Constanza F. Schibber September 14, 2017
There are several possible ways of creating a vector or matrix object.
- A vector of specific values: e.g.,
$\mathbf{a}=[3,4,5]$
a <- c(3,4,5)
a## [1] 3 4 5
# You can check whether an object is a vector
is.vector(a)## [1] TRUE
# You can also get the length of a vector
length(a)## [1] 3
- A vector of a repeated number: E.g.,
$\mathbf{b}=[1,1,1,1,1]$
b <- rep(1, 5)
b## [1] 1 1 1 1 1
Class Question. Is an object
- A vector of consecutive numbers: E.g.,
$\mathbf{c} = [1,2,3,4,5]$
c<- c(1:5)
c## [1] 1 2 3 4 5
- A vector with an incremental sequence: E.g.,
$\mathbf{d} = [-1, -0.5, 0, 0.5, 1]$
d <- seq(-1, 1, by=0.5)
d## [1] -1.0 -0.5 0.0 0.5 1.0
- A matrix filled with the same constant: E.g., $\mathbf{A}= \begin{bmatrix}2 & 2 & 2 \2 & 2 & 2 \ 2 & 2 & 2 \2 & 2 & 2\2 & 2 & 2 \end{bmatrix}$
A <- matrix(2, ncol=3, nrow=5, byrow=FALSE)
A## [,1] [,2] [,3]
## [1,] 2 2 2
## [2,] 2 2 2
## [3,] 2 2 2
## [4,] 2 2 2
## [5,] 2 2 2
# You can check whether an object is a matrix
is.matrix(A)## [1] TRUE
# You can also get the dimension of a matrix
dim(A)## [1] 5 3
- A matrix by treating several vectors as columns:
E.g., Create a matrix$\mathbf{B}$ , whose columns are$\mathbf{c}$ and$\mathbf{d}$
B <- cbind(c,d)
B## c d
## [1,] 1 -1.0
## [2,] 2 -0.5
## [3,] 3 0.0
## [4,] 4 0.5
## [5,] 5 1.0
Class Question. Create a matrix
That is,
$\mathbf{C}= \begin{bmatrix}1 & 1 & -1.0 \1 & 2 & -0.5 \ 1 & 3 & 0.0 \1 & 4 & 0.5 \1 & 5 & 1.0 \end{bmatrix}$.
Test if an object
- A matrix by treating several vectors as rows: E.g., Create a matrix
$\mathbf{R}$ , whose rows are vectors$\mathbf{c}$ and$\mathbf{d}$
R<-rbind(c, d)
R## [,1] [,2] [,3] [,4] [,5]
## c 1 2.0 3 4.0 5
## d -1 -0.5 0 0.5 1
# Note R is the transpose of a matrix B.
t(B) # B Transpose ## [,1] [,2] [,3] [,4] [,5]
## c 1 2.0 3 4.0 5
## d -1 -0.5 0 0.5 1
- A matrix by listing all element, filling either column-by-column or
row-by-row.
E.g., $\mathbf{W} =\begin{bmatrix}1 & 3 \2 & 4 \end{bmatrix}$
W <- matrix(c(1,2,3,4), ncol=2, nrow=2)
W## [,1] [,2]
## [1,] 1 3
## [2,] 2 4
E.g., $\mathbf{Z} =\begin{bmatrix}1 & 2 & 3 \4 & 5 & 6 \7 & 8 & 9 \end{bmatrix}$
Z <- matrix(c(1:9), ncol=3, nrow=3, byrow=TRUE)
Z## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 4 5 6
## [3,] 7 8 9
- You can ask to print a particular element of a matrix using subscripting:
#view an element of a vector a
a## [1] 3 4 5
a[3]## [1] 5
#view a column of a matrix
B## c d
## [1,] 1 -1.0
## [2,] 2 -0.5
## [3,] 3 0.0
## [4,] 4 0.5
## [5,] 5 1.0
B[,2] # the second column of a matrix B## [1] -1.0 -0.5 0.0 0.5 1.0
#use column name to call a matrix column
B[,"d"]## [1] -1.0 -0.5 0.0 0.5 1.0
#view a row of a matrix
B[1,] # the first row of a matrix B## c d
## 1 -1
- Another way to create a matrix is to begin with a blank matrix and then reasign element by element.
