Eigenspaces-symmetric matrix (2x2)

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SUMMARY

A 2x2 symmetric matrix with real entries and two distinct eigenvalues has eigenspaces that are perpendicular lines through the origin in R². The characteristic polynomial's discriminant must be greater than zero, specifically expressed as (a-d)² + 4b² > 0, indicating that a cannot equal d and b cannot equal zero simultaneously. The orthogonality of eigenvectors can be established using the properties of symmetric matrices, specifically that the eigenvectors corresponding to distinct eigenvalues are orthogonal.

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  • Understanding of 2x2 symmetric matrices
  • Knowledge of eigenvalues and eigenvectors
  • Familiarity with characteristic polynomials
  • Concept of orthogonality in vector spaces
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  • Study the properties of symmetric matrices and their eigenvalues
  • Learn how to compute the characteristic polynomial of a matrix
  • Explore the concept of orthogonality in R²
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*best&sweetest*
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I have a tough time proving that if a 2x2 symmetric matrix with real entries has two distinct eigenvalues than the eigenspaces corresponding to those eigenvalues are perpendicular lines through the origin in R^2.



All I have is symmetric matrix

a b
b d

and I know that since it has distinct eigenvalues it must be that the discriminant of the characteristic polynomial must be greater than zero, i.e

(a-d)^2 + 4b^2 > 0

so it cannot be that a=d and b=0 at the same time.

Now, I have no clue what to do next, and I have a feeling that this might be on my final exam, so any help would be greatly appreciated. Thanks!
 
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You know a lot more than this. You know there are two eigenvectors and two eigenvalues. Now, what is the only way we have of deciding when two vectors are orthogonal? Remember, symmetrice means A=A^t, and what have you been taught about matrices and the only method you know for deciding when two vectors are orthogonal.
 
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