# Finding an eigenvector of 3x3 matrix

• enc08
In summary, the conversation is about finding an eigenvector of a matrix with a known eigenvalue of λ = 1. The matrix is given in a rotation form and the method used to compute the eigenvector is discussed. The conversation also touches upon potential rounding errors and the use of more precise values to obtain a more accurate result.
enc08
Hi,

I'm trying to find an eigenvector of a matrix. I know that λ = 1, so my matrix (A - λI) is
$$[-0.5253, 0.8593, -0.1906; -0.8612, -0.5018, 0.1010; 0.1817, 0.1161, -0.0236]$$

And from rows 2 and 3 I get these simultaneous equations

$$-0.8612t_{1}-0.5018t_{2}+0.1010t_{3}=0$$
$$0.1817t_{1}+0.1161t_{2}+0.0236t_{3}=0$$

I eliminate to find $t_{2} = -4.02t_{3}$ and $t_{1}=-2.23t_{3}$

Thus the eigenvector is

t=$k [-2.23, -4.02, 1]$

But using an online solver gives the eigenvector as (-0.016, 0.206, 0.978).

Thanks for any pointers.

Last edited:
Before you do anything else, check to see if it is an eigenvector (At = λt ?).
Also how do know λ = 1?

Hi,

The reason I'm assuming λ = 1 is because A is a rotation matrix (I didn't mention this in the original post), and I'm looking for the equivalent axes eigenvector.

As a sanity check, I have checked the eigenvalues and λ = 1 is one of them.

I realized I typo in my original matrix (matrix term 3,3 in the first post should be negative). I've correct it.

I get a closer answer, but my first term appears to be incorrect:

My answer: (-0.0137, 0.225, 1)
Given answer: (-0.0088, 0.216, 1)

Since your matrix is a rotation matrix, the entries are sin and cos values. I would try using more precision than the 4 decimal places you show, and see if that makes a difference in your resulting eigenvector.

I've just been given the numbers as is, and told that it's a rotation matrix.

Can you tell if my method for computing the eigenvectors is correct?

My approach is

λ = 1, so (A - λI) is
$$[-0.5253, 0.8593, -0.1906; -0.8612, -0.5018, 0.1010; 0.1817, 0.1161, -0.0236]$$

From rows 2 and 3:

$$-0.8612t_{1}-0.5018t_{2}+0.1010t_{3}=0$$

$$0.1817t_{1}+0.1161t_{2}-0.0236t_{3}=0$$

Eliminate to find $$t_{2}=0.225t_{3}$$ and $$t_{1}=-0.0137t_{3}$$

Thus the eigenvector is

t=$$k (-0.0137, 0.225, 1)$$

But the actual answer is given as (-0.0088, 0.216, 1).

enc08 said:
I've just been given the numbers as is, and told that it's a rotation matrix.

Can you tell if my method for computing the eigenvectors is correct?

My approach is

λ = 1, so (A - λI) is
[-0.5253, 0.8593, -0.1906; -0.8612, -0.5018, 0.1010; 0.1817, 0.1161, -0.0236]
Here's your matrix in a nicer form. To see what I did, right-click the matrix to view the underlying script.
$$A - I =\begin{bmatrix} -0.5253 & 0.8593 & -0.1906 \\ -0.8612 & -0.5018 & 0.1010 \\ 0.1817 & 0.1161 & -0.0236\end{bmatrix}$$

To check your work, I multiplied A - I above times your eigenvector x. That multiplication should produce a zero vector, but what I got was off by a little. My result was approximately < .0099, .0088, .00003>.

When you were row-reducing A - I, if you weren't careful with your arithmetic, you could have introduced a certain amount of imprecision in your results.

enc08 said:
From rows 2 and 3:

$$-0.8612t_{1}-0.5018t_{2}+0.1010t_{3}=0$$

$$0.1817t_{1}+0.1161t_{2}-0.0236t_{3}=0$$

Eliminate to find $$t_{2}=0.225t_{3}$$ and $$t_{1}=-0.0137t_{3}$$

Thus the eigenvector is

t=$$k (-0.0137, 0.225, 1)$$

But the actual answer is given as (-0.0088, 0.216, 1).
This is easy to check. Just multiply (A - I) times <-0.0088, 0.216, 1>.

Thanks Mark44. I have used more precise values, and using your technique for testing the eigenvector, I get an answer much closer to a zero vector.

That you've confirmed my method is correct will do, especially since the discrepancies seem only due to rounding errors. I would have much preferred the question to use round numbers, but never mind :)

## 1. What is an eigenvector of a 3x3 matrix?

An eigenvector of a 3x3 matrix is a vector that remains in the same direction after the matrix is multiplied by it. In other words, the eigenvector is only scaled by a constant value, known as the eigenvalue.

## 2. Why is finding an eigenvector important in matrix calculations?

Finding an eigenvector allows for easier computation of certain matrix operations, such as diagonalization and power calculation. It also helps to identify important characteristics of a matrix, such as its stability and behavior over time.

## 3. How do you find an eigenvector of a 3x3 matrix?

To find an eigenvector of a 3x3 matrix, you first need to find the eigenvalues by solving the characteristic equation. Then, substitute each eigenvalue into the original matrix and solve for the corresponding eigenvector. This can be done using various methods, such as row reduction or the inverse matrix method.

## 4. Can a 3x3 matrix have more than one eigenvector?

Yes, a 3x3 matrix can have multiple eigenvectors for the same eigenvalue. This is known as an eigenspace and represents all the possible eigenvectors for that particular eigenvalue.

## 5. What are some real-world applications of finding eigenvectors of 3x3 matrices?

Eigenvectors are used in a variety of fields, such as engineering, physics, and economics. Some examples include analyzing the stability of a building structure, understanding the behavior of a vibrating system, and predicting the growth of a population in economics.

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