#create a blank matrix
BLANK <- matrix(NA, ncol=3, nrow=3)
BLANK## [,1] [,2] [,3]
## [1,] NA NA NA
## [2,] NA NA NA
## [3,] NA NA NA
#fill in values of the blank matrix piecewise
BLANK[1,1] <- 2
BLANK[2,1] <- pi
BLANK[,2] <- a # use a vector a to fill in the second column
#check our progress
BLANK## [,1] [,2] [,3]
## [1,] 2.000000 3 NA
## [2,] 3.141593 4 NA
## [3,] NA 5 NA
#You can create row and column names for a matrix
rownames(BLANK)<-c('ID1','ID2','ID3') # add row names
colnames(BLANK)<-c("var1", "var2", "var3") # add column names
colnames(BLANK)[1]<-"one" # change the first column name
#check our progress
BLANK## one var2 var3
## ID1 2.000000 3 NA
## ID2 3.141593 4 NA
## ID3 NA 5 NA
Class Question. Fill in the values and make a
E.g., $\mathbf{D}=\begin{bmatrix} 1 & 0 & 0 & 0\0 & 2 & 0 & 0 \0 & 0 & 3 & 0 \0 & 0 & 0 & 4 \end{bmatrix}$
D <- diag(c(1:4),4,4) #diag(x,nrow,ncol)
D## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 2 0 0
## [3,] 0 0 3 0
## [4,] 0 0 0 4
#You can ask R to print the diagonal elements of a sqaure matrix
diag(D)## [1] 1 2 3 4
#To calculate the trace of D you could do
sum(diag(D))## [1] 10
E.g., $\mathbf{I}=\begin{bmatrix} 1 & 0 & 0 & 0 & 0\0 & 1 & 0 & 0 & 0\0 & 0 & 1 & 0 & 0\0 & 0 & 0 & 1 & 0 \0 & 0 & 0 & 0 & 1 \end{bmatrix}$
I <- diag(1,5,5)
I## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 0 0 0
## [2,] 0 1 0 0 0
## [3,] 0 0 1 0 0
## [4,] 0 0 0 1 0
## [5,] 0 0 0 0 1
E.g., $\mathbf{J}=\begin{bmatrix} 1 & 1 & 1 & 1 & 1\1 & 1 & 1 & 1 & 1\1 & 1 & 1 & 1 & 1\1 & 1 & 1 & 1 & 1 \1 & 1 & 1 & 1 & 1 \end{bmatrix}$
J <- matrix(1, ncol=5, nrow=5)
J## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 1 1 1 1
## [2,] 1 1 1 1 1
## [3,] 1 1 1 1 1
## [4,] 1 1 1 1 1
## [5,] 1 1 1 1 1
E.g., Given a matrix $\mathbf{Z} =\begin{bmatrix}1 & 2 & 3 \4 & 5 & 6 \7 & 8 & 9 \end{bmatrix}$, the lower triangular matrix is $\mathbf{Z_{LT}} =\begin{bmatrix}1 & 2 & 3 \0 & 5 & 6 \0 & 0 & 9 \end{bmatrix}$ and the upper triangular matrix is $\mathbf{Z_{UT}} =\begin{bmatrix}1 & 0 & 0 \4 & 5 & 0 \7 & 8 & 9 \end{bmatrix}$.
# e.g., given a matrix Z, create the lower trinagular matrix
lower.tri(Z, diag=FALSE)## [,1] [,2] [,3]
## [1,] FALSE FALSE FALSE
## [2,] TRUE FALSE FALSE
## [3,] TRUE TRUE FALSE
Z_lt <- Z
Z_lt[lower.tri(Z)] <- 0
Z_lt## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 0 5 6
## [3,] 0 0 9
# e.g., given a matrix Z, create the upper trinagular matrix
upper.tri(Z, diag=FALSE)## [,1] [,2] [,3]
## [1,] FALSE TRUE TRUE
## [2,] FALSE FALSE TRUE
## [3,] FALSE FALSE FALSE
Z_ut <- Z
Z_ut[upper.tri(Z)] <- 0
Z_ut## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] 4 5 0
## [3,] 7 8 9
As shown above, length( ) gives the length of a vector, dim( ) gives
the dimension of a matrix. The transposition of a matrix is t( ). You
can also ask to calculate important characteristics of a matrix, the
trace and determinant.
#the trace of a matrix
sum(diag(Z)) #tr(Z)## [1] 15
#the determinant of a matrix
det(W)## [1] -2
det(Z)## [1] 6.661338e-16
The inverse of a matrix is calculated using solve( ).
P <- matrix(c(5,6,7,8), ncol=2, nrow=2)
P ## [,1] [,2]
## [1,] 5 7
## [2,] 6 8
solve(P) ## [,1] [,2]
## [1,] -4 3.5
## [2,] 3 -2.5
# Checking whether we get the identity matrix back
P%*%solve(P) # You can see that the result has some rounding issues## [,1] [,2]
## [1,] 1 3.552714e-15
## [2,] 0 1.000000e+00
round(P%*%solve(P)) # With rounding, you can now see the identity matrix## [,1] [,2]
## [1,] 1 0
## [2,] 0 1
But not all matrices are invertible. The matrix
#Not all square matrix are invertible
Z## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 4 5 6
## [3,] 7 8 9
solve(Z) ## Error in solve.default(Z): system is computationally singular: reciprocal condition number = 2.59052e-18
Class Question. Compute the inverse of a matrix
# STUDENTS: FILL IN ANSWER HERE| Operator | Description |
|---|---|
a * b |
Inner Product of Vectors |
A + B |
Matrix Addition |
A - B |
Matrix Subtraction |
A * s |
Scalar Multiplication |
A / s |
Scalar Division |
A %*% B |
Matrix multiplication. |
kronecker(A,B) |
Kronecker product. |
A %o% B |
Outer product. |
solve(A) |
Inverse of A, where A is a sqaure matrix. |
In the table,
Class Question. Answer the following questions using R.
Perform some math arithmetic using the following vectors and matrices.
$\mathbf{A} =\begin{bmatrix} -1\ 3\ 4 \end{bmatrix}$
$\mathbf{B} = \begin{bmatrix} 3\ -2\ 1 \end{bmatrix}$
$\mathbf{C} = \begin{bmatrix} 3 & 2 & -4\ -8 & 0 & 6 \end{bmatrix}$
$\mathbf{D} = \begin{bmatrix} 6 & -2\ -1 & 3\ -3 & 8 \end{bmatrix}$
$3\mathbf{A}-2\mathbf{B}$ $\mathbf{CA}$ $\mathbf{BD}$ $\mathbf{B}\otimes \mathbf{D}$ $\mathbf{CD}$ $\mathbf{C}^\prime \mathbf{C}$
A <- matrix(c(-1,3,4), ncol=1, nrow=3, byrow=TRUE)
B <- matrix(c(3,-2,1), ncol=1, nrow=3, byrow=TRUE)
C <-matrix(c(3,2,-4,
-8,0,6), ncol=3, nrow=2, byrow=TRUE)
D <-matrix(c(6,-2,
-1,3,
-3,8), ncol=2, nrow=3, byrow=TRUE)