Positive Definite Matrix

2013-05-18 4:48 am

A symmetric matrix A is positive definite if x(transpose)Ax > 0 for any non-zero matrix x.
Also, A is positive definite if and only if all its eigenvalues are positive.

I have already proved that if A is positive definite, all its eigenvalues are positive.
How can I prove "If all the eigenvalues of symmetric A are positive, then A is positive definite"? i.e. Given that lambda > 0, how to show x(transpose)Ax > 0 for non-zero x?

更新1:

Are you using the diagonalisability of A to prove this? Can I prove it without using the diagonalisability of A? i.e. ∵ Ax = λx, xTAx = xTλx = λxTx = λ(x1^2 + x2^2 + ... + xn^2) > 0 as x = [x1 x2 ... xn]T

回答 (1)

用戶9112

2013-05-18 6:50 am

✔ 最佳答案

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2013-05-18 10:30:42 補充：
http://i1090.photobucket.com/albums/i376/Nelson_Yu/int-775_zps54c8290f.png

2013-05-20 20:58:36 補充：
Ax = λx only when x is an eigenvector of A and not in general

2013-05-20 21:00:02 補充：
The finite-dimensional spectral theorem says that any symmetric matrix whose entries are real can be diagonalized by an orthogonal matrix.

2013-05-20 21:00:14 補充：
More explicitly: For every symmetric real matrix A there exists a real orthogonal matrix Q such that D = QTAQ is a diagonal matrix. Every symmetric matrix is thus, up to choice of an orthonormal basis, a diagonal matrix.

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