Why is the degrees of freedom for a matched pairs $t$-test the number of pairs minus 1? I am used to knowing "degrees of freedom" as $n - r$, where you have the linear model $$\mathbf{y} = \mathbf{X}\boldsymbol{\beta} + \boldsymbol{\epsilon}$$
with $\mathbf{y} \in \mathbb{R}^n$, $\mathbf{X} \in M_{n \times p}(\mathbb{R})$ the design matrix with rank $r$, $\boldsymbol{\beta} \in \mathbb{R}^p$, $\boldsymbol{\epsilon} \in \mathbb{R}^n$ with $\boldsymbol{\epsilon}  \sim \mathcal{N}(\mathbf{0}, \sigma^2 \mathbf{I}_n)$, $\sigma^2 > 0$.
From what I recall of elementary statistics (i.e., pre-linear models with linear algebra), the degrees of freedom for the matched-pairs $t$-test is the number of differences minus $1$. So this would entail $\mathbf{X}$ having rank 1, perhaps. Is this correct? If not, why is $n-1$ the degrees of freedom for the matched-pairs $t$-test?
To understand the context, suppose I have a mixed-effects model
$$y_{ijk} = \mu_i + \text{ some random effects} + e_{ijk}$$
where $i = 1, 2$, $j = 1, \dots, 8$, and $k = 1, 2$. There is nothing special about $\mu_i$ other than that it's a fixed effect, and $e_{ijk} \overset{iid}{\sim}\mathcal{N}(0, \sigma^2_e)$. I'm assuming that the random effects are irrelevant to this problem, since we only care about the fixed effects in this case.
I would like to provide a confidence interval for $\mu_1 - \mu_2$.
I have already shown that $\bar{d}_\cdot = \dfrac{1}{8}\sum d_j$ is an unbiased estimator of $\mu_1 - \mu_2$, where $d_j = \bar{y}_{1j\cdot} - \bar{y}_{2j\cdot}$, $\bar{y}_{1j\cdot} = \dfrac{1}{2}\sum_{k}y_{1jk}$, and $\bar{y}_{21\cdot}$ is defined similarly. The point estimate $\bar{d}_{\cdot}$ has been computed.
I have already shown that $$s^2_d = \dfrac{\sum_{j}(d_j - \bar{d}_{\cdot})^2}{8-1}$$
is an unbiased estimator of the variance of $d_j$, and thus, 
$$\sqrt{\dfrac{s^2_d}{8}}$$
is the standard error of $\bar{d}_{\cdot}$. This has been computed.
Now the last part is figuring out the degrees of freedom. For this step, I usually try to find the design matrix - which obviously has rank 2 - but I have the solution to this problem, and it says that the degrees of freedom is $8-1$.
In the context of finding the rank of a design matrix, why are the degrees of freedom $8-1$?
Edited to add: Perhaps helpful in this discussion is how the test statistic is defined. Suppose I have a parameter vector $\boldsymbol{\beta}$. In this case, $$\boldsymbol{\beta} = \begin{bmatrix}
\mu_1 \\
\mu_2
\end{bmatrix}$$
(unless I'm missing something entirely). We are essentially performing the hypothesis test $$\mathbf{c}^{\prime}\boldsymbol{\beta} = 0$$
where $\mathbf{c}^{\prime} = \begin{bmatrix}
1 & -1 
\end{bmatrix}$. Then, the test statistic is given by
$$t = \dfrac{c^{\prime}\hat{\boldsymbol{\beta}}}{\sqrt{\hat{\sigma}^2c^{\prime}(\mathbf{X}^{\prime}\mathbf{X})^{-1}\mathbf{c}}}$$
which would be tested against a central $t$-distribution with $n - r$ degrees of freedom, where $\mathbf{X}$ is the design matrix as above, and 
$$\hat{\sigma}^2 = \dfrac{\mathbf{y}^{\prime}(\mathbf{I}-\mathbf{P}_{\mathbf{X}})\mathbf{y}}{n-r}$$
where $\mathbf{P}_{\mathbf{X}} = \mathbf{X}(\mathbf{X}^{\prime}\mathbf{X})^{-1}\mathbf{X}^{\prime}$.
 A: The matched-pairs $t$-test with $n$ pairs is actually just a one-sample $t$-test with a sample of size $n$.  You have $n$ differences $d_1,\ldots,d_n$, and these are i.i.d. and normally distributed.
$$
\begin{array}{ccccc}
\begin{bmatrix} d_1 \\  \vdots \\ d_n \end{bmatrix} & = &  \begin{bmatrix} \bar d \\  \vdots \\  \bar d \end{bmatrix} & + & \begin{bmatrix} d_1 - \bar d \\  \vdots \\  d_1 - \bar d \end{bmatrix} \\[10pt]
n \text{ d.f.} & & 1 \text{ d.f.} & & (n-1) \text{ d.f.}
\end{array}
$$
The first column after $\text{“}{=}\text{''}$ has $1$ degree of freedom because of the linear constraint that says all entries are equal; the second has $n-1$ degrees of freedom because of the linear constraint that says the sum of the entries is $0$.
A: Many, many thanks to Michael Hardy for answering my question.
