Say we've got all possible unit vectors in $\mathbb R_2$ drawn out. Their endpoints would trace out a unit circle.
If we were to multiply every unit vector composing that circle by a symmetric matrix, it would turn the circle into an ellipse – this is due to symmetric matrices having orthogonal eigenvectors, aka stretching space in perpendicular directions.
The easiest example to think of is with a diagonal matrix
$\begin{bmatrix}
3 & 0 \\
0 &
4
\end{bmatrix} $
Which would turn our circle of unit vectors…:
($x^2 +y^2 =1$)
…into an ellipse:
$(\frac{x}{3})^2 + (\frac{y}{4})^2=1$
For a symmetric matrix, it's easy to find in which direction the axes of that ellipse point after we apply the transformation – the axes of the ellipse will point in the direction of the eigenvectors of the symmetric matrix, since those are the directions in which the symmetric matrix stretched space.
In this case, the major and minor axes are on the $x$ and $y$ axes, since the eigenvectors of this matrix lay on the $x$ and $y$ axes in the first place.
But, let's say our matrix isn't symmetric to begin with. For example…
$\begin{bmatrix}
4 & 2 \\
3 &
7
\end{bmatrix} $
And we multiply every unit vector composing the original circle by this matrix.
How would we be able to find where the new axes of this transformed circle (of the ellipse) would be?
Thanks!
Best Answer
If $A$ is the inverse of the transformation matrix, then the equation of the transformed circle is: $$ x^TA^TAx=1, $$ hence you must look for the eigenvectors of $A^TA$, which is a symmetric matrix.
In your example: $$ A=\pmatrix{4 & 2\\ 3 & 7}^{-1}=\pmatrix{7/22 & -1/11\\ -3/22 & 2/11}; \quad A^TA=\pmatrix{29/242 & -13/242\\ -13/242 & 5/121}, $$ with eigenvectors $(-19\pm\sqrt{1037}, 26)$.