Proof of The Standard Basis of the Complex Coordinate Space is an Orthonormal Basis

Throughout, two elements of $\mathbb{C}^{n}$ are equal exactly when all their components agree, since by the definition of the complex coordinate space an element of $\mathbb{C}^{n}$ is an ordered $n$-tuple of complex numbers. We also record that for $j,k\in[n]$ the $j$-th component of $e_{k}$ is $1$ if $j=k$ and $0$ otherwise, by the definition of the standard basis vectors. By the definition of a tuple, an $n$-tuple in a set $X$ is a map from $[n]$ to $X$, so the finite sums below are formed from maps on $[n]$ as required.

Claim 1. Fix $k\in[n]$. By the definition of the standard inner product,

where $(e_{k})_{j}$ denotes the $j$-th component of $e_{k}$. The $n$-tuple in $\mathbb{C}$ whose $j$-th component is $\overline{(e_{k})_{j}}u_{j}$ takes the value $\overline{0}\,u_{j}=0$ at every $j\in[n]$ with $j\ne k$, and at $j=k$ it takes the value $\overline{1}\,u_{k}=u_{k}$; here $\overline{0}=0$ and $\overline{1}=1$ because $0$ and $1$ are real numbers and a complex number is fixed by conjugation exactly when it is real, by claim 1 of Properties of Complex Conjugation and Modulus. By claim 7 of Properties of Finite Sums the finite sum equals $u_{k}$.

Claim 2. Fix $j\in[n]$ and let $\pi_{j}:\mathbb{C}^{n}\to\mathbb{C}$ send an element to its $j$-th component. Since addition and scalar multiplication on $\mathbb{C}^{n}$ are defined componentwise in The Complex Coordinate Space, we have $\pi_{j}(x+y)=\pi_{j}(x)+\pi_{j}(y)$ and $\pi_{j}(\lambda x)=\lambda\pi_{j}(x)$ for all $x,y\in\mathbb{C}^{n}$ and $\lambda\in\mathbb{C}$, so $\pi_{j}$ is a linear map from $\mathbb{C}^{n}$ to $\mathbb{C}$, the latter regarded as a complex vector space over itself. By claim 4 of Properties of Finite Sums of Vectors,

The $n$-tuple in $\mathbb{C}$ whose $k$-th component is $u_{k}(e_{k})_{j}$ vanishes at every $k\ne j$ and equals $u_{j}\cdot1=u_{j}$ at $k=j$, so by claim 7 of Properties of Finite Sums the sum is $u_{j}$. Thus $\sum_{k=1}^{n}u_{k}e_{k}$ has $j$-th component $u_{j}$ for every $j\in[n]$, that is, it equals $u$.

Claim 3. We first check that $e$ is orthonormal. Let $k\in[n]$. By claim 1 applied to $u=e_{k}$ we get $\langle e_{k},e_{k}\rangle=(e_{k})_{k}=1$, so $\lVert e_{k}\rVert^{2}=1$ by the definition of the induced norm. Since $0\le\lVert e_{k}\rVert$ and also $0\le1$ with $1^{2}=1$, the uniqueness of nonnegative square roots in Existence and Uniqueness of the Nonnegative Square Root gives $\lVert e_{k}\rVert=1$, so $e_{k}$ is a unit vector. If $j,k\in[n]$ with $j\ne k$, then claim 1 applied to $u=e_{k}$ gives $\langle e_{j},e_{k}\rangle=(e_{k})_{j}=0$, so $e_{j}$ and $e_{k}$ are orthogonal.

Now let $u\in\mathbb{C}^{n}$. By claim 1 the $n$-tuples in $\mathbb{C}$ with components $u_{k}$ and $\langle e_{k},u\rangle$ coincide, so claim 2 reads $u=\sum_{k=1}^{n}\langle e_{k},u\rangle e_{k}$. Since this holds for every $u\in\mathbb{C}^{n}$, claim 1 of Orthonormal Expansion and Parseval's Identity in Finite Dimensions shows that $e$ is an orthonormal basis of $\mathbb{C}^{n}$.

Claim 4. By claim 3 and claim 2 of Orthonormal Expansion and Parseval's Identity in Finite Dimensions,

Replacing $\langle e_{k},u\rangle$ by $u_{k}$ and $\langle e_{k},v\rangle$ by $v_{k}$ using claim 1 gives the two stated identities.