Graphical intuition for the eigenvalue of a linear map by looking at the unit sphere

Question

Let's suppose we're in a finite dimensional normed real vector space $V$, where $\dim(V) = n$. Let's say we have a linear map $T: V \to V$, which transforms the unit sphere, $\{v \in V: ||v|| = 1\}$ which we can understand as a circle (looking at its two dimensional analogue) into an ellipse. The claim I'm struggling with is that the vector that lies on the major axis (i.e. the "longest" vector in this ellipse) is an eigenvector of $T$. I don't really understand why this is the case, so I'd appreciate some graphical intuition.

Answer 1

First of all, it's important to know that vector that maximizes $\|Tv\|$, i.e. the "direction of the major axis", is the eigenvector of the maximal eigenvalue of $T^*T$. These facts are usually proven by applying the Rayleigh-Ritz theorem and noting that $$ \|Tx\|^2 = \langle Tx, Tx \rangle = \langle T^*Tx,x \rangle. $$ As far as graphical intuition though, the polar decomposition of a matrix is useful. In particular, polar decomposition tells us that any transformation $T$ can be written in the form $UP$ where $U$ is unitary and $P$ is positive definite (i.e. self-adjoint with positive eigenvalues). More specifically, $P$ is given by $P = \sqrt{T^*T}$.

Because $P$ is a positive definite matrix, it is a "pure squish/stretch" of space along perpendicular axes (as guaranteed by the spectral theorem). So when $P$ acts on the unit circle, the major axis of the ellipsoidal image will point in the direction of the eigenvector $v$ of $P$ associated with the largest eigenvalue.

The $U$ that is applied afterwords rotates and reflects but does not distort the resulting ellipsoid, so $v$ is still the vector that got stretched out the most. Keep in mind, however, that $v$ will not necessarily be an eigenvector of $T = UP$ since the rotation/reflection might make it so that the major axis no longer points in the direction of $v$.

So, in a nutshell: the directions along which $A$ "stretches" and "squishes" are encoded by the eigenvectors of $P = \sqrt{T^*T}$, which are the same as the eigenvectors of $T^*T$.