Rank and Nullity

http://algebra.math.ust.hk/vector_space/10_rank/lecture1.shtml

Simple SAS codes to understand set loop and order duplicated cases.

data try;
input pid value;
datalines;
1 1
1 3
1 5
2 1
2 2
3 1
3 2
3 6
3 8
;
run;
data try2;
set try;
by pid;
episode+1;
if first.pid then episode=1;
run;

Spectral decomposition

A<-matrix(c(2,1,1,2),nrow=2,ncol=2)
A
#     [,1] [,2]
#[1,]    2    1
#[2,]    1    2
E<-eigen(A)
E
#$values
#[1] 3 1

#$vectors
#          [,1]       [,2]
#[1,] 0.7071068 -0.7071068
#[2,] 0.7071068  0.7071068


c1<-c(E$vectors[1,1],E$vectors[2,1])
c1
#[1] 0.7071068 0.7071068

c2<-c(E$vectors[1,2],E$vectors[2,2])
c2
#[1] -0.7071068  0.7071068

Q1<-c1%*%t(c1)
Q2<-c2%*%t(c2)

3*Q1+1*Q2
#     [,1] [,2]
#[1,]    2    1
#[2,]    1    2

Use R to solve linear euqations

https://cran.r-project.org/web/packages/matlib/vignettes/linear-equations.html

Michael Friendly

2016-09-16

This vignette illustrates the ideas behind solving systems of linear equations of the form Ax=b$\mathbf{A}\mathbf{x}\mathbf{=}\mathbf{b}$ where

• A$\mathbf{A}$ is an m×n$m\times n$ matrix of coefficients for m$m$ equations in n$n$ unknowns
• x$\mathbf{x}$ is an n×1$n\times 1$ vector unknowns, x1,x2,…xn$x_1,x_2,\dots x_n$
• b$\mathbf{b}$ is an m×1$m\times 1$ vector of constants, the “right-hand sides” of the equations

The general conditions for solutions are:

• the equations are consistent (solutions exist) if r(A|b)=r(A)$r(\mathbf{A}|\mathbf{b})=r(\mathbf{A})$
  • the solution is unique if r(A|b)=r(A)=n$r(\mathbf{A}|\mathbf{b})=r(\mathbf{A})=n$
  • the solution is underdetermined if r(A|b)=r(A)<n$r(\mathbf{A}|\mathbf{b})=r(\mathbf{A})<n$
• the equations are inconsistent (no solutions) if r(A|b)>r(A)$r(\mathbf{A}|\mathbf{b})>r(\mathbf{A})$

We use c( R(A), R(cbind(A,b)) ) to show the ranks, and all.equal( R(A), R(cbind(A,b)) ) to test for consistency.

library(matlib)   # use the package

Equations in two unknowns

Each equation in two unknowns corresponds to a line in 2D space. The equations have a unique solution if all lines intersect in a point.

Two consistent equations

A <- matrix(c(1, 2, -1, 2), 2, 2)
b <- c(2,1)
showEqn(A, b)

## 1*x1 - 1*x2  =  2 
## 2*x1 + 2*x2  =  1

c( R(A), R(cbind(A,b)) )          # show ranks

## [1] 2 2

all.equal( R(A), R(cbind(A,b)) )  # consistent?

## [1] TRUE

Plot them:

plotEqn(A,b)

##   x1   - x2  =  2 
## 2*x1 + 2*x2  =  1

Three consistent equations

For three (or more) equations in two unknowns, r(A)≤2$r(\mathbf{A})\le 2$, because r(A)≤min(m,n)$r(\mathbf{A})\le min(m,n)$. The equations will be consistent if r(A)=r(A|b)$r(\mathbf{A})=r(\mathbf{A}\mathbf{|}\mathbf{b})$, which means that whatever linear relations exist among the rows of A$\mathbf{A}$ are the same as those among the elements of b$\mathbf{b}$.

A <- matrix(c(1,2,3, -1, 2, 1), 3, 2)
b <- c(2,1,3)
showEqn(A, b)

## 1*x1 - 1*x2  =  2 
## 2*x1 + 2*x2  =  1 
## 3*x1 + 1*x2  =  3

c( R(A), R(cbind(A,b)) )          # show ranks

## [1] 2 2

all.equal( R(A), R(cbind(A,b)) )  # consistent?

## [1] TRUE

Plot them:

plotEqn(A,b)

##   x1   - x2  =  2 
## 2*x1 + 2*x2  =  1 
## 3*x1   + x2  =  3

Three inconsistent equations

A <- matrix(c(1,2,3, -1, 2, 1), 3, 2)
b <- c(2,1,6)
showEqn(A, b)

## 1*x1 - 1*x2  =  2 
## 2*x1 + 2*x2  =  1 
## 3*x1 + 1*x2  =  6

c( R(A), R(cbind(A,b)) )          # show ranks

## [1] 2 3

all.equal( R(A), R(cbind(A,b)) )  # consistent?

## [1] "Mean relative difference: 0.5"

An approximate solution is sometimes available using a generalized inverse.

x <- MASS::ginv(A) %*% b
x

##      [,1]
## [1,]    2
## [2,]   -1

Plot the equations. You can see that each pair of equations has a solution, but all three do not have a common, consistent solution.

plotEqn(A,b, xlim=c(-2, 4))

##   x1   - x2  =  2 
## 2*x1 + 2*x2  =  1 
## 3*x1   + x2  =  6

points(x[1], x[2], pch=15)

Equations in three unknowns

Each equation in three unknowns corresponds to a plane in 3D space. The equations have a unique solution if all planes intersect in a point.

