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I have identity matrix which can be generated via diag(5)
[,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
I want to convert it to the matrix wherein series starts after 1. For example 1st column, values 1 through 5. Second column - values 1 through 4.
Desired Output
[,1] [,2] [,3] [,4] [,5]
[1,] 1 0 0 0 0
[2,] 2 1 0 0 0
[3,] 3 2 1 0 0
[4,] 4 3 2 1 0
[5,] 5 4 3 2 1
711
The identity matrix An important matrix is the identity matrix: It transforms a point to itself: P1=P2=I·P1. This can be interpreted as. - translation with (0,0) - rotation with 0 degrees, since cos (0)=1 and sin (0) =0.
The identity transform is a data transformation that copies the source data into the destination data without change. The identity transformation is considered an essential process in creating a reusable transformation library.
The identity map idV :V V, idV v v v V, is a linear transformation whose matrix is the identity matrix I with respect to any single choice of basis B for V. 1. Page 2. Similar matrices. Now let V be a finite-dimensional vector space, with basis B, and let F :V V be a linear transfor- mation.
When the transformation matrix [a,b,c,d] is the Identity Matrix (the matrix equivalent of "1") the [x,y] values are not changed: Changing the "b" value leads to a "shear" transformation (try it above): And this one will do a diagonal "flip" about the x=y line (try it also):
The identity transform is a data transformation that copies the source data into the destination data without change. The identity transformation is considered an essential process in creating a reusable transformation library. By creating a library of variations of the base identity transformation,...
For example: The above is 2 × 4 matrix as it has 2 rows and 4 columns. Let’s multiply the 2 × 2 identity matrix by C. Hence proved. 3) We always get an identity after multiplying two inverse matrices. If we multiply two matrices which are inverses of each other, then we get an identity matrix.
An identity matrix is a square matrix in which each of the elements of its principal diagonal is a 1 and each of the other elements is a 0. It is also known as the unit matrix. We represent an identity matrix of order n × n (or n) as I n. Sometimes we denote this simply as I. The mathematical definition of an identity matrix is,
Try the code below (given m <- diag(5))
> (row(m) - col(m) + 1)*lower.tri(m,diag = TRUE)
[,1] [,2] [,3] [,4] [,5]
[1,] 1 0 0 0 0
[2,] 2 1 0 0 0
[3,] 3 2 1 0 0
[4,] 4 3 2 1 0
[5,] 5 4 3 2 1
Another option is using apply + cumsum
> apply(lower.tri(m, diag = TRUE), 2, cumsum)
[,1] [,2] [,3] [,4] [,5]
[1,] 1 0 0 0 0
[2,] 2 1 0 0 0
[3,] 3 2 1 0 0
[4,] 4 3 2 1 0
[5,] 5 4 3 2 1
1) If d <- diag(5) is the identity matrix then:
pmax(row(d) - col(d) + 1, 0)
giving:
[,1] [,2] [,3] [,4] [,5]
[1,] 1 0 0 0 0
[2,] 2 1 0 0 0
[3,] 3 2 1 0 0
[4,] 4 3 2 1 0
[5,] 5 4 3 2 1
2) This alternative is slightly longer (though still a one-liner) but also works if the columns of d are rearranged and/or some columns are missing. For example,
dd <- d[, 4:1] # test data
pmax(outer(1:nrow(dd) + 1, max.col(t(dd)), `-`), 0)
giving the same result for d and this for dd:
[,1] [,2] [,3] [,4]
[1,] 0 0 0 1
[2,] 0 0 1 2
[3,] 0 1 2 3
[4,] 1 2 3 4
[5,] 2 3 4 5
A solution based on nested cumsum:
n <- 5
m <- diag(n)
apply(m, 2, function(x) cumsum(cumsum(x)))
#> [,1] [,2] [,3] [,4] [,5]
#> [1,] 1 0 0 0 0
#> [2,] 2 1 0 0 0
#> [3,] 3 2 1 0 0
#> [4,] 4 3 2 1 0
#> [5,] 5 4 3 2 1
One option could be:
x <- 1:5
embed(c(rep(0, length(x) - 1), x), length(x))
[,1] [,2] [,3] [,4] [,5]
[1,] 1 0 0 0 0
[2,] 2 1 0 0 0
[3,] 3 2 1 0 0
[4,] 4 3 2 1 0
[5,] 5 4 3 2 1
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