If $T$ is self-adjoint and $\langle T(u),u \rangle > 0$ for every $u$ in $K$, $u\neq0$ then $T = R^2$ for some self-adjoint non-singular operator $R$

Question

it´s my first question here and it´s because I can´t find the way to stablish the equivalence between those statements. I don´t see how the inner product or the fact that T is a self-adjoint operator could give me the operator R that i need. My attemp was trying to use the matrix representation of T and show that that matrix is positive definite and diagonalizable, then I could extract the square root matrix and then argument that that matrix is the representation of a non-singular linear operator (since T is non-singular) and that is self-adjoint. but I can´t propose the first matrix. Thanks in advance.