Expressing a summation using matrix algebra Consider the $r \times n$ matrix 
$$\begin{pmatrix}
X_{11} & X_{12} & \cdots & X_{1n} \\
X_{21} & X_{22} & \cdots & X_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
X_{r1} & X_{r2} & \cdots & X_{rn}
\end{pmatrix}\text{.}$$
Define 
$$\begin{align*}
&\bar{X} = \dfrac{\sum\limits_{i=1}^{r}\sum\limits_{j=1}^{n}X_{ij}}{nr} \\
&\bar{X}_{i} = \dfrac{\sum\limits_{j=1}^{n}X_{ij}}{n}\text{.}
\end{align*}$$
I am interested in knowing if there is a possible way to write the summations
$$\begin{align*}
\hat{v}^{S} &= \sum\limits_{i=1}^{r}\sum\limits_{j=1}^{n}\left(X_{ij}-\bar{X}_{i}\right)^{2} \\
\hat{a}^{S} &= \sum\limits_{i=1}^{r}\left(\bar{X}_i - \bar{X}\right)^{2}
\end{align*}$$
in terms of matrix operations (anything one would learn in a first course in linear algebra, such as multiplication of matrices, inverses of matrices, determinants, eigenvalues, etc.).
The reason why is because I have to memorize these formulas for the actuarial exam I will be taking soon, and I am not interested in memorizing summations if there is a way to express them in matrix form. 
There may not be an answer to what I seek, and I might just have to memorize these summations as is, but I thought I would ask in case there is.
ETA: I did pass this exam (at least 93% scored) but am still interested in knowing if there is a solution to this problem.
 A: There is a way to express them in matrix form, but it makes less sense than intuition from statistics.
Let
$$
A_m := 
\begin{pmatrix}1/m \\ 1/m \\ \vdots \\ 1/m\end{pmatrix}
\ \text{(an}\ 1×m\ \text{column vector)}
$$
for any $m∈{\mathbb N}$.
Then:
$$\bar{X}_i = X\,A_n;\quad
\text{Likewise, }
\bar{X} = \bar{X}_i^{\mathsf T}\,A_r = A_r^{\mathsf T}\,X\,A_n\,.
$$
Obviously, the matrix $X_{ij} - \bar{X}_i$ (let denote it by $W$) can be expressed as
$$W =
X - n\cdot\bar{X}_i\,A_n^{\mathsf T} =
X - n\cdot X\,A_n\,A_n^{\mathsf T} =
X\,(I_n - n\,A_n\,A_n^{\mathsf T})\,,
$$
where $I_n$ is n × n identity matrix.
Sum of squares is trickier. For an n × 1 row vector $\mathbf w$, we know that
$$\sum\limits_{j=1}^{n} (w_j)^2 = {\mathbf w}\,{\mathbf w}^{\mathsf T}\,,$$
so if $r=1$, then
$$\hat v^S =
\sum\limits_{j=1}^{n} (X_{1j} - \bar{X}_1)^2 =
W\,W^{\mathsf T} =
X\,(I_n - n\,A_n\,A_n^{\mathsf T})^2\,X^{\mathsf T}.
$$
What is $W\,W^{\mathsf T}$ for general $r$? It is easy to get that it’s an r × r matrix with diagonal elements containing squares of respective rows of $W$. Their sum (known as matrix trace, “tr”) gives $\hat v^S$:
$$\hat v^S =
\operatorname{tr}(W\,W^{\mathsf T}) =
\operatorname{tr}(X\,(I_n - n\,A_n\,A_n^{\mathsf T})^2\,X^{\mathsf T}).
$$
And, finally,
$$\hat a^S =
\bar{X}_i^{\mathsf T}\,(I_r - r\,A_r\,A_r^{\mathsf T})^2\,\bar{X}_i =\quad\\=
A_n^{\mathsf T}\,X^{\mathsf T}\,(I_r - r\,A_r\,A_r^{\mathsf T})^2\,X\,A_n\,.
$$

Update: it can be found that $I_m - m\,A_m\,A_m^{\mathsf T}$ matrix (or “$\mathbf{I}_{m} - \frac{1}{m}\mathbf{1}_{m}\mathbf{1}_{m}^{T}$” in m.a.’s notation) is idempotent (actually, it is orthogonal projection on a hyperplane). Hence, there is no difference between $(I_n - n\,A_n\,A_n^{\mathsf T})^2$ and $\mathbf{I}_{n} - \frac{1}{n}\mathbf{1}_{n}\mathbf{1}_{n}^{\mathsf T}$.
