Homework due Class 28
1. Check whether the following vectors in R5 are linearly independent. If they are not linearly independent, find one of them which is a linear combinationof the others.
1. (1,2, 3, 4, 5), (2,3,4,5,6), (1,0,0,0,1), (2, 5, 7, 9, 10);
2. (2,1,1,1,1), (2,1,4,4,4), (3,3,3,1,1).
2. Are these functions linearly independent as elements of C[0,1]: f(x)=x, g(x)=ex, h(x) = e2x.
1. Check whether the following vectors in R5 are linearly independent. If they are not linearly independent, find one of them which is a linear combination of the others.
1. (1,2, 3, 4, 5), (2,3,4,5,6), (1,0,0,0,1), (2, 5, 7, 9, 10);
The set of vectors in R5 are linearly independent if the only solution to the equation below is the trivial (all zeros) solution:
a(1,2,3,4,5)+b(2,3,4,5,6)+c(1,0,0,0,1)+d(2,5,7,9,10)=(0,0,0,0,0)
or
a+2b+c+2d = 0
2a+3b+5d = 0
3a+4b+7d = 0
4a+5b+9d = 0
5a+6b+c+10d = 0
>with(linalg);
Warning: new definition for norm
Warning: new definition for trace
>A:=matrix([[1,2,1,2,0],[2,3,0,5,0],[3,4,0,7,0],[4,5,0,9,0],[5,6,1,10,0]]);
| A := |
[ 1 |
2 |
1 |
2 |
0 ] |
| [ 2 |
3 |
0 |
5 |
0 ] |
| [ 3 |
4 |
0 |
7 |
0 ] |
| [ 4 |
5 |
0 |
9 |
0 ] |
| [ 5 |
6 |
1 |
10 |
0 ] |
>gaussjord(A);
| [ 1 |
0 |
0 |
1 |
0 ] |
| [ 0 |
1 |
0 |
1 |
0 ] |
| [ 0 |
0 |
1 |
-1 |
0 ] |
| [ 0 |
0 |
0 |
0 |
0 ] |
| [ 0 |
0 |
0 |
0 |
0 ] |
The solutions of this system are:
a = -d
b = -d
c = d
d = d
So the system has nontrivial solutions. Thus set of vectors in R5 form a linearly dependent set.
Let us try to form (1,2,3,4,5) as linear combination of (2,3,4,5,6),(1,0,0,0,1) and (2,5,7,9,10). These vectors must then sastify the system of linear equation below:
e(2,3,4,5,6)+f(1,0,0,0,1)+g(2,5,7,9,10)=(1,2,3,4,5)
or
2e+f+2g = 1
3e+5g = 2
4e+7g = 3
5e+9g = 4
6e+f+10g = 5
>B:=matrix([[2,1,2,1],[3,0,5,2],[4,0,7,3],[5,0,9,4],[6,1,10,5]]);
| B := |
[ 2 |
1 |
2 |
1 ] |
| [ 3 |
0 |
5 |
2 ] |
| [ 4 |
0 |
7 |
3 ] |
| [ 5 |
0 |
9 |
4 ] |
| [ 6 |
1 |
10 |
5 ] |
>gaussjord(B);
| [ 1 |
0 |
0 |
-1 ] |
| [ 0 |
1 |
0 |
1 ] |
| [ 0 |
0 |
1 |
1 ] |
| [ 0 |
0 |
0 |
0 ] |
| [ 0 |
0 |
0 |
0 ] |
the solutions of this system are:
e =- 1
f = 1
g = 1
So (1,2,3,4,5) is a linear combination of the others:
(1,2,3,4,5) = -(2,3,4,5,6) + (1,0,0,0,1) + (2,5,7,9,10).
1. Check whether the following vectors in R5 are linearly independent. If they are not linearly independent, find one of them which is a linear combinationof the others.
2. (2,1,1,1,1), (2,1,4,4,4), (3,3,3,1,1).
similarly, we construct of the system of equation:
x1(2,1,1,1,1)+x2(2,1,4,4,4)+x3(3,3,3,1,1)=(0,0,0,0)
or
2x1+2x2+3x3=0
x1+ x2+3x3=0
x1+4x2+3x3=0
x1+4x2+ x3=0
x1+4x2+ x3=0
>A:=matrix([[2,2,3,0],[1,1,3,0],[1,4,3,0],[1,4,1,0],[1,4,1,0]]);
| A := |
[ 2 |
2 |
3 |
0 ] |
| [ 1 |
1 |
3 |
0 ] |
| [ 1 |
4 |
3 |
0 ] |
| [ 1 |
4 |
1 |
0 ] |
| [ 1 |
4 |
1 |
0 ] |
>gaussjord(A);
| [ 1 |
0 |
0 |
0 ] |
| [ 0 |
1 |
0 |
0 ] |
| [ 0 |
0 |
1 |
0 ] |
| [ 0 |
0 |
0 |
0 ] |
| [ 0 |
0 |
0 |
0 ] |
the solution of the system are:
x1=0
x2=0
x3=0
Thus the system has only one solution. Therefore, the vectors(2,1,1,1,1),(2,1,4,4,4) and (3,3,3,1,1)form a linearly independent set.
2. Are these functions linearly independent as elements of C[0,1]: f(x)=x, g(x)=ex, h(x) = e2x.
These functions are linearly independent if only the trivial solution sastifies the system below:
k1f(x)+k2g(x)+k3h(x)=0
or
k1x+k2*ex+k3*e2x=0
By substituting for specific value of x, we can generate a system of equations which could help show that these functions are linearly independent.
At x=0: k2 + k3 = 0
At x=1: k1 + k2*e + k3*e2 = 0
At x=0.2: 0.2*k1 + k2*e0.2 + k3*e0.4 = 0
At x=0.3: 0.3*k1 + k2*e0.3 + k3*e0.6 = 0
Represented as a matrix:
>A:=matrix([[0,1,1,0],[1,evalf(E^1),evalf(E^2),0],[0.2,evalf(E^0.2),evalf(E^0.4),0],[0.3,evalf(E^0.3),evalf(E^0.6),0]]);
| A := |
[ 0 |
1 |
1 |
0 ] |
| [ 1 |
2.718281828 |
7.389056096 |
0 ] |
| [ .2 |
1.221402758 |
1.491824698 |
0 ] |
| [ .3 |
1.349858808 |
1.822118800 |
0 ] |
>gaussjord(A);
| [ 1 |
0 |
0 |
0 ] |
| [ 0 |
1 |
0 |
0 ] |
| [ 0 |
0 |
1 |
0 ] |
| [ 0 |
0 |
0 |
0 ] |
As we can see, even for these limited values of x, the only solution is the trivial solution, so the equations are linearly independent.