Understanding Eigenvectors and Eigenvalues in Linear Algebra

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Homework Help Overview

The discussion revolves around understanding eigenvectors and eigenvalues in the context of a specific matrix in linear algebra. Participants are exploring the properties of eigenvectors, particularly regarding scalar multiples and their validity as eigenvectors.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between different scalar multiples of eigenvectors and question whether two given vectors are equivalent eigenvectors. They also reference the definition of eigenvectors and how to verify them using matrix multiplication.

Discussion Status

There is an ongoing exploration of the properties of eigenvectors, with some participants providing clarifications regarding the nature of scalar multiples. Guidance has been offered on how to check if a vector is an eigenvector using the definition, but no consensus has been reached on further implications or applications.

Contextual Notes

Participants express varying levels of familiarity with linear algebra concepts, indicating that some foundational knowledge may be assumed while others are still developing their understanding.

TheSpaceGuy
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I am trying to get an eigenvector for the following matrix, I am up to the final step.
4 1
0 0

I got it to be
-1
4

is this the same as
1
-4



sorry I am pretty new to linear algebra.
 
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TheSpaceGuy said:
I am trying to get an eigenvector for the following matrix, I am up to the final step.
4 1
0 0

I got it to be
-1
4

is this the same as
1
-4
sorry I am pretty new to linear algebra.

Of course, they aren't the same vector. But if x is an eigenvector then c*x is also an eigenvector for any constant c. In your example the c is (-1). Both of those are fine eigenvectors.
 
As Dick said, any scalar multiple of an eigenvector is an eigenvector- in fact, any linear combination of eigenvectors is an eigenvector.

Of course, a good way to check if any vector is any eigenvector is to use the definition of "eigenvector". If v is an eigenvector of A, corresponding to eigenvalue \lambda, then Av= \lambda v.

Here,
\begin{bmatrix}4 & 1 \\ 0 & 0\end{bmatrix}\begin{bmatrix}-1 \\ 4\end{bmatrix}= \begin{bmatrix}-4+ 4 \\ 0\end{bmatrix}= \begin{bmatrix}0 \\ 0\end{bmatrix}= 0\begin{bmatrix}-1 \\ 4\end{bmatrix}
and
\begin{bmatrix}4 & 1 \\ 0 & 0\end{bmatrix}\begin{bmatrix} 1 \\ -4\end{bmatrix}= \begin{bmatrix}4- 4 \\ \end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}= 0\begin{bmatrix}1 \\ -4\end{bmatrix}

So these are both eigenvectors corresponding to eigenvalue 0.
 
Last edited by a moderator:
Thanks to the both of you. That really clears things up for me!
 

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