Can anyone please check my work and answer for this math problem?

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Homework Help Overview

The discussion revolves around determining whether a given vector is an eigenvector based on a specific mathematical condition involving scalar multiples. The subject area is linear algebra, particularly focusing on eigenvalues and eigenvectors.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • The original poster seeks confirmation on their conclusion regarding the vector being an eigenvector. Some participants explore the relationship between the given vectors and question whether one is a scalar multiple of the other.

Discussion Status

The discussion includes attempts to clarify the concept of eigenvectors and scalar multiples. Some participants provide insights into the mathematical reasoning behind the relationship between the vectors, while others express uncertainty about the original poster's conclusion.

Contextual Notes

There is a mention of humor that may have caused confusion, and the original poster expresses uncertainty about their answer, indicating a need for further clarification on the topic.

Math100
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Homework Statement
Is [-1+sqrt(2), 1] an eigenvector of [2, 1; 1, 4]? If so, find the eigenvalue.
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Can anyone please confirm and check my work with answer because I'm not 100% sure if my work with answer is 100% correct. The question is asking if the given vector is an eigenvector and I've answered no. What's the correct answer for this problem? It's Yes or No?
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I think you need to write a little bit larger. Got any A3 nearby?
 
Sorry, I don't understand. What do you mean about A3 nearby?
 
Nevermind, it was just a bad attempt at humour. As for your question, is the vector ##
\begin{pmatrix}
2\sqrt{2} - 1\\
\sqrt{2} +3
\end{pmatrix}## a scalar multiple of the vector ##
\begin{pmatrix}
\sqrt{2} - 1\\
1
\end{pmatrix}##?
 
I think the answer is Yes, because (√2-1)(√2+3)=2√2-1. Therefore, it's scalar multiple. Am I right?
 
Right. And and if the result is a scalar multiple of the first vector, as in ##Av = \lambda v##, then ##v## is an eigenvector. Does that answer your question? :smile:
 
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Yes. Thank you so much for the help.
 
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