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12. Let A be n×n symmetric matrix. Prove:

(a) The eigenvectors of A from different eigenspaces are orthogonal. (5%)
(b) If all the eigenvalues of A are positive then A is positive definite . (5%)

User Juan Ayala
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Final answer:

a) The eigenvectors of a symmetric matrix A from different eigenspaces are orthogonal.

b) If all the eigenvalues of A are positive, then A is positive definite.

Step-by-step explanation:

(a) The eigenvectors of A from different eigenspaces are orthogonal:

Let v and w be eigenvectors of A corresponding to different eigenvalues λ and μ, respectively. We need to show that v · w = 0.

Given that A is a symmetric matrix, it has real eigenvalues and its eigenvectors can be chosen to be orthogonal. Therefore, we can assume that v and w are orthogonal.

The dot product of orthogonal vectors is 0, so v · w = 0, which proves that the eigenvectors of A from different eigenspaces are orthogonal.


(b) If all the eigenvalues of A are positive then A is positive definite:

To prove that A is positive definite, we need to show that x · Ax > 0 for any non-zero vector x.

Let λ be an eigenvalue of A and v be the corresponding eigenvector. Since λ is positive, we have v · Av = v · (λv) = λ(v · v) > 0, as v · v > 0 (the dot product of a vector with itself is always non-negative and is positive only when the vector is non-zero).

Since A can be diagonalized with a basis of eigenvectors, any vector x can be written as a linear combination of eigenvectors, x = c₁v₁ + c₂v₂ + ... + cₙvₙ, where c₁, c₂, ..., cₙ are the coefficients and v₁, v₂, ..., vₙ are the normalized eigenvectors of A.

Therefore, we have x · Ax = (c₁v₁ + c₂v₂ + ... + cₙvₙ) · A(c₁v₁ + c₂v₂ + ... + cₙvₙ) = c₁²(v₁ · Av₁) + c₂²(v₂ · Av₂) + ... + cₙ²(vₙ · Avₙ) > 0, as each term in the summation is positive.

User Swagrov
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