Proof of Existence and Uniqueness of the Positive Semi-Definite Square Root

Write $n=\dim V$ for the dimension of $V$ and let $[n]$ be the initial segment determined by $n$. Sums of vectors are finite sums in $V$, and all order relations between real numbers are those of the ordered field $\mathbb{R}$; real numbers are complex numbers by condition 1 of The Complex Numbers, so they act as scalars on $V$.

Setting up. By Spectral Theorem for a Self-Adjoint Operator in Finite Dimensions there are an orthonormal basis $e\in V^{n}$ of $V$, with components $e_{k}$, and an $n$-tuple $\lambda\in\mathbb{R}^{n}$ of real numbers, with components $\lambda_{k}$, such that

Since $T$ is positive semi-definite, An Operator with an Orthonormal Eigenbasis is Positive Semi-Definite Exactly When its Eigenvalues are Nonnegative gives $0\le\lambda_{k}$ for every $k\in[n]$. For each $k\in[n]$ let $\mu_{k}$ be the unique real number with $0\le\mu_{k}$ and $\mu_{k}^{2}=\lambda_{k}$, which exists and is unique by Existence and Uniqueness of the Nonnegative Square Root of a Nonnegative Real Number; this defines an $n$-tuple $\mu$ with components $\mu_{k}$.

Existence. Let $R$ be the map from $V$ to $V$ given by

By claims 1 and 2 of Operators Diagonal in an Orthonormal Basis, $R$ is a self-adjoint linear operator on $V$ with $R(e_{k})=\mu_{k}e_{k}$ for every $k\in[n]$. Applying An Operator with an Orthonormal Eigenbasis is Positive Semi-Definite Exactly When its Eigenvalues are Nonnegative to $R$ and the tuple $\mu$, whose components are real and satisfy $0\le\mu_{k}$, shows that $R$ is positive semi-definite.

Let $RR$ be the product of Operations on Linear Operators, that is, the map sending $x\in V$ to $R(R(x))$; it is a linear operator on $V$ by claim 3 of Sums, Scalar Multiples, Composites and the Identity are Linear Operators. For every $k\in[n]$, condition 2 of Linear Map and condition 5 of Vector Space over a Field give

So $RR$ and $T$ are linear operators on $V$ that both send $e_{k}$ to $\lambda_{k}e_{k}$ for every $k\in[n]$, and claim 2 of Action of an Operator with an Orthonormal Eigenbasis, applied with the tuple $\lambda$, gives

Thus $R$ is self-adjoint, positive semi-definite and satisfies the required identity.

Uniqueness. Let $R'$ be any linear operator on $V$ that is self-adjoint, positive semi-definite and satisfies $R'(R'(x))=T(x)$ for every $x\in V$. Fix $k\in[n]$. Then

with $\lambda_{k}$ a real number satisfying $0\le\lambda_{k}$ and with $\mu_{k}$ its nonnegative square root. Hence A Positive Semi-Definite Square Root Acts on Eigenvectors by the Nonnegative Square Root, applied to $R'$ with $v=e_{k}$, gives $R'(e_{k})=\mu_{k}e_{k}$.

So $R'$ and $R$ are linear operators on $V$ that both send $e_{k}$ to $\mu_{k}e_{k}$ for every $k\in[n]$, and claim 2 of Action of an Operator with an Orthonormal Eigenbasis, applied with the tuple $\mu$, gives $R'(x)=R(x)$ for every $x\in V$, that is, $R'=R$.

Therefore $R$ is the unique linear operator on $V$ that is self-adjoint, positive semi-definite and satisfies $R(R(x))=T(x)$ for every $x\in V$.