If a self adjoint operator over an $n$ dimensional (real/complex) inner product space V has all its eigenvalues positive then the operator is positive definite.
Definition: Operator $T$ is positive definite if inner product of $Ax$ and $x$ is always positive.
I am comfortable when $T$ is real symmetric. But don't know how to proceed for general self adjoint operator.
Best Answer
Since $A$ is self-adjoint, then $A$ is orthogonally diagonalizable. So any vector $x$ can express as the linear combination of eigenvectors $\{x_i\}_{i=1}^n$, say $x=\sum_{i=1}^n a_ix_i$. Note that we can assume that $\{x_i\}_{i=1}^n$ form a orthonormal basis and $Ax_i=\lambda_i x_i$ for some $\lambda_i>0$. Then
$(Ax,x)=(\sum_{i=1}^n a_iAx_i,\sum_{j=1}^n a_jx_j)=\sum_{i,j}a_i \overline{a_j}(Ax_i, x_j)=\sum_{i,j}a_i \overline{a_j}(\lambda_ix_i, x_j)$
by $\{x_i\}_{i=1}^n$ form a orthonormal basis $\sum_{i,j}a_i \overline{a_j}(\lambda_ix_i, x_j)=\sum_{i}|a_i|^2\lambda_i(x_i, x_i)=\sum_{i}|a_i|^2\lambda_i||x_i||^2>0$.