Deviance for a Normal linear model This is an excerpt from the second edition of An Introduction to Generalized Linear Models by Annette J. Dobson:


The author seems to assume that $(I - H)/\sigma^2 = V^-$, where $V$ is the variance-covariance matrix for $y$, and $V^-$ the generalized inverse, it's not clear how he arrived there.
The author referred to section 1.4.2 part 8, it's also listed here:

 A: The OP is right in concluding that the author, for internal consistency, has to treat $\frac 1{\sigma^2}\mathbf M=\frac 1{\sigma^2}(\mathbf I - \mathbf H)$ as a generalized inverse of the variance covariance matrix of $\mathbf y$, to obtain the result the author arrives at (of zero non-centrality parameter).  
One page previously in the book (we are at section 5.6.2), the distribution of $\mathbf y$ is assumed as 
$$\mathbf y \sim N(\mathbf X\beta, \sigma^2\mathbf I)$$
so the question is: does $\frac 1{\sigma^2}\mathbf M$ satisfy the required properties so that it is a generalized inverse of $\sigma^2\mathbf I$?
These properties are four (see for example here), and it is easily seen that $\frac 1{\sigma^2}\mathbf M$, due to it being symmetric and idempotent, satisfies the 2nd, 3d and 4th -but it does not satisfy the 1st. 
We have:
1st : $\left(\sigma^2\mathbf I\right)\left(\frac 1{\sigma^2}\mathbf M\right)\left(\sigma^2\mathbf I\right) = \sigma^2 \mathbf M \neq \sigma^2 \mathbf I \qquad \text{not satisfied}$  
2nd : $\left(\frac 1{\sigma^2}\mathbf M\right)\left(\sigma^2\mathbf I\right)\left(\frac 1{\sigma^2}\mathbf M\right) = \frac 1{\sigma^2} \mathbf M \mathbf M = \frac 1{\sigma^2}  \mathbf M \qquad \text{satisfied}$  
3d : $\left(\sigma^2\mathbf I\frac 1{\sigma^2}\mathbf M\right)^T   = \left(\frac 1{\sigma^2}\mathbf M^T\right)\left(\sigma^2\mathbf I^T\right) = \left(\sigma^2\mathbf I\right)\left(\frac 1{\sigma^2}\mathbf M\right)  \qquad \text{satisfied}$
4th : $\left(\frac 1{\sigma^2}\mathbf M\sigma^2\mathbf I\right)^T   = \left(\sigma^2\mathbf I^T\right)\left(\frac 1{\sigma^2}\mathbf M^T\right) = \left(\frac 1{\sigma^2}\mathbf M\right)\left(\sigma^2\mathbf I\right)  \qquad \text{satisfied}$
The problem is that the 1st condition is the fundamental property in order for a matrix to be a generalized inverse (in fact, if just the 1st is satisfied we have a generalized inverse -the rest of the conditions are needed to successively obtain generalized inverses endowed with more properties, leading to the Moore-Penrose unique pseudo-inverse that satisfies all four conditions).  
So it appears that $\frac 1{\sigma^2}\mathbf M$ cannot be treated as a generalized inverse of $\frac 1{\sigma^2}\mathbf I$, and so the "zero non-centrality parameter" result cannot follow from section 1.4.2 part 8 of the book as the author claims. Now, A. Dobson is not usually wrong, so maybe I am missing something, but I am posting this answer so that perhaps somebody else can settle this better...
(ADDENDUM)
...and @whuber just did!  We have (regressors are deterministic here)
$$\mathbf y \sim N(\mathbf X\beta, \sigma^2\mathbf I) \Rightarrow \mathbf z=\mathbf M\mathbf y\sim N(\mathbf 0, \sigma^2\mathbf M)$$
Note the variance-covariance matrix of $\mathbf z$ is $V(\mathbf z) = \mathbf MV(\mathbf y)\mathbf M^{T} = \sigma^2 \mathbf M $
where we have used the properties of $\mathbf M$, $\mathbf M\mathbf X=\mathbf 0$ and $\mathbf M\mathbf M^T = \mathbf M\mathbf M=\mathbf M$.
Then, we have that $\mathbf z^T(\sigma^2\mathbf M)^-\mathbf z$ follows a central chi-square. But 
$$\mathbf z^T(\sigma^2\mathbf M)^-\mathbf z = \mathbf y^T\mathbf M^T(\sigma^2\mathbf M)^-\mathbf M\mathbf y$$
and $\mathbf M$ can be used as a generalized inverse of its own self because $\mathbf M\mathbf M \mathbf M =\mathbf M$. So 
$$\mathbf z^T(\sigma^2\mathbf M)^-\mathbf z = \mathbf y^T\mathbf M^T\frac 1 {\sigma^2} \mathbf M\mathbf M\mathbf y = \mathbf y^T\left(\frac 1 {\sigma^2}\mathbf M\right)\mathbf y $$
Therefore $D =\mathbf y^T\left(\frac 1 {\sigma^2}\mathbf M\right)\mathbf y$ follows a central chi-squared.
