Simple eigenvector question - please evaluate

In summary, the conversation involves a request for someone to check a problem solution involving vectors and a matrix, specifically the calculation of (A - I)x for two different vectors. The person asking for help has already verified their method and is seeking numerical confirmation. They also mention that they don't want to clutter up the conversation thread and suggest checking both vectors to see if either works.
  • #1
enc08
42
0
Hi,

[tex]A - I =\begin{bmatrix} -0.5253 & 0.8593 & -0.1906 \\ -0.8612 & -0.5018 & 0.1010 \\ 0.1817 & 0.1161 & -0.0236\end{bmatrix}[/tex]

My eigenvector answer is

t= k(−0.0137,0.225,1)

My solution sheet's answer is

t = k(-0.0088, 0.216, 1)

Could I please ask that somebody checks this by hand? (not using an online solver as that's part of the problem).

A slightly awkward request but I appreciate anyone's answer.

I'm just seeking numerical confirmation. Thanks :)

(To save cluttering up this thread, I have already verified that I'm using the correct method here: https://www.physicsforums.com/showthread.php?t=668624)
 
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  • #2
enc08 said:
Hi,

[tex]A - I =\begin{bmatrix} -0.5253 & 0.8593 & -0.1906 \\ -0.8612 & -0.5018 & 0.1010 \\ 0.1817 & 0.1161 & -0.0236\end{bmatrix}[/tex]

My eigenvector answer is

t= k(−0.0137,0.225,1)

My solution sheet's answer is

t = k(-0.0088, 0.216, 1)

Could I please ask that somebody checks this by hand? (not using an online solver as that's part of the problem).

A slightly awkward request but I appreciate anyone's answer.

I'm just seeking numerical confirmation. Thanks :)

(To save cluttering up this thread, I have already verified that I'm using the correct method here: https://www.physicsforums.com/showthread.php?t=668624)

I really don't want to go through all the arithmetic to hand-check your work, but you can check both purported vectors to see if either (or neither) works.

Let's call your vector xenc08 = <−0.0137, 0.225, 1>, and the one from the solution sheet xanswer = <-0.0088, 0.216, 1>

Calculate (A - I)x for both vectors. If one of the vectors is an eigenvector (or close to it), the result should be close to <0, 0, 0>.

Since both vectors have a component of 1, it can't be the case that your vector is a scalar multiple of the one shown in the answer sheet.
 
  • #3
enc08 said:
Could I please ask that somebody checks this by hand?

Since it is your problem, wouldn't it make more sense that you check it by hand??
 

FAQ: Simple eigenvector question - please evaluate

1. What is an eigenvector?

An eigenvector is a vector that, when multiplied by a matrix, results in a scalar multiple of itself. In other words, the direction of the eigenvector remains unchanged after the matrix transformation, but its length may be scaled.

2. How do you find eigenvectors?

To find eigenvectors, you must first find the eigenvalues of a given matrix. This can be done by solving the characteristic equation det(A-λI) = 0, where A is the matrix and λ is the eigenvalue. Once the eigenvalues are found, you can plug them back into the equation (A-λI)x = 0 and solve for x to find the corresponding eigenvectors.

3. What is the significance of eigenvectors?

Eigenvectors are important in many areas of mathematics and science, particularly in linear algebra and quantum mechanics. They are used to describe the behavior of linear systems and can be used to simplify complex calculations.

4. Can there be more than one eigenvector for a given eigenvalue?

Yes, it is possible to have multiple eigenvectors for a single eigenvalue. In fact, an eigenvalue can have an infinite number of eigenvectors, as long as they are linearly independent.

5. How are eigenvectors used in data analysis?

In data analysis, eigenvectors are used in principal component analysis (PCA) to reduce the dimensionality of a dataset. This allows for easier visualization and interpretation of the data. Eigenvectors are also used in machine learning algorithms, such as linear regression and clustering.

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