

I think people may be trying to give you too "sophisticated" answers when your questions are really very simple. I must say I found your original question very peculiar! you had done the hard part, found that the eigenvalue equation was $(\lambda- 1)((\lambda- 5)^2- 36)= 0$ and apparently were asking how to solve that equation! If you are doing linear algebra you should have learned how to solve polynomial equations, especially those that are already factored, long ago! In order to have that product, $(\lambda- 1)((\lambda- 5)^2- 36)$ equal to 0, you must have either $\lambda- 1= 0$ or $(\lambda-5)^2- 36$ equal to 0. From $\lambda- 1= 0$ we get, of course, $\lamba= 1$. From $(\lambda- 5)^2- 36= 0$ we get $(\lambda- 5)^2= 36$ so $\lambda- 5= \pm 6$. $\lambda= 5+ 6= 11$ or $\lambda= 5- 6= -1$.

Now to find the corresponding eigenvectors, just do the computation:
For $\lambda= 1$, for example,
$$\left[\begin{array}{ccc}1 & 2 & 3 \\0 & 5 & 6 \\ 0 & 6 & 5\end{array}\right]\left[\begin{array}{c}a \\ b \\ c\end{array}\right]= \left[\begin{array}{c} 1*a \\ 1*b \\ 1*c\end{array}\right]$$

Multiply that out and you get 3 equations for a, b, c:
a+ 2b+ 3c= a, 5b+ 6c= b, and 6b+ 5c= c, or 2b+ 3c= 0, 4b+ 6c= 0, 6b+ 4c= 0. Three independent equations in 3 variables will have ONE solution: obviously a= b= c= 0 is a solution. Since 1 is an eigenvalue, this must have non-trivial solutions and so an infinite number of solutions: the equations must not be independent. In fact, a has disappeared from the equations so a can be anything. Notice that 4b+ 6c= 2(2b+ 3c) so 2b+ 3c= 0 and 4b+ 6c= 0 are really the same equation. But we can still solve 2b+ 3c= 0 and 6b+ 4c= 0, to get b= c= 0. Any eigenvector is of the form [a 0 0]= a[1 0 0].

Similarly for the other eigenvalues. If $\lamba= -1$ then the matrix equation becomes
$$\left[\begin{array}{ccc}1 & 2 & 3 \\0 & 5 & 6 \\ 0 & 6 & 5\end{array}\right]\left[\begin{array}{c}a \\ b \\ c\end{array}\right]= \left[\begin{array}{c} -1*a \\ -1*b \\ -1*c\end{array}\right]$$
Multiplying that out gives a+ 2b+ 3c= -a, 5b+ 6c= -b, and 6b+ 5c= -c, or 2a+ 2b+ 3c= 0, 6b+ 6c= 0, 6b+ 6c= 0. (Was that where you got that matrix before?) Obviously the last two equations are the same: b= -c. Putting that into the first equation, 2a+ 2b- 3b= 0 gives b= 2a. Any eigenvector, corresponding to eigenvalue -1 is of the form [a, 2a, -2a]=
a[1, 2, -2].

For $\lambda= 11$ it is
$$\left[\begin{array}{ccc}1 & 2 & 3 \\0 & 5 & 6 \\ 0 & 6 & 5\end{array}\right]\left[\begin{array}{c}a \\ b \\ c\end{array}\right]= \left[\begin{array}{c} 11*a \\ 11*b \\ 11*c\end{array}\right]$$
Multiplying, a+ 2b+ 3c= 11a, 5b+ 6c= 11b, and 6b+ 5c= 11c, or -10a+ 2b+ 3c= 0, -6b+ 6c= 0, 6b- 6c= 0. Again, the last two equations are equivalent and give b= c. Putting that into the first equation, -10a+ 2b+ 3b= 0 gives -10a+ 5b= 0 or b= -2a. Any eigenvector corresponding to $\lambda= 11$ must be of the form [a, -2a, -2a]= a[1, -2, -2].

To answer your last question, "normalizing" a vector means reducing it to unit length: find the length of the vector and divide each component by that. [1 0 0] already has length 1 and is already normalized. [1 2 -2] has length $\sqrt{1^2+ 2^2+(-2)^2}= \sqrt{9}= 3$. The unit vector in the direction of [1 2 -2] is $\left[\begin{array}{ccc}\frac{1}{3} & \frac{2}{3} & -\frac{2}{3}\end{array}\right]$. [1 -2 -2] also has length 3 and a unit vector in its direction is $\left[\begin{array}{ccc}\frac{1}{3} & -\frac{2}{3} & -\frac{2}{3}\end{array}\right]$.

 

[embury]

The above answer is pretty good. But if you preform row operations on the matrix:
[0 2 3 : 0]
[0 4 6 : 0]
[0 6 4 : 0]
you get:
[0 2 3 : 0]
[0 0 0 : 0]
[0 0 -5 : 0]
from this you get:
-5c = 0
c=0

2b+3c=0
b=0

x is a free variable, so a = s

you then get:

s x [1]
[0]
[0]
The vector:
[1]
[0]
[0]

is a bisis for the eigenspace of A corresponding to lamda=1

 

[HallsofIvy]

Thanks, embury. Yes, you can do directly on the matrices the operations I did to solve the equations, rather than converting from matrices to equations and back again. Either way works. Some people, just beginning to work with matrices, find it easier to think in terms of equations even though that way is longer.