Test 3. Fall 1996. Solutions



1. Prove that if a linear operator T: Rn --> Rn is invertible then the rank of its standard matrix is n.

Proof. Let A be the standard matrix of T. Then A is a square matrix of order n. Since T is invertible, T is surjective, so Range(T) is equal to Rn. Since the range of T is the column space of its standard matrix, the columns of this matrix span Rn. By the theorem about dimension, n vectors spanning an n-dimensional vector space must be linearly independent, so the columns of A are linearly independent, and the (column) rank of A is n.

2. Check if the following map T from C[0,1] to C[0,1] is a linear transformation:

T(f(x))=f2(x)+2f(2x)


Solution. Let f(x)=x, g(x)=1. Then T(f)=x2+4x, T(g)=1+2=3. But T(f+g)=T(x+1)=(x+1)2+2(2x+1)=x2+6x+3. Thus T(f+g) is not equal to T(f)+T(g) and T is not a linear transformation.

3. Check if the following subset of M2,2 is a subspace: V consists of all matrices 


              a b 
              
              c d
such that a+b=c+d.

Solution. Let A and B be two matrices from V: 
            a b         x y
         A=     ,   B = 
            c d         z t

Then a+b=c+d and x+y=z+t. The sum A+B is equal to 

             a+x  b+y
             
             c+z  d+t

Notice that a+x+b+y=(a+b)+(x+y)=(c+d)+(z+t)=c+z+d+t. So A+B also belongs to V.

Now take a number k. Then kA is equal to 

             ka  kb
 
             kc  kd

Notice that ka+kb=k(a+b)=k(c+d)=kc+kd. Thus kA also belongs to V. So V is a subspace.

4. Find the rank, the null space and the basis of the column space of the following matrix: 

           1 3 4 5
           2 6 5 1
           3 9 2 4

Solution. The reduced row echelon form of this matrix is 

           1 3 0 0
           0 0 1 0
           0 0 0 1

The number of leading 1's is 3, so the rank of the matrix is 3. The solution of the homogeneous system Av=0 is 

          x=-3t
          y=t
          z=0
          t=0

So the nullspace is spanned by the vector (-3,1,0,0).

The leading 1's are in the first, third and fourth columns, so the column space is spanned by the first, third and fourth columns of the original matrix: (1,2,3), (4,5,2), (5,1,4).

5. What is the kernel and the range of the following linear transformation from C[0,1] to R2:

T(f)=(f(0)+2f(1), f(0)-f(1))


Solution. The kernel of T consists of all functions f such that f(0)+2f(1)=0 and f(0)-f(1). It is easy to see that this can happen if and only if f(1)=f(0)=0.

The range of T is a subspace of R2. So the dimension of the range cannot exceed 2. Let f(x)=x. Then T(f)=(2,-1). Let f(x)=x+1. Then T(f)=(5,-1). So the vectors (2,-1) and (5,-1) belong to the range of T. Since these vectors are not proportional, they are linearly indepenent, so the dimension of the range is at least 2. Thus the dimension of the range is 2 and the range coincides with R2.

6. Find the least squares solution of the following system of equations: 


                 x-y=5
                 x+y=6
                 x-2y=1

Solution. Let A be the matrix of coefficients of this system of equations: 

  
                 1  -1
             A = 1   1
                 1  -2
Let b be the vector of the right sides: b=(5,6,1). Then the least squares solution of our system is the solution of the system ATA v=ATb: 
                 3x - 2y = 12
                -2x + 6y =-1

The solution is x=5, y=3/2. This is the least squares solution of our system.