Three consistent equations

A <- matrix(c(2, 1, -1,
             -3, -1, 2,
             -2,  1, 2), 3, 3, byrow=TRUE)
colnames(A) <- paste0('x', 1:3)
b <- c(8, -11, -3)
showEqn(A, b)

##  2*x1 + 1*x2 - 1*x3  =    8 
## -3*x1 - 1*x2 + 2*x3  =  -11 
## -2*x1 + 1*x2 + 2*x3  =   -3

Are the equations consistent?

c( R(A), R(cbind(A,b)) )          # show ranks

## [1] 3 3

all.equal( R(A), R(cbind(A,b)) )  # consistent?

## [1] TRUE

Solve for x$\mathbf{x}$.

solve(A, b)

## x1 x2 x3 
##  2  3 -1

solve(A) %*% b

##    [,1]
## x1    2
## x2    3
## x3   -1

inv(A) %*% b

##      [,1]
## [1,]    2
## [2,]    3
## [3,]   -1

Another way to see the solution is to reduce A|b$\mathbf{A}\mathbf{|}\mathbf{b}$ to echelon form. The result is I|A−1b$\mathbf{I}\mathbf{|}{\mathbf{A}}^{\mathbf{-}\mathbf{1}}\mathbf{b}$, with the solution in the last column.

echelon(A, b)

##      x1 x2 x3   
## [1,]  1  0  0  2
## [2,]  0  1  0  3
## [3,]  0  0  1 -1

echelon(A, b, verbose=TRUE, fractions=TRUE)

## 
## Initial matrix:
##      x1  x2  x3     
## [1,]   2   1  -1   8
## [2,]  -3  -1   2 -11
## [3,]  -2   1   2  -3
## 
## row: 1 
## 
##  exchange rows 1 and 2 
##      x1  x2  x3     
## [1,]  -3  -1   2 -11
## [2,]   2   1  -1   8
## [3,]  -2   1   2  -3
## 
##  multiply row 1 by -1/3 
##      x1   x2   x3       
## [1,]    1  1/3 -2/3 11/3
## [2,]    2    1   -1    8
## [3,]   -2    1    2   -3
## 
##  multiply row 1 by 2 and subtract from row 2 
##      x1   x2   x3       
## [1,]    1  1/3 -2/3 11/3
## [2,]    0  1/3  1/3  2/3
## [3,]   -2    1    2   -3
## 
##  multiply row 1 by 2 and add to row 3 
##      x1   x2   x3       
## [1,]    1  1/3 -2/3 11/3
## [2,]    0  1/3  1/3  2/3
## [3,]    0  5/3  2/3 13/3
## 
## row: 2 
## 
##  exchange rows 2 and 3 
##      x1   x2   x3       
## [1,]    1  1/3 -2/3 11/3
## [2,]    0  5/3  2/3 13/3
## [3,]    0  1/3  1/3  2/3
## 
##  multiply row 2 by 3/5 
##      x1   x2   x3       
## [1,]    1  1/3 -2/3 11/3
## [2,]    0    1  2/5 13/5
## [3,]    0  1/3  1/3  2/3
## 
##  multiply row 2 by 1/3 and subtract from row 1 
##      x1   x2   x3       
## [1,]    1    0 -4/5 14/5
## [2,]    0    1  2/5 13/5
## [3,]    0  1/3  1/3  2/3
## 
##  multiply row 2 by 1/3 and subtract from row 3 
##      x1   x2   x3       
## [1,]    1    0 -4/5 14/5
## [2,]    0    1  2/5 13/5
## [3,]    0    0  1/5 -1/5
## 
## row: 3 
## 
##  multiply row 3 by 5 
##      x1   x2   x3       
## [1,]    1    0 -4/5 14/5
## [2,]    0    1  2/5 13/5
## [3,]    0    0    1   -1
## 
##  multiply row 3 by 4/5 and add to row 1 
##      x1   x2   x3       
## [1,]    1    0    0    2
## [2,]    0    1  2/5 13/5
## [3,]    0    0    1   -1
## 
##  multiply row 3 by 2/5 and subtract from row 2 
##      x1 x2 x3   
## [1,]  1  0  0  2
## [2,]  0  1  0  3
## [3,]  0  0  1 -1

Plot them. plotEqn3d uses rgl for 3D graphics. If you rotate the figure, you’ll see an orientation where all three planes intersect at the solution point, x=(2,3,−1)$\mathbf{x}=(2,3,-1)$

plotEqn3d(A,b, xlim=c(0,4), ylim=c(0,4))

Three inconsistent equations

A <- matrix(c(1,  3, 1,
              1, -2, -2,
              2,  1, -1), 3, 3, byrow=TRUE)
colnames(A) <- paste0('x', 1:3)
b <- c(2, 3, 6)
showEqn(A, b)

## 1*x1 + 3*x2 + 1*x3  =  2 
## 1*x1 - 2*x2 - 2*x3  =  3 
## 2*x1 + 1*x2 - 1*x3  =  6

Are the equations consistent? No.

c( R(A), R(cbind(A,b)) )          # show ranks

## [1] 2 3

all.equal( R(A), R(cbind(A,b)) )  # consistent?

## [1] "Mean relative difference: 0.5"