Proof of An Operator with an Orthonormal Eigenbasis is Positive Semi-Definite Exactly When its Eigenvalues are Nonnegative

Let $\lVert\cdot\rVert$ be the norm induced by the inner product, let $\overline{z}$ denote the complex conjugate of a complex number $z$ and $|z|$ its modulus, and write $\mathbb{R}$ for the set of real numbers; for $x$ in a field, $x^{2}$ abbreviates $x\cdot x$. Since $e$ is orthonormal, each $e_{j}$ is a unit vector, so $\langle e_{j},e_{j}\rangle=\lVert e_{j}\rVert^{2}=1$ by Norm Induced by a Complex Inner Product.

Necessity. Suppose $T$ is positive semi-definite and fix $j\in[n]$. Condition 3 of Complex Inner Product Space gives

By Positive Semi-Definite Operator the left-hand side is a real number $r$ with $0\le r$. Hence $\lambda_{j}$ is a real number with $0\le\lambda_{j}$.

Sufficiency. Suppose every $\lambda_{k}$ is a real number with $0\le\lambda_{k}$, and let $u\in V$. Claim 1 of Action of an Operator with an Orthonormal Eigenbasis gives

the sum being a finite sum in $V$. The first identity of claim 6 of Properties of Finite Sums of Vectors, applied with the coefficients $\lambda_{k}\langle e_{k},u\rangle$ and the vectors $e_{k}$, gives

a finite sum in $\mathbb{C}$. Condition 1 of Complex Inner Product Space gives $\langle u,e_{k}\rangle=\overline{\langle e_{k},u\rangle}$, and claim 3 of Properties of Complex Conjugation and Modulus gives $\langle e_{k},u\rangle\overline{\langle e_{k},u\rangle}=\bigl|\langle e_{k},u\rangle\bigr|^{2}$. Writing $r_{k}=\bigl|\langle e_{k},u\rangle\bigr|$ and using associativity of multiplication in $\mathbb{C}$, we obtain

By Modulus of a Complex Number each $r_{k}$ is a real number with $0\le r_{k}$, so condition 2 of Ordered Field gives first $0\le r_{k}^{2}$ and then $0\le\lambda_{k}r_{k}^{2}$.

By condition 1 of The Complex Numbers, $\mathbb{R}\subseteq\mathbb{C}$ and the addition and multiplication of $\mathbb{C}$ restrict on $\mathbb{R}$ to those of $\mathbb{R}$; consequently each product $\lambda_{k}r_{k}^{2}$ is the same whether formed in $\mathbb{R}$ or in $\mathbb{C}$, so every summand in the display above is a real number, and it is the nonnegative real number just exhibited. Moreover the finite sum of these real numbers formed in $\mathbb{C}$ and the one formed in $\mathbb{R}$ satisfy the same recursion, namely the one recorded in claim 1 of Properties of Finite Sums: both take the value $\lambda_{1}r_{1}^{2}$ at index $1$, and both pass from the index $m$ to the index $S(m)$ by adding $\lambda_{S(m)}r_{S(m)}^{2}$, the addition being the same in either field. Induction on the index therefore shows that the two sums are equal. Hence $\langle u,T(u)\rangle$ is a real number, and claim 5 of Properties of Finite Sums, applied to the nonnegative real summands $\lambda_{k}r_{k}^{2}$, gives $0\le\langle u,T(u)\rangle$.

Since $u\in V$ was arbitrary, $T$ satisfies the condition of Positive Semi-Definite Operator and is positive semi-definite.