## 3.3 Representing Summation Using Matrix Notation

Consider the sum: $\sum_{k=1}^{n}x_{k}=x_{1}+\cdots+x_{n}.$ Let $$\mathbf{x}=(x_{1},\ldots,x_{n})^{\prime}$$ be an $$n\times1$$ vector and $$\mathbf{1}=(1,\ldots,1)^{\prime}$$ be an $$n\times1$$ vector of ones. Then, $\mathbf{x}^{\prime}\mathbf{1}=\left[\begin{array}{ccc} x_{1} & \ldots & x_{n}\end{array}\right]\cdot\left[\begin{array}{c} 1\\ \vdots\\ 1 \end{array}\right]=x_{1}+\cdots+x_{n}=\sum_{k=1}^{n}x_{k},$ and, $\mathbf{1}^{\prime}\mathbf{x}=\left[\begin{array}{ccc} 1 & \ldots & 1\end{array}\right]\cdot\left[\begin{array}{c} x_{1}\\ \vdots\\ x_{n} \end{array}\right]=x_{1}+\cdots+x_{n}=\sum_{k=1}^{n}x_{k}.$ Next, consider the sum of squared $$x$$ values, $\sum_{k=1}^{n}x_{k}^{2}=x_{1}^{2}+\cdots+x_{n}^{2}.$ This sum can be conveniently represented as, $\mathbf{x}^{\prime}\mathbf{x}=\left[\begin{array}{ccc} x_{1} & \ldots & x_{n}\end{array}\right]\cdot\left[\begin{array}{c} x_{1}\\ \vdots\\ x_{n} \end{array}\right]=x_{1}^{2}+\cdots+x_{n}^{2}=\sum_{k=1}^{n}x_{k}^{2}.$ Last, consider the sum of cross products, $\sum_{k=1}^{n}x_{k}y_{k}=x_{1}y_{1}+\cdots x_{n}y_{n}.$ This sum can be compactly represented by, $\mathbf{x}^{\prime}\mathbf{y}=\left[\begin{array}{ccc} x_{1} & \ldots & x_{n}:\end{array}\right]\cdot\left[\begin{array}{c} y_{1}\\ \vdots\\ y_{n} \end{array}\right]=x_{1}y_{1}+\cdots x_{n}y_{n}=\sum_{k=1}^{n}x_{k}y_{k}.$ Note that $$\mathbf{x}^{\prime}\mathbf{y=y}^{\prime}\mathbf{x}$$.

Example 2.17 (Computing sums in R)

In R, summing the elements in a vector can be done using matrix algebra.

# create vector of 1's and a vector of x
onevec = rep(1,3)
onevec
## [1] 1 1 1
xvec = c(1,2,3)
# sum elements in x
t(xvec)%*%onevec
##      [,1]
## [1,]    6

The functions crossprod() and sum() are generally computationally more efficient:

crossprod(xvec,onevec)
##      [,1]
## [1,]    6
sum(xvec)
## [1] 6

Sums of squares are best computed using:

crossprod(xvec)
##      [,1]
## [1,]   14
sum(xvec^2)
## [1] 14

The dot-product two vectors can be conveniently computed using the crossprod() function:

yvec = 4:6
xvec
## [1] 1 2 3
yvec
## [1] 4 5 6
crossprod(xvec,yvec)
##      [,1]
## [1,]   32
crossprod(yvec,xvec)
##      [,1]
## [1,]   32

$$\blacksquare$$