How do I find the eigenvalues and eigenvectors of a 3x3 matrix A?

In summary, to find the eigenvalues and eigenvectors of the given matrix A, we use the determinant method to obtain the polynomial (s^2 - 2)(1 - s) = 0, giving us eigenvalues of 1 and +/-sqrt(2). We then solve for the corresponding eigenvectors by substituting the eigenvalues into (A-sI) and setting the resulting matrix equal to [x,y,z] multiplied by z, where x, y, and z are the corresponding components of the eigenvector. We can check the accuracy of our eigenvectors by multiplying them by the original matrix A and seeing if we get the eigenvalue times the eigenvector.
  • #1
karnten07
213
0

Homework Statement



Find the eigenvalues and eigenvectors of A

1 1 a
1 -1 b
0 0 1

you can assume a and b are not equal to zero

Homework Equations






The Attempt at a Solution



Using det(A-sI) = 0 where s is the eigenvalue

i get (s^2 - 2)(1 - s) = 0

therefore giving s1 = 1 and s2/3 = +/- 2

Id be very grateful if someone could check this is correct because I am a little uncertain.

From here i struggle to find the eigenvectors, i substitute the eigenvalues to (A-sI) to get

0 1 a
1 -2 b
0 0 0

Then i saw a method where i equated the sum of this matrix with (x, y, z) - a vector written vertically, to (0,0,0) also written vertically.

From that i get y + za = 0, x - 2y + bz = 0

But then i get lost in the method from here because I am unsure i can apply it to this case. In fact i think i may have gone about finding the vectors in the complete wrong way, or I've missed a trick or something. I've searched on the internet for examples and methods but their seems to be so many and i can't seem to decipher which ones i can use or even how to use them. So any help at all would be greatly appreciated, preferably before tuesday morning as that is when it is due. Many thanks
 
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  • #2
I think the roots of that polynomial are 1 and +/-sqrt(2), right? You are on the right track for the eigenvector of 1. The first equation tells you y=za. Put that into the second equation and get x in terms of z also. Now write the [x,y,z] vector all in terms of z. Do you see the eigenvector?
 
  • #3
Dick said:
I think the roots of that polynomial are 1 and +/-sqrt(2), right? You are on the right track for the eigenvector of 1. The first equation tells you y=za. Put that into the second equation and get x in terms of z also. Now write the [x,y,z] vector all in terms of z. Do you see the eigenvector?

I get y = -za, and writing the x,y,z all in z i get -za, -2za -bz, z

im not sure how i get the eigenvector from this?
 
  • #4
[-za,-2za-bz,z]=z*[-a,-2a-b,1]. Remember eigenvectors are only defined up to an multiplicative constant.
 
  • #5
Dick said:
[-za,-2za-bz,z]=z*[-a,-2a-b,1]. Remember eigenvectors are only defined up to an multiplicative constant.

Ok i see that now, but for s = +/-sqrt(2) I am finding even harder to do it in this way. It seems very lengthy and I am wondering if there is an easier way to do it for this case or even if the result i am getting is right.

When i use the eigenvalue of sqrt(2) i again try to write it all in terms of z because i thought this looked easiest as one of them is already written in terms of z.

So i get from subsitution

x(1 - sqrt(2)) + y +az = 0 (1)
x + y(1 - sqrt(2)) + bz = 0 (2)
z(1- sqrt(2)) = 0 (3)

Originally i did think that this meant z = 0 but then i substituted that into 1 and 2 but then ended up getting 2 different equations for x and y eg. x = -y(1-sqrt(2)) and
x = -y/(sqrt(2)), so instead i got x from 1, substituted it back into 1 then rearranged to get y in terms of z. Using y i got x in terms of z from 1.

x = bz((-1 - sqrt(2))/2)
y = -bz/2

Subsituting all into 1, 2 and 3 in terms of z i get:

az = 0
0 = 0
z(1 - sqrt(2)) = 0

so i get eigenvector of z*(a, 0, 1 - sqrt(2))

Is there a way i can check this is right? Many thanks
 
  • #6
Dick said:
[-za,-2za-bz,z]=z*[-a,-2a-b,1]. Remember eigenvectors are only defined up to an multiplicative constant.

