Let's denote the sphere by $S^2 = \{x \in \Bbb{R}^3: |x| = 1\}$
Let $T$ be a symmetric matrix with orthonormal eigenpairs $\{\sigma_i, v_i\}$, $i = 1,2,3$ such that $\sigma_1 > \sigma_2 > \sigma_3$. I want to show that the maximum of the real function $$x \mapsto |Tx - (Tx \cdot x)x|$$ on $S^2$ is $\frac{1}{2}(\sigma_1-\sigma_3)$ and this value is attained at $x = \pm \frac{1}{\sqrt{2}}(v_1 \pm v_3)$.
Attempt: That function squared is $$|Tx - (Tx\cdot x)x|^2 = |Tx|^2 - 2(Tx\cdot x)^2 +(Tx \cdot x)^2 |x|^2 = x \cdot T^TTx - (Tx\cdot x)^2 $$ Using the method of Lagrange multiplier, the auxillary function $\mathcal{L}$ is $$x \cdot T^2x - (Tx\cdot x)^2 -\lambda(x \cdot x -1) $$ Using the fact that $\frac{\partial}{\partial x}(x \cdot Ax) = (A+A^T)x$, $\partial_x \mathcal{L} = 2T^2x -2(Tx \cdot x)(2Tx)- 2\lambda x $. Setting $\partial_x \mathcal{L} = 0,$ we have $[T^2-2(Tx \cdot x)T-\lambda I]x = 0$. Setting $\partial_\lambda \mathcal{L} = 0$ gives our original constraint $|x| = 1$...
Background: I wish to solve this problem because this is the maximum shearing stress at a point under a triaxial stress state. The above result is written in most mechanics of material/continuum mechanics textbook, but most probably without a proof.
In mechanics, every body is associated with a Cauchy stress tensor field $\mathcal{T}$. Given a point $p$, the value $\mathcal{T}(p)$ of that (1,1)-tensor field $\mathcal{T}$ is called the stress state at $p$, which can be regarded as a symmetric matrix $T$. The Cauchy's fundamental theorem states that for any surface $S$ having the unit normal vector $x$ at point $p$, the traction on $S$ is given by $Tx$. Now we can orthogonally decompose $$Tx = \sigma x + v$$ for some $v \perp x$. Then the unique values $\sigma$ and $|v|$ are called the normal stress and shear stress on $S$ at $x$ respectively. With the Lagrange multiplier method, it can be shown that $\max_{x \in S^2}\sigma = \max\{|\sigma_1|, |\sigma_3|\}$ and it is attained at either $v_1$ or $-v_3$. Butseems we need more than Lagrange multiplier to maximize $\tau$...