Column vectors of U and V in singular value decomposition

Say I have a $3 \times 3$ matrix $C$ with singular value decomposition $$C = USV^T$$ where $U = [u_1 \hspace{1mm} u_2 \hspace{1mm} u_3], V = [v_1 \hspace{1mm} v_2 \hspace{1mm} v_3]$ are orthogonal matrices ($u_i,v_i$ column vectors) and $S$ is the singular values diagonal matrix.

I did an experiment using Matlab and found that the matrices $u_iv_i^T$ has norm ($\|\cdot\|_2$) approximately equal to $1$ for each $i=1,2,3$.

My question is that, mathematically, are the matrix 2-norms of $u_iv_i^T$ indeed equal to $1$?

Let $\mathbf{u}_i = \begin{bmatrix}u_{1i} \\ u_{2i} \\ u_{3i}\end{bmatrix}$. What you are computing is $$\mathbf{u}_i \mathbf{v}_j^T = \begin{bmatrix}u_{1i}\mathbf{v}_j^T \\ u_{2i}\mathbf{v}_j^T \\ u_{3i}\mathbf{v}_j^T\end{bmatrix},$$ which is a rank-1 matrix with only one non-zero singular value. Consequently, $$||\mathbf{u}_i \mathbf{v}_j^T||_2 = ||\mathbf{u}_i \mathbf{v}_j^T||_{F} = 1.$$
Note that $||\cdot ||_F$ denotes the Frobenius norm of a matrix. Also, $$||\mathbf{u}_i \mathbf{v}_j^T||_{F} = \sqrt{|u_{1i}|^2||\mathbf{v}_j||^2+|u_{2i}|^2||\mathbf{v}_j||^2+|u_{3i}|^2||\mathbf{v}_j||^2}=\sqrt{|u_{1i}|^2+|u_{2i}|^2+|u_{3i}|^2}=1.$$