How can you subtract a value from an operator/matrix? I'm currently following Quantum Computation and Quantum Information by Nielsen & Chuang. I'm struggling to understand the derivation of The Heisenberg Uncertainty Principle in Box 2.4 page 89. I understand how they arrive to the following equation:
$$|\langle v|[A,B]|v\rangle|^2 \leq 4\langle v|A^2|v\rangle\langle v|B^2|v\rangle.$$
But the problem I have is with the following step. After this, they set $A = C - \langle C\rangle$  and $B = D - \langle D\rangle$ to arrive at 
$$\Delta(C)\Delta(D) \geq \frac{|\langle v|[C,D]|v\rangle|}{2}.$$
What I don't understand is how you can subtract C, a matrix, with $\langle C\rangle$, a value, and set it equal to A, a Hermitian. Earlier in the book they define 
$\langle M \rangle = \langle v|M|v\rangle$ which is an inner product and thus a value and not a matrix. How can you subtract a value from an operator/matrix?
Edit: I now see that $\langle C \rangle$ is $\langle C \rangle 1$, but am still confused about how this substitution gets us from the first inequality to the 2nd inequality.
 A: I'll write down the whole calculation for completion. Firstly we define 
$$A = C-\langle C\rangle \qquad B = D-\langle D \rangle$$
We firstly evaluate the following 
$$\frac{\langle v|A^2|v\rangle}{\langle v | v \rangle} = \frac{\langle v|C^2-2C\langle C\rangle+\langle C\rangle^2|v\rangle}{\langle v | v \rangle} = \frac{\langle v|C^2|v\rangle}{\langle v | v \rangle}-2\langle C\rangle\frac{\langle v|C|v\rangle}{\langle v | v \rangle}+\left(\frac{\langle v|C|v\rangle}{\langle v | v \rangle} \right)^2\\
= \frac{\langle v|C^2|v\rangle}{\langle v | v \rangle} -2\left(\frac{\langle v|C|v\rangle}{\langle v | v \rangle} \right)^2+\left(\frac{\langle v|C|v\rangle}{\langle v | v \rangle} \right)^2\\
 = \frac{\langle v|C^2|v\rangle}{\langle v | v \rangle} -\left(\frac{\langle v|C|v\rangle}{\langle v | v \rangle} \right)^2 = \Delta A^2
$$
which is precisely the definition of standard deviation. The same also works for $B$ clearly.
We now define two states
$$|\alpha\rangle = A|v\rangle \qquad |\beta\rangle = B|v\rangle $$
notice that
$$\langle\alpha|\alpha\rangle = \langle v|A^2|v\rangle = \Delta A^2\langle v|v\rangle$$
and the same for $|\beta\rangle$. What we want to find is an inequality on the product of the standard deviation
$$\Delta A^2\Delta B^2 = \frac{\langle v|A^2|v\rangle}{\langle v|v\rangle}\frac{\langle v|B^2|v\rangle}{\langle v|v\rangle}\geq\frac{|\langle v |B^\dagger A|v \rangle|^2}{|\langle v|v\rangle|^2} = \frac{|\langle v |B A|v \rangle|^2}{|\langle v|v\rangle|^2}$$
where we used Schwartz's inequality and the hermiticiy of the operators (remember that we're talking about measurable quantities here).
We invoke now the following equality
$$BA = \frac{\{A,B\}}{2}+\frac{[A,B]}{2}$$
Plugging this back in we get
$$\Delta A^2\Delta B^2 \geq \frac{\left|\left\langle v \left|\frac{\{A,B\}}{2}+\frac{[A,B]}{2}\right|v \right\rangle\right|^2}{|\langle v|v\rangle|^2}\\
=\frac{1}{4}\frac{|\langle v |[A,B]|v \rangle|^2}{|\langle v|v\rangle|^2}+\frac{1}{4}\frac{|\langle v |\{A,B\}|v \rangle|^2}{|\langle v|v\rangle|^2}
$$
Since now both terms are positive and real, clearly omitting one of them does not affect the inequality, so we discard the anticommutator one and get the result 
$$\Delta A^2 \Delta B^2\geq\frac{1}{4}\frac{|\langle v |[A,B]|v \rangle|^2}{|\langle v|v\rangle|^2}$$
or, by taking the square root of both members
$$\Delta A\Delta B\geq\frac{1}{2}\frac{|\langle v |[A,B]|v \rangle|}{|\langle v|v\rangle|}$$
In your result is implied that $\langle v|v\rangle = 1$, so that the states are normalized.
