Let v1= (1,1,2,1)
v2= (0,1,3,3)
v3= (1,-1,-4,-5)
v4= (1,0,-2,-4)
a) Let U=span{v1,v2,v3,v4}. Find the dimension of U
b) Is span {v1,v2,v3,v4}=R4?
c) Find a basis for U
Very Much Appreciated!!!
If U=span(v1,v2,v3,v4), then check the rank of the matrix
1 1 2 1
0 1 3 3
1 -1 -4 -5
1 0 -2 -4
(the v's are in rows)
The rank of the matrix is the dimension of U.
This can be done by reducing the matrix to the echelon form. The number of non-zero pivots is the rank of the matrix.
Do you need help with the reduction to the echelon form?
a) To find the dimension of U, we need to find the number of linearly independent vectors in the set {v1, v2, v3, v4} that span U. We can do this by performing row reduction on the augmented matrix [v1 | v2 | v3 | v4] and counting the number of non-zero rows.
Working through the row reduction:
1. Write the augmented matrix:
[1 0 1 1 | 1]
[1 1 -1 0 | 1]
[2 3 -4 -2 | 2]
[1 3 -5 -4 | 4]
2. Perform row operations to get zeroes below the leading entries:
R2 = R2 - R1,
R3 = R3 - 2R1,
R4 = R4 - R1
[1 0 1 1 | 1]
[0 1 -2 -1 | 0]
[0 3 -6 -4 | 0]
[0 3 -6 -5 | 3]
3. Perform row operations to get zeroes in the third column:
R3 = R3 - 3R2,
R4 = R4 - 3R2
[1 0 1 1 | 1]
[0 1 -2 -1 | 0]
[0 0 0 2 | 0]
[0 0 0 -2 | 3]
4. Swap the third and fourth rows:
[1 0 1 1 | 1]
[0 1 -2 -1 | 0]
[0 0 0 -2 | 3]
[0 0 0 2 | 0]
5. Perform row operations to get zeroes in the fourth column:
R4 = R4 + R3/2
[1 0 1 1 | 1]
[0 1 -2 -1 | 0]
[0 0 0 -2 | 3]
[0 0 0 0 | 3]
The row-reduced echelon form has only three non-zero rows, which means there are three linearly independent vectors in the set. Therefore, the dimension of U is 3.
b) To check if span{v1, v2, v3, v4} = R4, we need to see if every vector in R4 can be expressed as a linear combination of v1, v2, v3, and v4. Since the set only has three linearly independent vectors, it cannot span R4, which has a dimension of 4. Therefore, span{v1, v2, v3, v4} ≠ R4.
c) To find a basis for U, we can choose the three linearly independent vectors obtained from the row reduction: v1, v2, and v4.
Therefore, a basis for U is {v1, v2, v4} which can be written as:
{(1, 1, 2, 1), (0, 1, 3, 3), (1, 0, -2, -4)}.