[step:Identify the successive maxima with the singular values of $C$]
Let $r := \operatorname{rank}(C)$. By the [Singular Value Decomposition](/theorems/???), choose orthonormal bases
\begin{align*}
u_1,\dots,u_p \in \mathbb{R}^p,
\qquad
v_1,\dots,v_q \in \mathbb{R}^q
\end{align*}
and numbers
\begin{align*}
\sigma_1 \geq \sigma_2 \geq \cdots \geq \sigma_s \geq 0
\end{align*}
such that $\sigma_i>0$ for $1 \leq i \leq r$, $\sigma_i=0$ for $r<i\leq s$, and the full diagonal action is
\begin{align*}
C v_i = \sigma_i u_i,
\qquad
C^{\top}u_i = \sigma_i v_i
\end{align*}
for $1 \leq i \leq s$. If $q>s$, then $C v_j=0$ for $s<j\leq q$; if $p>s$, then $C^{\top}u_i=0$ for $s<i\leq p$.
For unit vectors $\alpha \in \mathbb{R}^p$ and $\beta \in \mathbb{R}^q$, write
\begin{align*}
\alpha = \sum_{i=1}^{p} x_i u_i,
\qquad
\beta = \sum_{j=1}^{q} y_j v_j,
\end{align*}
where $\sum_{i=1}^{p}x_i^2=1$ and $\sum_{j=1}^{q}y_j^2=1$. Then
\begin{align*}
\alpha^{\top}C\beta
=
\sum_{i=1}^{s}\sigma_i x_i y_i.
\end{align*}
By the [Cauchy-Schwarz Inequality](/theorems/???) applied in $\mathbb{R}^s$ to the vectors $(\sigma_i x_i)_{i=1}^s$ and $(y_i)_{i=1}^s$,
\begin{align*}
\alpha^{\top}C\beta
\leq
\left(\sum_{i=1}^{s}\sigma_i^2 x_i^2\right)^{1/2}
\left(\sum_{i=1}^{s}y_i^2\right)^{1/2}
\leq
\sigma_1.
\end{align*}
Equality is attained at $\alpha=u_1$ and $\beta=v_1$, so $\rho_1=\sigma_1$.
For the $k$-th constrained maximisation, we choose the already constructed canonical directions to be the SVD pairs $(u_i,v_i)$ for $1 \leq i<k$. This choice is legitimate because canonical directions are only determined up to an orthonormal change of basis inside a repeated singular-value subspace, while the canonical correlations are the ordered constrained maximum values. With this SVD choice, the transformed orthogonality conditions are
\begin{align*}
\alpha \perp u_1,\dots,u_{k-1},
\qquad
\beta \perp v_1,\dots,v_{k-1}.
\end{align*}
Thus $x_i=y_i=0$ for $1 \leq i<k$, and the same computation gives
\begin{align*}
\alpha^{\top}C\beta
=
\sum_{i=k}^{s}\sigma_i x_i y_i
\leq
\sigma_k.
\end{align*}
Equality is attained at $\alpha=u_k$ and $\beta=v_k$. If a different orthonormal basis is chosen inside a repeated singular-value subspace, the constraints remove the same number of dimensions from that subspace, and the displayed SVD coordinate estimate on the remaining orthogonal complement gives the same next ordered value. Therefore
\begin{align*}
\rho_k=\sigma_k
\end{align*}
for every $1 \leq k \leq s$.
[/step]