Question

A =

100

1. Compute all four spaces of the matrix .

2. Find bases in them.

3. Find the matrix of A|c(A^T) in the bases you computed

Answer 1

1). The four fundamental subspaces are the column space and the null space of A and A^T. Also, Col (A^T) = Null(A)^⊥ and Null(A^T) = Col(A)^⊥. To determine these 4 subspaces, we will reduce A to its RREF as under:

1. Add -2 times the 2nd row to the 3rd row

2. Add -1 times the 2nd row to the 1st row

Then the RREF of A is

1	0	0
0	1	1
0	0	0

This implies that the first 2 columns of A are linearly independent and its 3^rd column is same as its 2^nd column. Therefore, Col(A) = span {(1,0,0)^T,(1,1,2)^T}.

Null(A) is the set of solutions to the equation AX = 0. If X = (x,y,z)^T, then the equation AX= 0 is equivalent to x = 0 and y+z = 0 or, y = -z. Then X = (0,-z,z)^T = z(0,-1,1)^T. This implies that every solution to the the equation AX= 0 is a scalar multiple of the vector(0,-1,1)^T. Hence, Null(A) = span {(0,-1,1)^T}.

Now, let X = (x,y,z)^T be an arbitrary vector in Col (A^T) = Null(A)^⊥.

Then (x,y,z)^T. (0,-1,1)^T = 0 or, -y+z = 0 or, y = z. Then X = (x,z,z)^T = x(1,0,0)^T+ z( 0,1,1)^T. This implies that every vector in Col (A^T) = Null(A)^⊥ is a linear combination of 2 linearly independent vectors (1,0,0)^T,( 0,1,1)^T. Hence, Col (A^T) = Null(A)^⊥ = span {(1,0,0)^T,( 0,1,1)^T }.

Further, let X = (x,y,z)^T be an arbitrary vector in Null(A^T) = Col(A)^⊥.

Then (x,y,z)^T. (1,0,0)^T = 0 or, x = 0 and (x,y,z)^T. (1,1,2)^T = 0 or,x+y+2z = 0 or, y+2z = 0 or, y = -2z. Then X = (0,-2z,z)^T = z(0,-2,1)^T. This implies that every vector in Null(A^T) is a scalar multiple of the vector(0,-2,1)^T. Hence, Null(A^T)   = Col(A)^⊥ =span{(0,-2,1)^T }.

2). The set {(1,0,0)^T,(1,1,2)^T} is a basis for Col(A).

The set {(0,-1,1)^T} is a basis for Null(A).

The set {(1,0,0)^T,( 0,1,1)^T } is a basis for Col (A^T) = Null(A)^⊥.

The set {(0,-2,1)^T } is a basis for Null(A^T) = Col(A)^⊥.

3). Please advise the meaning of the notation A|c(A^T).