Here are two n-dimensional vectors: $V_1$ and $V_2$

$V_1 (v_1,v_2, \dots ,v_n)$

$V_2 (v_1,v_2, \dots ,v_n)$

$V_1 \cos(\theta) + V_2 \sin(\theta)$ is an ellipse in the $n$-D space. (Its center is the origin.)

I want to find the directions and magnitudes of the major and minor axes.

I have a $2$-D example here: Ellipse with non-orthogonal minor and major axes?

People used SVD to find the minor and major axes of that $2$-D example, but I'm not familiar with SVD. So I'm having problem extending it to $n$-dimension or arbitrary $V_1$ and $V_2$.

Could anybody give a formula for it? Thank you very much in advance!

  • 1
    $\begingroup$ It's still possible to compute the singular value decomposition of the matrix whose columns are your two vectors; the singular values will then correspond to the two axes. $\endgroup$ Aug 23, 2012 at 17:52

1 Answer 1


For example, let's take the 3-dimensional vectors $$ V_1 = \pmatrix{1\cr 2\cr 1\cr},\ V_2 = \pmatrix{3\cr -3\cr 2\cr} $$ With these as columns we form the matrix $$A = \pmatrix{1 & 3\cr 2 & -3\cr 1 & 2\cr}$$ Now $$ A A^T = \pmatrix{1 & 3\cr 2 & -3\cr 1 & 2\cr} \pmatrix{1 & 2 & 1\cr 3 & -3 & 2\cr} = \pmatrix{10 & -7 & 9\cr -7 & 13 & 0\cr 9 & 0 & 13\cr}$$ which has eigenvalues $23, 13, 0$ corresponding to normalized eigenvectors $$ W_1 = \frac{1}{\sqrt{230}} \pmatrix{10\cr -7 \cr 9\cr},\ W_2 = \frac{1}{\sqrt{130}} \pmatrix{0 \cr 9\cr 7\cr},\ W_3 = \frac{1}{\sqrt{299}} \pmatrix{13\cr 7\cr -9\cr} $$ Note that these are orthogonal since $A A^T$ is symmetric. The major and minor axes have lengths $\sqrt{23}$ and $\sqrt{13}$ (the singular values, which are the square roots of the nonzero eigenvalues of $A A^T$) and directions $W_1$ and $W_2$ respectively.

  • $\begingroup$ Thank you so much. So for n-dimensional space (where n>3), Is it possible that we get more eigenvalues? Do we always pick the first two eigenvalues and corresponding eigenvectors as major and minor axes? $\endgroup$
    – Hi271
    Aug 24, 2012 at 1:55
  • $\begingroup$ If we're talking about an ellipse, $A$ has rank $2$, so $A A^T$ has rank $2$, and there are only $2$ nonzero eigenvalues. The largest eigenvalue gives you the major axis and the second gives you the minor axis. $\endgroup$ Aug 24, 2012 at 5:36
  • $\begingroup$ Thank you very much! I get it. $\endgroup$
    – Hi271
    Aug 24, 2012 at 14:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .