Homework due Class 36
1. Find the basis of the orthogonal complement of the subspace of R4 spanned by the vectors (1,2,3,4) and (2,3,1,1). Let us label the vector (1,2,3,4) w1 and the vector (2,3,1,1) w2. The space spanned by w1 and w2 is the same as the rowspace of the following matrix A:
A := [ 1 2 3 4 ]
[ 2 3 1 1 ]
The nullspace is the orthogonal complement of the row space in Rn with respect to the Euclidean inner product. Thus we need to find the basis for the nullspace to solve this problem. To do this we augment a column of zeros to matrix A.
A := [ 1 2 3 4 0 ]
[ 2 3 1 1 0 ]
Using Gauss-Jordan elimination we obtain the matrix below:
A := [ 1 0 -7 10 0 ]
[ 0 1 5 7 0 ]
In equation form this is 
                           x1 =  7r + 10s
                           x2 = -5r -  7s
                           x3 = r
                           x4 = s.
So the set of all vectors (x1,x2,x3,x4), which is the nullspace of A and thus the orthogonal complement to R4, is spanned by the vectors (7,-5,1,0) and (10,-7,0,1). These two vectors are not proportional and thus are linearly independent (the theorem about linearly independent sets guarantees this). So we have a set of linearly independent vectors spanning our nullspace; by definition, this is a basis of the nullspace, and thus of the orthogonal complement to the subspace described above. 2. For which values of t do the following matrices form a basis of M2,2:
[ t 2t ]
[ 2 3t ]
,
[ 1 2 ]
[ 2t 3 ]
,
[ 1 2t ]
[ t+1 t+2 ]
,
[ 1 t+1 ]
[ 2 2t+1 ]
In the WebNotes, it is written that the set of matrices
[ 1 0 ]
[ 0 0 ]
[ 0 1 ]
[ 0 0 ]
[ 0 0 ]
[ 1 0 ]
[ 0 0 ]
[ 0 1 ]
is a basis of M2,2. Thus, M2,2 has dimension 4. So we may immediately apply the theorem about dimension: Let V be M2,2, which we have just shown to be a 4-dimension space; let S be the set of the four matrices given in the problem. Then S is a basis of V if S spans V or if S is linearly independent; we only need to establish one of these conditions. The linear independence is by far the easiest. For linear independence of S, we need the matrix equation
x1 *
[ t 2t ]
[ 2 3t ]
+ x2 *
[ 1 2 ]
[ 2t 3 ]
+ x3 *
[ 1 2t ]
[ t+1 t+2 ]
+ x4 *
[ 1 t+1 ]
[ 2 2t+1 ]
=
[ 0 0 ]
[ 0 0 ]
to have only the trivial solution. This can be rewritten to make it easier to work with:
[ t 1 1 1 ]
[ 2t 2 2t t+1 ]
[ 2 2t 2t t+1 ]
[ 3t 3 t+2 2t+1 ]

[ x1 ]
[ x2 ]
[ x3 ]
[ x4 ]
=
[ 0 ]
[ 0 ]
[ 0 ]
[ 0 ]
This will have a single [trivial] solution if the matrix of coefficients is invertible; this is true if and only if the determinant of the matrix is nonzero. Let Maple do the work:
> B:=matrix(4,4,[t,1,1,1,2*t,2,2*t,t+1,2,2*t,t+1,2,3*t,3,t+2,2*t+1]);
B := [ t 1 1 1 ]
[ 2t 2 2t t+1 ]
[ 2 2t 2t t+1 ]
[ 3t 3 t+2 2t+1 ]

> det(");

12 t3 - 12 t - 6 t4 + 6

> solve("=0,t);

-1, 1, 1, 1


So, if t does not equal 1 or -1, the determinant of the matrix of coefficients is not zero. Then the equation has only the trivial solution, and S is linearly independent. So S is a basis of M2,2 when t does not equal 1 or -1.