Math 235 Solution of Final Exam Spring 1999
a) The rank of A is 3.
b) A basis for the null space of A: The reduced echelon form is The free variables are and . The general solution is
and the two column vectors on the right hand side are a basis for Null(A).
c) A basis for the column space of A: The first, third and fifth columns of A are its pivot columns. They are a basis for the column space of A.
d) A basis for the row space of A: Take the three non-zero rows of B.
a) . Thus, .
b) The dimension of the column space is equal to the dimension of the row space, which is 4.
The three eigenvalues are -1, 1 and 2. The -1-eigenspace is Null(A+I) and it is spanned by
The 1-eigenspace is Null(A-I) and it is spanned by
The 2-eigenspace is Null(A-2I) and it is spanned by
Hence, is a besis of eigenvectors of A.
It is spanned by Hence, is a unit vector in .
Subtract from its orthogonal projection to the line spanned by . The resulting vector
will be perpendicular to the line spanned by . Thus,
is the required decomposition of .
First Method: (using part 6a). The vector Ax will be in the subspace W. The point in W closest to b is . Solve . Since , then .
Second Method: We can calculate x directly as the solution of the equation
The value can be written as the dot product of with . Hence, the sum can be written as the square of the norm of the vector
Plugging in the coordinates of the three points given, we get
The problem reduces to part 6c with
The eigenvalue of is .3 because
So, and .