Module 4 Practice: Vector Spaces
10 problems on vectors, subspaces, span, linear independence, basis, and dimension.
Problem 1
Compute (2,1,-3) . (4,-2,1).
Solution
8 + (-2) + (-3) = 3.
Problem 2
Find ||(3,-4)||.
Solution
sqrt(9+16) = sqrt(25) = 5.
Problem 3
Is {(x,y,z): x+y+z=0} a subspace of R^3?
Solution
Yes. Contains 0 (0+0+0=0). Closed under addition and scalar multiplication. This is the null space of [1 1 1].
Problem 4
Is (5,3) in Span{(1,1),(2,1)}?
Solution
Solve c_1+2c_2=5 and c_1+c_2=3. Subtract: c_2=2, c_1=1. Yes: (5,3)=1(1,1)+2(2,1).
Problem 5
Are {(1,0,1),(0,1,1),(1,1,0)} linearly independent?
Solution
det([1 0 1;0 1 1;1 1 0]) = 1(0-1)-0+1(0-1) = -1-1=-2 (nonzero). Independent.
Problem 6
Find a basis for Nul([1 2 1; 2 4 2]).
Solution
RREF: [1 2 1; 0 0 0]. Free: x_2=s, x_3=t. x_1=-2s-t. Basis: {(-2,1,0),(-1,0,1)}.
Problem 7
What is the rank of a 3x5 matrix with RREF having 2 pivots?
Solution
Rank = 2. Nullity = 5-2 = 3.
Problem 8
Can 3 vectors in R^4 be a basis for R^4?
Solution
No. A basis for R^4 needs exactly 4 vectors (dim R^4 = 4). 3 vectors can span at most a 3-dimensional subspace.
Problem 9
If rank(A) = 3 for a 5x4 matrix, what is the dimension of Nul(A)?
Solution
nullity = 4 - 3 = 1.
Problem 10
Give a basis for the column space of A = [1 3; 2 6].
Solution
RREF: [1 3; 0 0]. One pivot in column 1. Basis for Col(A): {(1,2)} (column 1 of original A).