A: Here are some expressions, although not necessarily easier to memorize!
Let $\mathbf{1}_{n}$ denote the column vector of all ones and length $n$,
and $\mathbf{I}$ the $n\times n$ identity matrix. Then,
$$
\hat{v}^{S} = \text{Tr}\left( \mathbf{X}^{T}\mathbf{X} \left(\mathbf{I}_{n}- \frac{1}{n}\mathbf{1}_{n}\mathbf{1}_{n}^{T}\right)\right),
$$
where $\text{Tr}(\mathbf{M})$ denotes the trace of $\mathbf{M}$, and
\begin{align*}
\hat{a}^{S} 
&=
\frac{1}{n^{2}}
\left(
\mathbf{1}_{n}^{T}\mathbf{X}^{T} \mathbf{X}\mathbf{1}_{n} - \frac{1}{r}
(\mathbf{1}_{r}^{T}\mathbf{X}\mathbf{1}_{n})^{2}
\right).
\end{align*}
Moving things around, you can also alternatively write:
\begin{align*}
\hat{a}^{S} 
&=
\frac{1}{n^{2}}
\mathbf{1}_{n}^{T}\mathbf{X}^{T} \left( \mathbf{I} - \frac{1}{r}
\mathbf{1}_{r}\mathbf{1}_{r}^{T}
\right)
\mathbf{X}\mathbf{1}_{n}.
\end{align*}
It is useful to note that $\mathbf{1}_{n}\mathbf{1}_{n}^{T}$ is the $n \times n$ all-ones matrix, and $\mathbf{1}_{r}^{T}\mathbf{X}\mathbf{1}_{n}$ is the sum of all entries in $\mathbf{X}$.
Why are the above true?
Note that ${\overline{\mathbf{X}}}_{i}$ is the sum of all entries of the $i$th row  of $\mathbf{X}$ normalized by $n$, which can be written as $\frac{1}{n}\mathbf{X}_{i,:}\mathbf{1}_{n}$, where $\mathbf{X}_{i,:}$ is the $i$th row of $\mathbf{X}$.
Let $\mathbf{R}$ denote the column vector of length $r$ obtained by vertically stacking the ${\overline{\mathbf{X}}}_{i}$'s. Then,
$$
\mathbf{R} = \frac{1}{n}\mathbf{X}\mathbf{1}_{n}.
$$
Also, note that $\mathbf{R}\mathbf{1}_{n}^{T}$ is an $r \times n$ matrix whose entire $i$th row is equal to $R_{i} = {\overline{\mathbf{X}}}_{i}$.
