Final Test, Fall 1996

1. Formulate the theorem about the rank and the nullity of a matrix. Use this theorem to prove that if a subspace W of Rn has dimension m then the orthogonal complement Wc has dimension n-m.

Solution

The theorem says that the rank of an m by n matrix plus the nullity of this matrix is equal to n.

Let W be spanned by m linearly independent vectors (A1=a11,...,a1n), ..., Am=(am1,...,amn). Then the orthogonal complement Wc consists of vectors which are orthogonal to A1,...,Am. Thus this orthogonal complement coinsides with the set of all vectors v=(x1,...,xm) which are solutions of the system of equations Mv=0 where M is the matrix with rows A1,...,Am). The dimension of Wc is then the nullity of M. The rank of M is equal to m because the rows of this matrix are linearly independdent. Therefore by the theorem about the rank and the nullity, the dimension of the orthogonal complement is n-m.

2. Is it true that the map from C(0,1) to R2 which takes every function f(x) from C(0,1) to the vector (f(1)+f(0)2, f(1)+f(0)) is a linear transformation?

Solution.

No. Take f(x)=x+1, g(x)=x+2. Then T(f+g) is not equal to T(f)+T(g).

3. What is the standard matrix of the rotation of R2 through the angle Pi/4 counterclockwise?

Solution. 
                   [cos(Pi/4) -sin(Pi/4)]
                   [sin(Pi/4)  cos(Pi/4)]

4. Represent the following matrix as a product of elementary matrices. 

                              [-7   4]
                              [-4   2]


Solution.


                              [-7   4] =[1 2]*[1 0]*[ 1 0]
                              [-4   2]  [0 1] [0 2] [-2 1]


5. (i) For which values of a is the following matrix A invertible.
(ii) In the case when it is invertible, find the inverse.
(iii) What is the rank of this matrix depending on a?
(iv) For which values of a the linear operator in R3 with standard matrix A is surjective?
(v) What is the dimension of the null space of this matrix depending on a?
(vi) What is the basis of the column space of this matrix depending on a? 

                              [a-1 1   1 ]
                         A =  [0   a   1 ]
                              [0   0  a-2]


Solution

(i) The matrix is invertible if and only if its determinant is not 0. The determinant of our matrix is (a-1)a(a-2). So the matrix is invertible if and only if a is not equal to 0, 1 and 2.

(ii) 

                       [1/(a-1)   -1/a(a-1)  -1/a(a-2)]
                 A-1=  [0          1/a       -1/a(a-2)]
                       [0          0          1/(a-2) ]

(iii) If a is 1, 0 or 2 then the rank is 2. Otherwise the rank is 3.

(iv) An operator in Rn is surjective if and only if it is invertible if and only if its standard matrix is invertible. So our operator is surjective if and only if a is not equal to 0, 1, 2.

(v) By the theorem about the rank and the nullity, the dimension of the null space is 0 if a is not 0, 1, 2. Otherwise it is equal to 1.

(vi) If a=1 then the basis is (1,1,0), (1,1,-1), if a=0 then the basis is (-1,0,0), (1,1,-2). If a=2 then the basis is (1, 0, 0), (1,2,0). In all other cases, the column space coincides with R3, so the basis is {i,j,k}.

6. For which values of parameter a do the following vectors form a basis in R2: (a2, 2a), (1,1).

Solution

Two vectors in R2 form a basis if and only if they are not proportional. Our vectors are proportional if and only if a2=b, 2a=b for some number b. This gives a=0, 2. Thus our vectors form a basis if and only if a is not 0 or 2.

7. For which values of k is the matrix 

        [1 -2]
        [k  0]
a linear combination of the following two matrices: 
        [ 3  0]           [ 2 -1]
        [-2  0],          [-5  0]

Solution.
Let the first matrix be A(k), and the other two matrices be B, C. We need to find k for which there exist numbers x, y such that A(k)=xB+yC This leads to the following system of equations: 

        1 = 3x +2y
       -2 =    - y
        k =-2x -5y

The augmented matrix of this system is 
         3  2  1
         0 -1 -2 
        -2 -5  k
Using the Gauss-Jordan procedure we get the matrix 
         1 0 -5-(1/2)k
         0 1     2
         0 0 12+(3/2)k
Thus the system has a solution if and only if 12+(3/2)k=0, that is k=-8. If k=-8 then x=-1, y=2.

8. Find the determinant of the following n by n matrix: 

                      [1 1 1 1.... 1]
                      [0 1 1 1.... 1]
                      [0 0 1 1.....1]  
                      ....................
                      [1 0 0 0 ....1]


Solution

If n=1 then the determinant is 1, if n=2 then the determinant is 0. Suppose that n>2. Expand along the first column. One of the two matrices that you get is upper triangular with determinant 1, and the second matrix has two equal rows, so its determinant is 0. Thus the determinant of our matrix is 1.

9. (i) Find the orthogonal basis in the subspace of R4 spanned by the following 3 vectors: (1,1,1,1), (1,-1,1,1), (1,1,-1,1).
(ii) Find the projection of the vector i=(1,0,0,0) onto this subspace.

Solution.

(i) The Gram-Schmidt method gives us three vectors: a=(1,1,1,1), b=(1/2, -3/2, 1/2, 1/2), c=(2/3, 0, -4/3, 2/3). This is the answer to part (i).

(ii) The projection is (< i,a>/< a,a>)a+(< i,b> /< b,b>)b+(< i,c>/< c,c>)c= 1/4(1,1,1,1)+1/6(1/2,-3/2,1/2,1/2)-1/2(2/3,0,-4/3,2/3)= (1/2, 0, 0, 1/2).

10. Let T be the linear operator in R2 with the following standard matrix: 

                [ 3 2]
                [-1 0]

(i) Find the eigenvalues of this operator.
(ii) Find a basis B of R2 in which the matrix of this operator is a diagonal.
(iii) Find the transition matrix from the standard basis {i, j} to the basis B.

Solution.
(i) The characteristic polynomial is (x-3)x+2. The roots are 1, 2. So 1 and 2 are the eigenvalues.

(ii) An eigenvector, corresponding to 1 is (1,-1), the eigenvector, corresponding to 2 is (2,-1). these two vectors for the basis Bwhere the matrix of our operator is diagonal with 1 and 2 on the main diagonal.


(iii) The transition matrix from B to the standard basis is 

              [ 1  2]
              [-1 -1]

The transition matrix from the standard basis to B is the inverse of this matrix: 

              [-1  -2]
              [ 1   1]