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I had this problem when I read this paper. It states that(around eq.(1)) for all the vectors $\vec{x}$ in the unit sphere, $T\vec{x}$ will lead the unit sphere into an ellipsoid if $T$ has full rank where $T$ stands for a $3\times 3$ matrix. And the problem is, the authors stated that the orientation of the semiaxes is given by the eigenvectors of $TT^T$, but I can't figure out why it is so?

So far, I can only solve the special case when $T$ is a positive matrix, that is we have the spectral decomposition of $T$ as $\sum_{i=1}^3 p_i \vec{a_i}\vec{a_i}^T$. Then $\vec{x}$ can be written in the basis of $\vec{a_i}$, i.e., $\vec{x}=\sum_{i=1}^3\alpha_i\vec{a_i}$ and have $\sum_i\alpha_i^2=1$ because $\vec{x}$ stands for the vector on unit sphere. Then $T\vec{x}$ equal to $\sum_i\tilde{\alpha_i}\vec{a_i}$ with $\tilde{\alpha_i}=p_i\alpha_i$. So we have $\sum_i\frac{\tilde{\alpha_i}^2}{p_i^2}=1$, which is the standard formula describing an ellipsoid. But I can't figure out the general case when $T$ is not a positive matrix, the orientation of the semiaxes is given by the eigenvectors of $TT^T$.

narip
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1 Answers1

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The unit sphere is given by

$ x^T x = 1 $

where $x$ is the position vector of a point on the unit sphere.

Now the image of $x$ is $y = A x $, so that $x = A^{-1} y $

Substitute this into the equation of the unit sphere, you get

$ y^T A^{-T} A^{-1} y = 1 $

where $A^{-T} = \left( A^{-1} \right)^T $

The matrix $Q = A^{-T} A^{-1} $ can be diagonalized into $Q = R D R^T $

and the lengths of the semi-axes of the resulting ellipsoid are the reciprocals of the square roots of the eigenvalues of $Q$, which are the diagonal entries of the diagonal matrix $D$, while the direction of the respective axes are given by the columns of the rotation matrix $R$. Finally we note that the eigenvectors of $Q$ and those of $Q^{-1} = A A^T $ are the same.

Hosam Hajeer
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