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