Then, $\hat{v}^{S}$ is the sum of the squared entries of the matrix
$$
 \mathbf{X} - \mathbf{R}\mathbf{1}_{n}^{T},
$$
which is also called the squared Frobenius norm of the matrix, denoted by $\|\|_{F}^{2}$,
that is,
\begin{align*}
\hat{v}^{S} 
&= \| \mathbf{X} - \mathbf{R}\mathbf{1}_{n}^{T} \|_{F}^{2} \\
%= \sum\limits_{i=1}^{r}\sum\limits_{j=1}^{n}\left(X_{ij}-\bar{X}_{i}\right)^{2}
&= \| \mathbf{X} - \frac{1}{n}\mathbf{X}\mathbf{1}_{n}\mathbf{1}_{n}^{T} \|_{F}^{2} \\
&= \| \mathbf{X} \left(\mathbf{I}_{n}- \frac{1}{n}\mathbf{1}_{n}\mathbf{1}_{n}^{T}\right) \|_{F}^{2}\\
&= \text{Tr}\left( \mathbf{X} \left(\mathbf{I}_{n}- \frac{1}{n}\mathbf{1}_{n}\mathbf{1}_{n}^{T}\right)\left(\mathbf{I}_{n}- \frac{1}{n}\mathbf{1}_{n}\mathbf{1}_{n}^{T}\right)^{T}\mathbf{X}^{T}\right)\\
&= \text{Tr}\left( \mathbf{X}^{T}\mathbf{X} \left(\mathbf{I}_{n}- \frac{2}{n}\mathbf{1}_{n}\mathbf{1}_{n}^{T} +\frac{1}{n^{2}}\mathbf{1}_{n}\mathbf{1}_{n}^{T}\mathbf{1}_{n}\mathbf{1}_{n}^{T}\right)\right)\\
&= \text{Tr}\left( \mathbf{X}^{T}\mathbf{X} \left(\mathbf{I}_{n}- \frac{2}{n}\mathbf{1}_{n}\mathbf{1}_{n}^{T} +\frac{1}{n}\mathbf{1}_{n}\mathbf{1}_{n}^{T}\right)\right)\\
&= \text{Tr}\left( \mathbf{X}^{T}\mathbf{X} \left(\mathbf{I}_{n}- \frac{1}{n}\mathbf{1}_{n}\mathbf{1}_{n}^{T}\right)\right).
\end{align*}
For the second part, $\overline{\mathbf{X}}$ is a scalar, equal to the sum of all entries in $\mathbf{X}$ normalized by $rn$, i.e.,
$$
\overline{\mathbf{X}}
= 
\frac{1}{rn}
\mathbf{1}_{r}^{T}
\mathbf{X}
\mathbf{1}_{n}.
$$
Then,
\begin{align*}
\hat{a}^{S} &= \sum\limits_{i=1}^{r}\left(\bar{X}_i - \bar{X}\right)^{2}\\
&=
\|\mathbf{R} - \overline{\mathbf{X}} \cdot \mathbf{1}_{r} \|_{2}^{2}\\
&=
\|\frac{1}{n}\mathbf{X}\mathbf{1}_{n} - \frac{1}{rn}
\mathbf{1}_{r}^{T}
\mathbf{X}
\mathbf{1}_{n} \cdot \mathbf{1}_{r} \|_{2}^{2} \\
&=
\frac{1}{n^{2}}
\|\mathbf{X}\mathbf{1}_{n} - \frac{1}{r}
(\mathbf{1}_{r}^{T}\mathbf{X}\mathbf{1}_{n}) 
\cdot \mathbf{1}_{r} \|_{2}^{2} \\
&=
\frac{1}{n^{2}}
\left(
\mathbf{1}_{n}^{T}\mathbf{X}^{T} \mathbf{X}\mathbf{1}_{n} - \frac{2}{r}
(\mathbf{1}_{r}^{T}\mathbf{X}\mathbf{1}_{n}) 
\cdot \mathbf{1}_{n}^{T}\mathbf{X}^{T}\mathbf{1}_{r}
+
\frac{1}{r^{2}}(\mathbf{1}_{r}^{T}\mathbf{X}\mathbf{1}_{n})^{2}
\mathbf{1}_{r}^{T}\mathbf{1}_{r}
\right)\\
&=
\frac{1}{n^{2}}
\left(
\mathbf{1}_{n}^{T}\mathbf{X}^{T} \mathbf{X}\mathbf{1}_{n} - \frac{2}{r}
(\mathbf{1}_{r}^{T}\mathbf{X}\mathbf{1}_{n})^{2}
+
\frac{1}{r}(\mathbf{1}_{r}^{T}\mathbf{X}\mathbf{1}_{n})^{2}
\right)\\
&=
\frac{1}{n^{2}}
\left(
\mathbf{1}_{n}^{T}\mathbf{X}^{T} \mathbf{X}\mathbf{1}_{n} - \frac{1}{r}
(\mathbf{1}_{r}^{T}\mathbf{X}\mathbf{1}_{n})^{2}
\right),
\end{align*}
which completes the proof.
