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The value of the dot product of the eigenvectors corresponding to any pair of different eigenvalues of a 4 - by - 4 symmetric positive definite matrix is ___________
Correct answer is '0'. Can you explain this answer?
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The value of the dot product of the eigenvectors corresponding to any ...
This is because eigen vectors corresponding to DIFFERENT eigen values of a REAL symmetric matrix are ORTHOGONAL to each other.
However, same eigen values they may not be.
And Dot -product of orthogonal vectors(perpendicular vectors ) is 0 (ZERO)
However, same eigen values they may not be.
And Dot -product of orthogonal vectors(perpendicular vectors ) is 0 (ZERO)
Most Upvoted Answer
The value of the dot product of the eigenvectors corresponding to any ...
Explanation:
Let A be a 4-by-4 symmetric positive definite matrix with eigenvalues λ1, λ2, λ3 and λ4.
Let v1, v2, v3 and v4 be the corresponding eigenvectors with unit length.
We know that eigenvectors corresponding to different eigenvalues are orthogonal to each other.
Proof:
Let λ1 and λ2 be two distinct eigenvalues of A, with corresponding eigenvectors v1 and v2. Then we have:
Av1 = λ1v1 and Av2 = λ2v2
Now take the dot product of v1 and v2:
v1 . v2 = (1/λ1)(Av1) . v2
= (1/λ1)v1 . (A^T)v2 (since A is symmetric)
= (1/λ1)v1 . Av2 (since v2 is an eigenvector of A)
= (1/λ1)λ2v1 . v2
= (λ2/λ1)(v1 . v2)
Since λ1 and λ2 are distinct, the fraction (λ2/λ1) is not equal to 1.
Therefore, v1 . v2 must be equal to 0, since v1 and v2 are orthogonal.
By the same argument, we can show that v1 . v3, v1 . v4, v2 . v3 and v2 . v4 are all equal to 0.
Therefore, the dot product of any pair of different eigenvectors is equal to 0.
Hence, the value of the dot product of the eigenvectors corresponding to any pair of different eigenvalues of a 4-by-4 symmetric positive definite matrix is 0.
Let A be a 4-by-4 symmetric positive definite matrix with eigenvalues λ1, λ2, λ3 and λ4.
Let v1, v2, v3 and v4 be the corresponding eigenvectors with unit length.
We know that eigenvectors corresponding to different eigenvalues are orthogonal to each other.
Proof:
Let λ1 and λ2 be two distinct eigenvalues of A, with corresponding eigenvectors v1 and v2. Then we have:
Av1 = λ1v1 and Av2 = λ2v2
Now take the dot product of v1 and v2:
v1 . v2 = (1/λ1)(Av1) . v2
= (1/λ1)v1 . (A^T)v2 (since A is symmetric)
= (1/λ1)v1 . Av2 (since v2 is an eigenvector of A)
= (1/λ1)λ2v1 . v2
= (λ2/λ1)(v1 . v2)
Since λ1 and λ2 are distinct, the fraction (λ2/λ1) is not equal to 1.
Therefore, v1 . v2 must be equal to 0, since v1 and v2 are orthogonal.
By the same argument, we can show that v1 . v3, v1 . v4, v2 . v3 and v2 . v4 are all equal to 0.
Therefore, the dot product of any pair of different eigenvectors is equal to 0.
Hence, the value of the dot product of the eigenvectors corresponding to any pair of different eigenvalues of a 4-by-4 symmetric positive definite matrix is 0.
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The value of the dot product of the eigenvectors corresponding to any pair of different eigenvalues of a 4 - by - 4 symmetric positive definite matrix is ___________Correct answer is '0'. Can you explain this answer?
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