Diagonalization versus s.d. product for non-commuting Hermitian matrices Although the application of the following is in quantum physics, the question per se is mathematical:

I have seen two characterizations of the problem in measuring a discrete variable of a state $\psi$ exactly with each of two non-commuting Hermitian operators $A$ and $B$:


The product of the standard deviations ( $=\sqrt{\langle \psi | A^2 | \psi \rangle - \langle \psi | A | \psi \rangle^2}$ and ditto for $B$) $\geq 1$.
One cannot simultaneously diagonalize the matrix representations of $A$ and $B$
(i.e., if $A = U^\dagger CU$ and $B = V^\dagger DV$, for unitary $U$ and $V$ and diagonal $C$ and $D$, with $\dagger$ denoting the adjoint), then $U \neq V$.


Where is the link between these two?
In quantum mechanics, the mean of an observable $A$ in the state $\psi$ is given by

$$\langle A \rangle_\psi = \langle \psi |A|\psi \rangle,$$

and from this the variance of $A$ in $\psi$ is usually defined by

$$\langle(\Delta A)^2 \rangle_\psi := \langle (A-\langle A \rangle_\psi)^2 \rangle_\psi = \| (A - \langle A \rangle_\psi) \psi \|^2 = \langle A^2 \rangle_\psi - \langle A \rangle_\psi^2.$$

The standard deviation is then given by taking the square root.


  Theorem: (Heisenberg's Uncertainty Principle)
  
  $$\langle (\Delta A)^2 \rangle_\psi \langle (\Delta B)^2 \rangle_\psi \geq \frac{1}{4} | \langle [A,B] \rangle_\psi |^2$$


Proof: Using Cauchy-Schwarz, we obtain
$$| \langle \psi | [A,B] \psi \rangle | = | \langle A \psi| B \psi \rangle - \langle B \psi| A \psi \rangle| \leq 2 \| A \psi \| \| B \psi \|.$$

Square and substitute $A \to A - \langle A \rangle_\psi$, $B \to B- \langle B \rangle_\psi$ to finish the proof. Q.E.D.

You can observe directly that the product of the standard deviations has lower bound $&gt;0$ if and only if $A$ and $B$ are non-commuting operators. This is statement (1).

From linear algebra, you might remember the following 


  Theorem: (without proof) A set of diagonalisable matrices commutes if and only if the set is simultaneously diagonalisable.


You can find the proof for that theorem in most linear algebra textbooks, in case you are interested.

Putting these two theorems together, you can see the connection between statement (1) and statement (2), assuming you have two non-commuting operators $A$ and $B$. I am however a bit confused because you claim the lower bound to be $1$, which is not the case in general. 

For example, taking $\mathcal{H} = L^2(\mathbb{R}^3)$ with the two operators 

$$x_k: \psi(x) \mapsto x_k \psi(x), \; p_k: \psi(x) \mapsto \frac{\hbar}{i} \frac{\partial \psi}{\partial x_k}$$

which are the components of position and momentum, you can compute

$$i [p_k,x_\ell] = \hbar \delta_{k \ell}$$

and thus

$$\langle (\Delta p_k)^2 \rangle \langle ( \Delta x_\ell )^2 \rangle \geq \frac{\hbar^2}{4} \delta_{k \ell}.$$
In quantum mechanics, the mean of an observable $A$ in the state $\psi$ is given by

$$\langle A \rangle_\psi = \langle \psi |A|\psi \rangle,$$

and from this the variance of $A$ in $\psi$ is usually defined by

$$\langle(\Delta A)^2 \rangle_\psi := \langle (A-\langle A \rangle_\psi)^2 \rangle_\psi = \| (A - \langle A \rangle_\psi) \psi \|^2 = \langle A^2 \rangle_\psi - \langle A \rangle_\psi^2.$$

The standard deviation is then given by taking the square root.


  Theorem: (Heisenberg's Uncertainty Principle)
  
  $$\langle (\Delta A)^2 \rangle_\psi \langle (\Delta B)^2 \rangle_\psi \geq \frac{1}{4} | \langle [A,B] \rangle_\psi |^2$$


Proof: Using Cauchy-Schwarz, we obtain
$$| \langle \psi | [A,B] \psi \rangle | = | \langle A \psi| B \psi \rangle - \langle B \psi| A \psi \rangle| \leq 2 \| A \psi \| \| B \psi \|.$$

Square and substitute $A \to A - \langle A \rangle_\psi$, $B \to B- \langle B \rangle_\psi$ to finish the proof. Q.E.D.

You can observe directly that the product of the standard deviations has lower bound $&gt;0$ if and only if $A$ and $B$ are non-commuting operators. This is statement (1).

From linear algebra, you might remember the following 


  Theorem: (without proof) A set of diagonalisable matrices commutes if and only if the set is simultaneously diagonalisable.


You can find the proof for that theorem in most linear algebra textbooks, in case you are interested.

Putting these two theorems together, you can see the connection between statement (1) and statement (2), assuming you have two non-commuting operators $A$ and $B$. I am however a bit confused because you claim the lower bound to be $1$, which is not the case in general. 

For example, taking $\mathcal{H} = L^2(\mathbb{R}^3)$ with the two operators 

$$x_k: \psi(x) \mapsto x_k \psi(x), \; p_k: \psi(x) \mapsto \frac{\hbar}{i} \frac{\partial \psi}{\partial x_k}$$

which are the components of position and momentum, you can compute

$$i [p_k,x_\ell] = \hbar \delta_{k \ell}$$

and thus

$$\langle (\Delta p_k)^2 \rangle \langle ( \Delta x_\ell )^2 \rangle \geq \frac{\hbar^2}{4} \delta_{k \ell}.$$