Let $f:\mathbb{R}^n \to \mathbb{R^n}$ be a linear application, we suppose that $f$ is symmetric ($\langle f(x),y\rangle=\langle x, f(y)\rangle$), without using spectral theorem how we can see that $f$ maps the unit ball into an ellipsoid ?
Or how can we prove spectral theorem geometrically (intrinsically) ?
Can expand this reasoning to see why not any square matrix is diagonalizable ?
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$An invertible linear operator $f$ on $\Reals^{n}$ maps the unit ball to an ellipsoid whether or not $f$ is symmetric (or even diagonalizable): One strategy is to write the unit ball as the locus of a quadratic inequality and perform a linear change of variables, concluding the image is compact and defined by a quadratic inequality, hence an ellipsoid (for some value of "hence").
In other words, knowing the image of the unit ball is an ellipsoid does not imply the spectral theorem, so if I understand what you're asking, the answer to your third question is "no".
As for proving the spectral theorem geometrically, one approach is to maximize the quadratic function $$ F(x) = \Brak{x, f(x)} $$ restricted to the unit sphere, i.e., subject to the constraint $g(x) = \|x\|^{2} = 1$. By Heine-Borel and the extreme value theorem, $F$ has an absolute maximum, $x_{0}$; Lagrange multipliers and symmetry of $f$ show that $$ 2f(x_{0}) = \nabla F(x_{0}) = \lambda \nabla g(x_{0}) = 2\lambda x_{0}; $$ that is, $x_{0}$ is an eigenvector of $f$, with eigenvalue equal to the maximum value of $F$. Now induct, restricting everything to the orthogonal complement of $x_{0}$, and iteratively constructing (for some value of "construct") an orthonormal $f$-eigenbasis of $\Reals^{n}$.