Please work out each of the given problems.  Credit will be based on the steps that you show towards the final answer.  Show your work. 

Problem 1

Let  L:  R2 R3   be a linear transformation such that 

        L (1,4) = (1,-1,3)   and     L (0,2) =  (2,1,4)

Find L(1,0)


The key to this is to find the coordinates of (1,0) with respect to the basis (1,4) and (0,2).  This is given by A-1(1,0)T  .We have

Matrix[(1,0),(4,2)]^-1 = Matrix[(1,0),(-2,1/2)]

Multiplying, gives the coordinates (1,-2).  Thus

L(1,0) = L[1(1,4) - 2(0,2)] 

=  L(1,4) - 2L(0,2)

=  (1,-1,3) + 2(2,1,4)

=  (5,1,11)


Problem 2

Suppose that you want to change the graphic file from the one on the left to the one on the right.  What is the matrix needed to adjust the pixels appropriately?  Assume the center of the picture is the origin.



This is a reflection across the y-axis.  The corresponding matrix is

A = Matrix[(-1,0),(0,1)]

Problem 3 For each of the two subsets below, determine if it is a subspace of Rn.  If it is a subspace prove that it is.  If it is not a subspace prove that it is not.

  1. {(x,y,z) in R3 | x - yz = 0}
  2. {(x,y,z,w) in R4 | x + 2y = z - w}


First, A. is not a subspace since (4,2,2) is in the set (4 - (2)(2) = 0), but 2(4,2,2) = (8,4,4) is not in the set (8 - (4)(4) = -8 is not 0).

For B. we need to prove the two statements.  Note that a vector in this set is defined by z = x + 2y + w

1.  If u = (a,b,a+2b+c,c) and v = (d,e,d+2e+f), then

u+v = (a+d,b+e,a+2b+c+d+2e+f,c+f) = (a+d,b+e,a+d+2(b+e)+c+f,c+f)

and the vector has the correct form

2.  ku = k(a,b,a+2b+c,c) = (ka,kb,ka+2kb+kc,kc)

and this vector has the correct form.  Therefore this is a subspace.

Problem 4 

Answer the following true or false and explain your reasoning.

A.     If A and B are n x n matrices and AB  =  0, then either A = 0 or B = 0.
False, Consider
    A = Matrix[(1,0),(0,0)], B = Matrix[(0,0),(0,1)]

B.     If A is a matrix with A2  =  In then either det(A)  =  1 or det(A)  =  -1.  
True:  det(A2) = det(AA) = det(A)det(A)  =  [det(A)]2 and det(I) = 1.  If x = det(A), this gives x2 = 1.  So that x = 1 or x = -1.

C.  If the columns of an mxn matrix A form a basis for Rm, then m = n.
True, since the dimension of Rm is m, there must be m vectors in any basis.  Thus the basis that consists of the columns must have m vectors, meaning there are m columns:  m = n.

D.  If the nullity of an 8x11 matrix A is 6 then the column space of A is a 5 dimensional subspace of R8.
True, the dimension of the column space is the rank of A.  We have:
     Nullity + Rank = n
     6 + Rank = 11
     Rank = 5.

E.  If the second column of a 4x4 matrix A does not contain a pivot, then A is not invertible.
True, if the second column does not contain a pivot, then by TFAE, A is not invertible.

F.  If (1,3,4,1)T and (2,5,0,6)T are both solutions of Ax = b, then the equation Ax = 0 has a nontrivial solution.
True.  If u = (1,3,4,1)T  and v = (2,5,0,6)T, then w = u-v is nontrivial, and
     Aw = A(u-v) = Au - Av = b - b = 0.


Problem 5 



A.     Find the rank and the nullity of A. 
The rank is equal to the number of pivots:  Rank A = 3.
The nullity is the number of free variables or number of columns - Rank A = 2.

B.     Find a basis for the Null Space of A.
To find the null space you can put the rref(A) into equation form to get:
x1 = 0, x2 = -x4 + x5, x3 = -2x4 -4x5, x4 = x4, x5 = x5
Now let x4 = 1 and x5 = 0 to get (0,-1,-2,1,0)
let x4 = 0 and x5 = 1 to get (0,1,-4,0,1)
These two vectors form a basis for the null space of A.

