Prove that if $A$ is a symmetric matrix with all eigenvalues greater than $0$, then it is positive definite.
If $A$ is symmetric then there exists an orthogonal matrix $S$, such that $S^TAS$ is a diagonal matrix. I'm not sure about this, but if a matrix is diagonalizable, then the eigenvalues of that matrix are equal to the corresponding entries in the diagonal of its diagonalized form, correct? So if $S^TAS$ is diagonal, then the entries in the diagonal equal to the eigenvalues?
Either way, even if they were and assuming they were positive, I could only show that $x^TAx>0$ if $x$ is an eigenvector of $A$, but how would one prove this for all $x \neq 0$?
Best Answer
Since a symmetric matrix is always diagonalizable, its eigenvectors form a basis. Moreover there there exists an orthonormal basis of eigen vectors corresponding to the eigen values. Let $\{e_1,e_2,..e_n\}$ be the ortonormal basis of eigen vectors corresponding to the eigenvalues $\{\lambda_1,\lambda_2,..\lambda_n\}$. For $x \ne 0$, $x =c_1e_1+c_2e_2+..c_ne_n$. So $x^{T}Ax=\{c_1e_1^{T}+...c_ne_n^T\}\{c_1Ae_1+..c_nAe_n\}=\{c_1e_1^{T}+...c_ne_n^T\}\{c_1\lambda_1e_1+..c_n\lambda_ne_n\}=c_1^2\lambda_1+...c_n^2\lambda_n \ge 0$