The idea is this: let 
$$\mathbf{y} = \begin{bmatrix}
d_1 \\
\vdots \\
d_n
\end{bmatrix}$$
and $\boldsymbol{\beta} = [\mu_1 - \mu_2]$. Then our linear model is then
$$\mathbf{y} = \mathbf{1}_{n \times 1}\boldsymbol{\beta} + \boldsymbol{\epsilon}$$
where $\mathbf{1}_{n \times 1}$ is the $n$-vector of all ones, and $$\boldsymbol{\epsilon} = \begin{bmatrix}
\epsilon_1 \\
\vdots \\
\epsilon_n
\end{bmatrix} \sim \mathcal{N}(\mathbf{0}, \sigma^2\mathbf{I}_n)\text{.}$$
Obviously $\mathbf{X} = \mathbf{1}_{n \times 1}$ has rank $1$, so then we have $n-1$ degrees of freedom.
How do we know to set $\boldsymbol{\beta}$ equal to $[\mu_1 - \mu_2]$? Recall that 
$$\mathbb{E}[\mathbf{y}] = \mathbf{X}\boldsymbol{\beta}$$
and as it can be easily seen, $\mathbb{E}[d_j] = \mu_1 - \mu_2$ for all $j$. Given our $\mathbf{X}$, it is obvious what $\boldsymbol{\beta}$ should be. This is because
$$\mathbb{E}[\mathbf{y}] = \mathbb{E}\left[\begin{bmatrix}
d_1 \\
\vdots \\
d_n
\end{bmatrix} \right] = \begin{bmatrix}
\mathbb{E}[d_1] \\
\vdots \\
\mathbb{E}[d_n]
\end{bmatrix} = \begin{bmatrix}
\mu_1 - \mu_2 \\
\vdots \\
\mu_1 - \mu_2
\end{bmatrix} = \mathbf{X}\boldsymbol\beta = \mathbf{1}_{n \times 1}\boldsymbol\beta = \begin{bmatrix}
1 \\
\vdots \\
1
\end{bmatrix}\boldsymbol\beta$$
so $\boldsymbol\beta$ should be a $1 \times 1$ matrix with $\boldsymbol\beta = [\mu_1 - \mu_2]$.
Set $\mathbf{c}^{\prime} = [1]$. Then our hypothesis test is $$H_0: \mathbf{c}^{\prime}\boldsymbol{\beta} = 0\text{.}$$
Our test statistic is thus
$$\dfrac{\mathbf{c}^{\prime}\hat{\boldsymbol{\beta}}}{\sqrt{\hat{\sigma}^2\mathbf{c}^{\prime}\left(\mathbf{X}^{\prime}\mathbf{X}\right)^{-1}\mathbf{c}}}\text{.}$$
We have
$$\hat{\sigma}^2 = \dfrac{\mathbf{y}^{\prime}(\mathbf{I}-\mathbf{P}_{\mathbf{X}})\mathbf{y}}{n-r(\mathbf{X})}\text{.}$$
After some work, it can be shown that
$$\mathbf{P}_\mathbf{X} = \mathbf{P}_{\mathbf{1}_{n \times 1}} = \mathbf{1}_{n \times 1}\left(\dfrac{1}{n}\right)\mathbf{1}^{\prime}\text{.}$$
It can also be shown that $\mathbf{I}-\mathbf{P}_{\mathbf{X}}$ is symmetric and idempotent. So,
$$\begin{align}
\hat{\sigma}^2 &= \dfrac{\mathbf{y}^{\prime}(\mathbf{I}-\mathbf{P}_{\mathbf{X}})\mathbf{y}}{n-r(\mathbf{X})} \\
&= \dfrac{\mathbf{y}^{\prime}(\mathbf{I}-\mathbf{P}_{\mathbf{X}})^{\prime}(\mathbf{I}-\mathbf{P}_{\mathbf{X}})\mathbf{y}}{n-r(\mathbf{X})} \\
&= \dfrac{\|(\mathbf{I}-\mathbf{P}_{\mathbf{X}})\mathbf{y}\|^{2}}{n-r(\mathbf{X})} \\
&=\dfrac{\left\|\left[\mathbf{I}-\mathbf{1}_{n \times 1}\left(\dfrac{1}{n}\right)\mathbf{1}^{\prime}\right]\mathbf{y} \right\|^2}{n-1} \\
&= \dfrac{\left\|\begin{bmatrix}
d_1 \\
\vdots \\
d_n
\end{bmatrix} - \begin{bmatrix}
\bar{d}_\cdot \\
\vdots \\
\bar{d}_\cdot
\end{bmatrix} \right\|^2}{n-1} \\
&= \dfrac{\sum_{i=1}^{n}(d_i-\bar{d}_{\cdot})^2}{n-1} \\
&= s^2_d
\end{align}$$
and
$$\mathbf{X}^{\prime}\mathbf{X} = \mathbf{1}_{n \times 1}^{\prime}\mathbf{1}_{n \times 1} = n$$
which obviously has inverse $1/n$, thus giving a test statistic
$$\dfrac{\hat\mu_1-\hat\mu_2}{\sqrt{s^2_d/n}}$$
which would be tested on a central $t$-distribution with $n - 1$ degrees of freedom as desired.
