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.