Uncertainty principle manifesting in $j(j+1)$ vs $j^2$ To motivate my question, please consider a system with total angular momentum $j$. The fact that the largest eigenvalue of $J_z$ is $j$, while $J^2 = J_x^2 + J_y^2 + J_z^2$ has all eigenvalues equal to $j(j+1)$ is often ascribed to the uncertainty principle.
For example, quoting page $51$ of the textbook "Models of Quantum Matter" by Hans-Peter Eckle,

"...However, the eigenvalue of $L^2$ is $l(l+1)$, larger than $l^2$.
This implies that the angular momentum operator $\bf{L}$ can never
align with certainty with $L_3$ and the uncertainty principle is
satisfied. If $\bf{L}$ could be aligned with $L_3$, then $L_1=L_2=0$
and we would have simultaneously sharp values of all three components
of the angular momentum operator, in contradiction to Heisenberg's
uncertainty relations..."


To what extent is this reasoning true in general? It is hard for me to formulate my question more precisely, but I will attempt to do so:
Consider some set of Hermitian operators $K_i$ which all pairwise do not commute but instead each commute with the sum of their squares, $K^2 = \sum_i K_i^2$. Is it guaranteed that $K^2$'s largest eigenvalue is strictly greater than any of the eigenvalues of the individual $K_i^2$? The strictly greater is key, as it is greater than or equal by this answer. I hope to see that the failure of the individual $K_i$ to commute amongst themselves imposes a stronger statement.
 A: The following two operators $A$ and $B$ have the properties that:

*

*$A$ and $B$ do not commute;

*$A^2 + B^2$ commutes with both $A$ and $B$; and

*the largest eigenvalue of $A^2$ equals the largest eigenvalue of $A^2 + B^2$:

$$
A = \begin{bmatrix} \lambda & 0 & 0 \\ 0 & 0 & 1/\sqrt{2} \\ 0 & 1/\sqrt{2} & 0 \end{bmatrix} \qquad B = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & i/\sqrt{2} \\ 0 & -i/\sqrt{2} & 0 \end{bmatrix} 
$$
for $\lambda \geq 1$. We have
$$
[A, B] = \begin{bmatrix} 0 & 0 & 0 \\ 0 & -i & 0 \\ 0 & 0 & i \end{bmatrix} \neq 0
$$
so the two operators do not commute.  However,
$$
A^2 = \begin{bmatrix} 
\lambda^2 & 0 & 0 \\ 0 & 1/2 & 0 \\ 0 & 0 & 1/2 \end{bmatrix} \quad \text{ and } \quad
 A^2 + B^2 = \begin{bmatrix} 
\lambda^2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.
$$
and we can see that for any $\lambda > 1$, the largest eigenvalues of both $A^2$ and $A^2 + B^2$ are $\lambda^2$.  It can also be easily shown that $A^2 + B^2$ commutes with both $A$ and $B$.
The "loophole" being exploited here is that $[A, B]$ has a non-trivial null space even though the commutator does not itself vanish.  Since $A$ and $B$ are simultaneously diagonalizable on this subspace, we're allowed to have a state with zero uncertainty of $A$ and $B$ lying within this subspace.
