Calculate Eigenvalues of a 2x2 Matrix | A, E, I | Eigenvectors | Eigenspace

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    Eigenvalue Matrix
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Homework Help Overview

The discussion revolves around calculating the eigenvalues of a 2x2 matrix, specifically the matrix E given as E = [[1, -1], [-1, 1]]. The original poster is attempting to understand the relationship between matrix E and the eigenvalue equation det(A - λI) = 0.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster questions whether matrix E is equivalent to matrix A and expresses confusion regarding the eigenvalue calculation. They attempt to derive the characteristic polynomial but seem uncertain about their results.

Discussion Status

Some participants have provided clarifications regarding the eigenvalue equation and pointed out potential errors in the original poster's determinant calculations. There appears to be an ongoing exploration of the correct eigenvalues, with different interpretations being discussed.

Contextual Notes

The original poster is working under the assumption that E might be the same as A, which raises questions about the definitions and relationships between the matrices involved. There is also a mention of the identity matrix I in the context of the eigenvalue equation.

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Homework Statement



Let A =
a b
c d
A characteristic value of A (often called an eigenvalue) is denoted by λ and satisfies the relation

det(A - λI) = 0

Obtain the characteristics values of E =
1 -1
-1 1


Homework Equations



Well I is the unit or identity matrix

1 0
0 1


The Attempt at a Solution



I don't understand how E can be of any relation to what the question is asking. Does E = A?

det(A - λI) = 0

=> a - λ, b
c, d - λ = 0

super. ad + λ² - λa - λd + bc = 0

Let's presume for a second that their asking me that A = E

that means a=1, b = -1,c = -1,d = 1

=> 1 + λ² - λ(1-1) + 1 = 0
=? λ² = -2

that canny be though can it?

Any suggestions are welcomed!

Thanks
Tom
 
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Yes, they want the solutions of the equation |E - [itex]\lambda[/itex]I| = 0

I get different eigenvalues, both real. Check your determinant work.
 
ahh, my bad

x² - 2x = 0

=> x(x-2)

therefore, x = 0, 2

Is that what you got?

Thanks :)
Tom
 
You lost your equation. x(x - 2) = 0, which allows you to say x = 0 or x = 2.
thomas49th said:
ahh, my bad

x² - 2x = 0

=> x(x-2)

therefore, x = 0, 2

Is that what you got?

Thanks :)
Tom
 

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