7C Positive Operators

Let $T \in \mathscr{L}(V)$. The following are equivalent. a) $T$ is a positive operator. b) $T$ is self-adjoint and all eigenvalues of $T$ are non-negative. c) $T$ has a diagonal matrix consisting of only non-negative diagonals with respect to some orthonormal basis. d) $T$ has a positive square root. e) $T$ has a self-adjoint square root. f) $T = R^*R$ for some $R \in \mathscr{L}(V)$.

a) $\Rightarrow$ b): To show that $T$ must have non-negative eigenvalues, we note that because $\Vert v \Vert > 0$ for every eigenvector $v,$ this means that $\langle T(v), v \rangle$ and $\lambda$ must have the same sign, as $\langle T(v), v \rangle = \lambda \Vert v \Vert^2.$ b) $\Rightarrow$ c): This follows from the spectral theorem, as $T$ being self-adjoint means there is diagonal matrix for $T$ with respect to an orthonormal basis. The non-negative eigenvalues are, of course, the diagonal entries. c) $\Rightarrow$ d): Since we have a diagonal matrix for $T$ with non-negative diagonal entries by hypothesis, we can construct a diagonal matrix with square roots of each of these entries on the diagonal. This is a square root for the diagonal matrix, so its associated operator is a square root of $T$. Furthermore, it is positive because its own conjugate transpose and every inner product of the form $\langle T(v), v \rangle$ is the standard dot product, which results in a sum of products of non-negative diagonal entries and squares, which are never negative. d) $\Rightarrow$ e): This follows because positive operators are already self-adjoint. Note that the converse, on the other hand, is proven in 5 steps in this proof. e) $\Rightarrow$ f): Setting $R$ to a self-adjoint square root of $T$ that exists by hypothesis, we have $T = R^2 = R^* R.$ f) $\Rightarrow$ a): Every operator of the form $R^* R$ is self-adjoint, even if $R$ is just a linear map. So we will check that $T$ is positive. Let $v \in V$. Then

Every positive operator on $V$ has a unique positive square root.

With existence proven in the last result, we will prove uniqueness. The strategy for the proof will be in showing that the behavior of R is uniquely determined on eigenvectors. Because $V$ has an orthonormal basis of eigenvectors of $T$, this will be sufficient. Suppose $T=R^2$ and $R$ are both positive operators, and suppose $v \in V$ is an eigenvector of $T$ with eigenvalue $\lambda.$ Let $(e_1, \ldots, e_n)$ be an orthonormal basis with respect to which $R$ has a diagonal matrix. So $R(v) = R(c_1 e_1 + \cdots c_n e_n) = \lambda_1 c_1 e_1 + \cdots \lambda_n c_n e_n,$ where the $\lambda_i$'s are the non-negative diagonal entries of $R$'s matrix, and by extension

Because we also have $T(v) = \lambda v$, this gives us a comparison of coefficients for linearly independent vectors (which must be the same):

For all terms where $\lambda \neq \lambda_i^2,$ this means $c_i = 0$, leaving us with only basis vectors that share the common eigenvalue $\sqrt{\lambda}.$ Therefore, $R$ must scale $v$ by the scalar $\sqrt{\lambda},$ uniquely determining $R$'s behavior on the eigenvectors of $T$ in $V$.