I'm reading Wikipedia's visual proof of the singular value decomposition. They say:
To get a more visual flavor of singular values and SVD factorization – at least when working on real vector spaces – consider the sphere $S$ of radius one in $\mathbb R^n$. The linear map $T$ maps this sphere onto an ellipsoid in $\mathbb R^m$. Non-zero singular values are simply the lengths of the semi-axes of this ellipsoid. Especially when $n = m$, and all the singular values are distinct and non-zero, the SVD of the linear map $T$ can be easily analyzed as a succession of three consecutive moves: consider the ellipsoid $T(S)$ and specifically its axes; then consider the directions in $\mathbb R^n$ sent by $T$ onto these axes. These directions happen to be mutually orthogonal.
Apply first an isometry $V^⁎$ sending these directions to the coordinate axes of $\mathbb R^n$. On a second move, apply an endomorphism $D$ diagonalized along the coordinate axes and stretching or shrinking in each direction, using the semi-axes lengths of $T(S)$ as stretching coefficients. The composition $D \circ V^⁎$ then sends the unit-sphere onto an ellipsoid isometric to $T(S)$. To define the third and last move $U$, apply an isometry to this ellipsoid so as to carry it over $T(S)$. As can be easily checked, the composition $U \circ D \circ V^⁎$ coincides with $T$.
Why should unit vectors that get linearly mapped to the semi-axes of an ellipse be orthogonal? Is there a neat mathematical argument for this?