In $\mathbb{R}^n$, given a linear transformation $A$ represented by a matrix $(A_{ij})$, it makes sense that $A: \mathbb{R}^n \to \mathbb{R}^n$ maps the unit $n$-sphere $x_1^2 + \cdots + x_n^2=1$ onto an $n$-dimensional ellipsoid. Is it possible to prove this, and, if so, is there a resulting general formula for the coefficients in the $n$-ellipsoid, $c_1x_1^2 + \cdots + c_nx_n^2=1$? Even for the $n=2$ case, the math quickly becomes unmanageable when I have tried to prove it.
In the $n=2$ case, $(x_1,x_2)$ is mapped by $A$ to $(A_{11}x_1 + A_{12}x_2,A_{21}x_1 + A_{22}x_2)$. When this vector is squared, we get $x_1^2(A_{11}^2+A_{21}^2)+x_2^2(A_{12}^2+A_{22}^2)+2x_1x_2(A_{11}A_{12}+A_{21}A_{22})$. Am I even on the right track here?
I would appreciate any and all help. Thank you.
$\endgroup$ 11 Answer
$\begingroup$What you are looking for is SVD decomposition. If $A$ is invertible, it will map the $n$-dimensional unit sphere onto an $n$-dimensional ellipsoid but the axes of the ellipsoid won't necessarily be the standard basis vectors (they can be rotated with respect to the standard axes) so the ellipsoid won't necessarily be of the form
$$ c_1 x_1^2 + \ldots + c_n x_n^2 = 1. $$
Using SVD decomposition, you can write $A$ as $A = UDV^T$ where $D = \operatorname{diag}(\sigma_1, \ldots, \sigma_n) \in M_n(\mathbb{R})$ is a diagonal matrix and $U,V \in M_n(\mathbb{R})$ are orthogonal matrices. Denoting by $u_i$ the columns of the matrix $U$, the image of the unit sphere will be
$$ \left \{ x_1 (\sigma_1 u_1) + \dots + x_n (\sigma_n u_n) \, \biggr\rvert \, \sum_{i=1}^n x_i^2 = 1 \right \} = \left \{ y_1 u_1 + \dots + y_n u_n \, \biggr\rvert \, \sum_{i=1}^n \frac{y_i^2}{\sigma_i^2} = 1 \right \} $$
which is an ellipsoid whose axes are $u_1, \ldots, u_n$ and whose lengths are $\sigma_i$. The numbers $\sigma_i$ are called the singular values of $A$ and given by the square roots of the eigenvalues of the positive-definite matrix $A^T A$ (or $A A^T$).
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