BTW, the x and y components are reversed here, you caught that right? There is an easy way to check an eigenvector. Multiply it by the original matrix and see if you get the eigenvalue times the eigenvector. If you do that with the sqrt(2) you'll see something has gone awry. I can see one thing already. The coefficient of y in the second equation should be -1-sqrt(2).
 
  • #7
Dick said:
BTW, the x and y components are reversed here, you caught that right? There is an easy way to check an eigenvector. Multiply it by the original matrix and see if you get the eigenvalue times the eigenvector. If you do that with the sqrt(2) you'll see something has gone awry. I can see one thing already. The coefficient of y in the second equation should be -1-sqrt(2).

oh no, i didnt see id put them the wrong way around, thanks for pointing it out. I will work on it today and see if i can work it out and report back here. Thanks again for all your help.
 
  • #8
Dick said:
BTW, the x and y components are reversed here, you caught that right? There is an easy way to check an eigenvector. Multiply it by the original matrix and see if you get the eigenvalue times the eigenvector. If you do that with the sqrt(2) you'll see something has gone awry. I can see one thing already. The coefficient of y in the second equation should be -1-sqrt(2).

ok so I am back on the eigenvalue of sqrt(2)

I substitute it into the matrix and get

(1- sqrt(2))x + y + az = 0
x + y(-1-sqrt(2)) + bz
(1-sqrt(2))z = 0

I don't see how to get the eigenvector from this stage, do i need to write x, y and z all in terms of just one term?
 
  • #9
Well, you do have z=0. So yes, now you just need to eliminate either x or y to get all of the vector components in terms of one variable.
 
  • #10
Dick said:
Well, you do have z=0. So yes, now you just need to eliminate either x or y to get all of the vector components in terms of one variable.

so by eliminating az and bz from 1 and 2, i get y = -x(1-sqrt(2))

so the eigenvector is x*(1, 1-sqrt(2), 0)

cool. th enext part of the question asks me to diagonalise the matrix by finding matrix B so that B = R^-1 A R is a diagonal matrix. I am just looking through my notes to see if i have anything on this, but any hints you can give me, thanks for all the help!
 
  • #11
Looks more to me like x*(1,sqrt(2)-1,0), but I think you have the general idea.
 
  • #12
Dick said:
Looks more to me like x*(1,sqrt(2)-1,0), but I think you have the general idea.


oh yes youre right again, well spotted. So for the diagonalization, it says that R is formed by the 3 eigenvectors as the columns of this new matrix. So i will make my first eigenvector corresponding to eigenvalue 1, in terms of y, so i can write out the diagonalized matrix all in terms of y.
 

FAQ: How do I find the eigenvalues and eigenvectors of a 3x3 matrix A?

1. What is a 3 x 3 matrix eigenvector?

A 3 x 3 matrix eigenvector is a vector that, when multiplied by a given 3 x 3 matrix, results in a scaled version of itself. In other words, the direction of the vector remains the same, but its magnitude changes.

2. How do you find the eigenvectors of a 3 x 3 matrix?

To find the eigenvectors of a 3 x 3 matrix, you first need to find the eigenvalues by solving the characteristic equation det(A - λI) = 0. Then, for each eigenvalue, you can find the corresponding eigenvector by solving the equation (A - λI)v = 0, where v is the eigenvector.

3. What is the significance of eigenvectors in linear algebra?

Eigenvectors are important in linear algebra because they represent the directions along which a matrix only stretches (or compresses) a vector without changing its direction. They are also used in many applications, such as solving systems of linear differential equations, image processing, and data analysis.

4. Can a 3 x 3 matrix have complex eigenvectors?

Yes, a 3 x 3 matrix can have complex eigenvectors. This occurs when the eigenvalues of the matrix are complex numbers. In this case, the eigenvectors will also be complex numbers, but their real and imaginary parts will still represent the direction and magnitude of the vector.

5. How many eigenvectors can a 3 x 3 matrix have?

A 3 x 3 matrix can have up to 3 linearly independent eigenvectors. This means that there can be at most 3 distinct eigenvectors for a given matrix, but there may be fewer depending on the eigenvalues and the properties of the matrix.

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