C.     Find a basis for the Column Space of A.
A basis for the Column Space of A can be found by using the columns of A that are pivots of A.  From rref(A), we see that the first three columns have pivots.  So the basis for the Column Space of A is (1,2,0,3,2)T, (2,6,-4,1,4)T, and (1,2,0,-2,2)T.


Problem 6 

  Let  S  =  {v1, v2, ..., vn} be a set of linearly independent vectors and let v be a vector in the span of S.  Prove that v can uniquely be written as a linear combination of elements of S.  That is that prove that if

      =  a1v1 +a2v2 + ... + anvn        and        =  b1v1 +b2v2 + ... + bnvn     


        a1  =  b2, a2  =  b2, ... , an  =  bn


Suppose that

      =  a1v1 +a2v2 + ... + anvn        and        =  b1v1 +b2v2 + ... + bnvn     

Now subtract the two equations to get:

v - v  =  a1v1 +a2v2 + ... + anvn -  (b1v1 +b2v2 + ... + bnv)

Combine like terms to get

0 = (a1 - b1)v1 + (a2 - b2)v2 + ... + (an + bn)vn

Now since S is linearly independent we have by definition that

a1 - b1 = 0,  + a2 - b2 = 0, ... , an + bn = 0

And the result follows.


Problem 7 Use the definition of span to prove that the vectors shown below span R2.

v = (3,5) and w = (2,3)


Let b = (b1,b2) be a vector in R2.  We need to show that there is a solution to

x1v + x2w = b

or that

x1(3,5) + x2(2,3) = (b1,b2)

This can be written as the matrix equation

Matrix[(3,2),(5,3)] Matrix[x1,x2] = Matrix[b1,b2]

This has a solution since rref(A) is the identity element, so the inverse exists and the solution is x = A-1b.



Problem 8 

Prove that if v and w are solutions to the matrix equation Ax = b and if r + s = 0, then

rv + sw is a solution to the homogeneous equation Ax = 0.


We have

A(rv + sw) = A(rv) = A(sw) = rAv + sAw

Since v and w are solutions to Ax = b, both Av and Aw are b.  We get that the above equals

rb + sb = (r+s)b = 1b = b


Problem 9

As a consultant for the South Tahoe ski marketing group, your task is to make predictions on the ski pass that locals hold.  You found that this year 55% of locals hold a pass for Heavenly, 30% hold a pass for Sierra-at-Tahoe, and 15% hold a pass for Kirkwood.  Each year, 20% of the Heavenly pass holders defect to Sierra-at-Tahoe and 5% defect to Kirkwood.  10% of the Sierra-at-Tahoe pass holders defect to Heavenly and 20% defect to Kirkwood.  5% of the Kirkwood pass holders defect to Heavenly and 25% defect to Sierra-at-Tahoe.  Assume that every pass holder remains a pass holder the following year and that everyone holds exactly one pass.  Determine what the distribution of pass holders will be in five years.

Solution:  We get the matrix:

A = Matrix[(.75,.10,.05),(.20,.70,.25),(.05,.20,.70)]

Since we want to find the distribution after five years we take

A^5 * Matrix[.55,.30,.15] = Matrix[.29,.42,.29]

So after 5 years, 29% will have Heavenly pass, 42% will have a Sierra-at-Tahoe pass, and 29% will have a Kirkwood pass.


Problem 10

Use the definition to find the determinant of the following matrix.  Do not use your calculator.

Matrix Rows:  (1,2,-1,0), (0,5,3,0), (-2,0,0,4), (0,6,-4,-3)

Expand by the second row.  Note that since the first and last entry of the second row are both 0, we only need the middle numbers.

5*Matrix[(1,-1,0),(-2,0,4),(0,-4,-3)] - 3*Matrix[(1,2,0),(-2,0,4),(0,6,-3)]

Now expand each of these by their first rows to get

5[Matrix[(0,4),(-4,-3)]+Matrix[(-2,4),(0,-3)]-3[Matrix[(0,4),(6,-3)] - 2*Marix[(-2,4),(0,-3)]]

= 5[16 + 6] - 3[-24 -12] = 110 + 108 = 218.

Extra Credit:  Write down one thing that your instructor can do to make the class better and one thing that you want to remain the same in the class.

(Any constructive remark will be worth full credit.)