Proof of A Positive Semi-Definite Square Root Acts on Eigenvectors by the Nonnegative Square Root

Write $0_{V}$ for the zero vector of $V$; for $y\in V$, $-y$ denotes the additive inverse of $y$ given by claim 2 of Elementary Identities in a Vector Space, and $x-y$ abbreviates $x+(-y)$.

Two preliminary facts.

(i) $\langle u,0_{V}\rangle=0$ for every $u\in V$. Indeed, claim 4 of Elementary Identities in a Vector Space gives $0_{V}=0\cdot 0_{V}$, so condition 3 of Complex Inner Product Space gives $\langle u,0_{V}\rangle=\langle u,0\cdot 0_{V}\rangle=0\,\langle u,0_{V}\rangle=0$.

(ii) $-(\nu y)=(-\nu)y$ for every complex number $\nu$ and every $y\in V$. Indeed, claim 5 of Elementary Identities in a Vector Space and condition 5 of Vector Space over a Field give $-(\nu y)=(-1)(\nu y)=\bigl((-1)\nu\bigr)y=(-\nu)y$.

An eigenvector equation. Put $w=R(v)-\mu v$, so that $w=R(v)+(-\mu)v$ by (ii). Since a linear operator is a linear map, conditions 1 and 2 of Linear Map give

On the other hand, conditions 7 and 5 of Vector Space over a Field give

since $(-\mu)(-\mu)=\mu^{2}=\lambda$ in the field of real numbers, by claim 2 of Zero Products and Elementary Identities in a Field. Comparing the two displays and using commutativity of addition in $V$ (condition 2 of Vector Space over a Field) gives

Consequence of positive semi-definiteness. Condition 3 of Complex Inner Product Space gives

Put $t=\langle w,w\rangle$; by condition 4 of Complex Inner Product Space, $t$ is a real number with $0\le t$, and $(-\mu)t=-(\mu t)$ by claim 2 of Zero Products and Elementary Identities in a Field. Since $R$ is positive semi-definite, $\langle w,R(w)\rangle$ is a real number with $0\le\langle w,R(w)\rangle$, that is,

Adding $\mu t$ to both sides, which preserves the order by condition 1 of Ordered Field, gives $\mu t\le0$. On the other hand $0\le\mu$ and $0\le t$ give $0\le\mu t$ by condition 2 of Ordered Field. The order of an ordered field is a total order and hence antisymmetric, so $\mu t=0$.

The vector $w$ is zero. Suppose first that $\mu\ne0$. Claim 3 of Zero Products and Elementary Identities in a Field applied to $\mu t=0$ gives $t=0$, that is $\langle w,w\rangle=0$, and condition 4 of Complex Inner Product Space gives $w=0_{V}$.

Suppose instead that $\mu=0$. Then $\lambda=\mu^{2}=0$ by claim 1 of Zero Products and Elementary Identities in a Field, and claim 3 of Elementary Identities in a Vector Space gives $\mu v=0v=0_{V}$ and $R(w)=(-\mu)w=0w=0_{V}$. Moreover $-0_{V}=0_{V}$, because $0_{V}+0_{V}=0_{V}$ and additive inverses are unique by claim 2 of Elementary Identities in a Vector Space; hence $w=R(v)-\mu v=R(v)+0_{V}=R(v)$. Since $R$ is self-adjoint,

by (i), so condition 4 of Complex Inner Product Space again gives $w=0_{V}$.

Conclusion. In both cases $R(v)+\bigl(-(\mu v)\bigr)=w=0_{V}$, so by commutativity of addition in $V$ the vector $R(v)$ satisfies $-(\mu v)+R(v)=0_{V}$. The vector $\mu v$ satisfies the same equation, since $-(\mu v)+\mu v=0_{V}$. By uniqueness of additive inverses, claim 2 of Elementary Identities in a Vector Space, applied to the vector $-(\mu v)$, we conclude $R(v)=\mu v$.