Proof of Any Two Orthonormal Bases of a Complex Inner Product Space Have the Same Size

Write $\lVert\cdot\rVert$ for the induced norm on $V$, and $|z|$ and $\overline{z}$ for the modulus and conjugate of a complex number $z$; $x^{2}$ abbreviates $x\cdot x$, and sums are finite sums of real numbers. For a natural number $p$ let $[p]$ be the initial segment it determines, and let $\sigma$ and the tuples $\mathbf{1}^{(p)}$ be as in The Sum of $n$ Ones is Strictly Increasing in $n$.

Since a modulus is a real number, there is an array $a\in(\mathbb{R}^{n})^{m}$ with

and we let $b\in\mathbb{R}^{m}$ and $c\in\mathbb{R}^{n}$ be formed from $a$ as in Interchange of a Finite Double Sum, taken over the field $\mathbb{R}$.

Rows. Fix $j\in[m]$. Claim 2 of Orthonormal Expansion and Parseval's Identity in Finite Dimensions, applied to the orthonormal basis $e$ with $u=f_{j}$, gives $b_{j}=\lVert f_{j}\rVert^{2}$. Since $f$ is orthonormal, its component $f_{j}$ is a unit vector, so $b_{j}=1$. Hence $b=\mathbf{1}^{(m)}$ and $\sum_{j=1}^{m}b_{j}=\sigma(m)$.

Columns. Fix $k\in[n]$. For every $j\in[m]$, condition 1 of Complex Inner Product Space gives $\langle f_{j},e_{k}\rangle=\overline{\langle e_{k},f_{j}\rangle}$, and claim 3 of Properties of Complex Conjugation and Modulus gives $|\overline{z}|=|z|$; hence $\bigl|\langle f_{j},e_{k}\rangle\bigr|^{2}=(a_{j})_{k}$. Claim 2 of Orthonormal Expansion and Parseval's Identity in Finite Dimensions, applied to the orthonormal basis $f$ with $u=e_{k}$, therefore gives $c_{k}=\lVert e_{k}\rVert^{2}$, which equals $1$ because $e_{k}$ is a unit vector. Hence $c=\mathbf{1}^{(n)}$ and $\sum_{k=1}^{n}c_{k}=\sigma(n)$.

Conclusion. By Interchange of a Finite Double Sum, $\sigma(m)=\sum_{j=1}^{m}b_{j}=\sum_{k=1}^{n}c_{k}=\sigma(n)$, so $m=n$ by claim 3 of The Sum of $n$ Ones is Strictly Increasing in $n$.