Linear Algebra Math 310
Spring 2012
Selected (*) Solutions Section 1.7

Even Numbered * Problems from Book

28.  How many pivot columns must 5 x 7 matrix have if its columns span R⁵?
SOLUTION:  There must be 5 pivot columns, which can be justified as follows.  Let the matrix in question be A.  The columns of A span R⁵ if and only if there is a pivot position in each row (Theorem 4, section 1.4).  Since there are 5 rows, that means there must be 5 pivot positions.  But each is in a different column, so that implies that there are 5 pivot columns.

30.  If A is an m × n matrix, then the columns of A are linearly independent if and only if A has n pivot columns.  
JUSTIFICATION:  This is an if and only if statement, so the justification requires two parts.  First, suppose that the columns are linearly independent.  That means that the only way to combine those columns to make a zero vector is to use a coefficient of 0 on each column.  But that is the same things as saying that the homogeneous equation Ax = 0 has only the trivial solution, and in particular, has a unique solution.  That in turn means that there are no free variables in the general solution of the equation, and hence no non-pivot columns in the RREF of A.  So, every column of A is a pivot column, and there are n of them.  This shows that if the columns are independent, then there are n pivot columns, and completes the first part of the justification.

For the second part, suppose that there are n pivot columns.  Then every column is a pivot column, and there are no non-pivot columns.  So, the equation Ax = 0 has no free variables.  Because this is a homogeneous equation, it is automatically consistent, and x = 0 will be one solution.  Now the absence of free variables tells us that this solution must be unique: the equation Ax = 0 has only the trivial solution.  By statement (3) page 66, we can conclude that the columns of A are linearly independent.  This shows that if there are n pivot columns, then the columns of A are linearly independent, and completes the second part of the justification.

40.  [Not included in revised assignment.] Suppose an m × n matrix A has n pivot columns.  Show that, for any b in R^m, the equation Ax = b has at most one solution.
SOLUTION:  Following the book's suggestion, it is enough to rule out the possibility of infinitely many solutions, for then there remain only the possibility of one solution or no solution.  We know that the case of infinitely many solutions only occurs when there are free varibles, and hence, only if the there are non-pivot columns in A.  But the given assumption is that all the columns of A are pivot columns.  So, with no non-pivot columns in A, there are no free variables, and can not be an infinite number of solutions.  This shows that any system of equations with coefficient matrix A can have at most one solution.

101. Try to complete the following statements of theorems and/or definitions without looking them up in the book:
1. A set of two vectors is linearly dependent if and only if one is a multiple of the other.
2. If a set of vectors in Rⁿ contains the zero vector then it is linearly dependent.
3. For a set of vectors, if the number of vectors in the set is more than the number of entries in each vector, then the set is linearly dependent.
4. The columns of a matrix A are linearly independent if and only if the matrix equation Ax = 0 has only the trivial solution (x = 0).
5. If S = { v₁ , ..., v_p } is a set of two or more vectors, then S is a linearly dependent set if and only if at least one element of S is a linear combination of the other elements.
   
   
102. (*) Prove: If {v₁,v₂, ..., v_n } is a dependent set of vectors, then so is {v₁,v₂, ..., v_n,v_n+1 } for any vector v_n+1.
     Solution:  We are given that {v₁,v₂, ..., v_n } is a dependent set of vectors.  That means there are coefficients c₁, c₂, ... c_n ,  such that c₁v_{1 +} c₂v₂+ ... + c_nv_n = 0, and not all of these coefficients equal 0.   Now let v_n+1 be any vector, and define one more coefficient, c_n+1 .  Furthermore, define this new coefficient to be 0.  Then the following two statements are true:
                      (a) c₁v_{1 +} c₂v₂+ ... + c_nv_n+ c_n+1v_n+1 = 0 ,  and
                      (b) not all of the coefficients in (a) equal 0.
     This shows that {v₁,v₂, ..., v_n, v_n+1 } is a dependent set of vectors , as desired.  
     
     
103. (*) Prove: If {v₁,v₂, ..., v_n } is an independent set of vectors, then so is any non-empty subset. To simplify the argument, you may assume that any non-empty subset can be expressed as {v₁,v₂, ..., v_k } for some k with 1 < k < n. Why is that valid?
     Solution:  Given any nonempty subset of the given set of vectors, it would be possible to relabel the original vectors so the subset consists of vectors 1 through k for some k between 1 and n inclusive.  The dependence or independence of a set of vectors does not depend on the labels we use to denote these vectors, so relabeling vectors in this way does not invalidate the proof.  So, we now want to show that {v₁,v₂, ..., v_k } is an independent set of vectors.  According to the definition of linear independence, we can do this by establishing that there is only a trivial solution to the homogeneous equation c₁v_{1 +} c₂v₂+ ... + c_kv_k = 0.  Now we argue by contradiction.  Suppose that there is a nontrivial solution.  That is, suppose there are coefficients c₁, c₂, ... c_k ,  such that c₁v_{1 +} c₂v₂+ ... + c_kv_k = 0, and not all of these coefficients equal 0.  Then we can define the additional coefficients c_k+1, c_k+2, ... c_n ,  and assign these each to equal 0.  Then the following two statements are true:
                      (a) c₁v_{1 +} c₂v₂+ ... + c_nv_n = 0 ,  and
                      (b) not all of the coefficients in (a) equal 0.
     This shows that {v₁,v₂, ..., v_n} is a dependent set of vectors.  But that is a contradiction of the starting assumption for the proof.  So, assuming that a nontrivial solution exists to c₁v_{1 +} c₂v₂+ ... + c_kv_k = 0 leads to a contradiction.  We conclude that this assumption must in fact be false.  Therefore, NO nontrivial solution exists to c₁v_{1 +} c₂v₂+ ... + c_kv_k = 0, and that establishes that {v₁,v₂, ..., v_k} is an independent set of vectors.
     
     
104. (*) Make up a definition for linear independence of functions (as opposed to vectors). According to your definition, are the functions f(x) = x and g(x) = x² linearly independent? What about the functions cos²x, sin²x, and 2-cos²x?
     Solution:  define the set of functions {f₁,f₂, ..., f_k} to be dependent if there is a set of coefficients c₁, c₂, ... c_k ,  not all zero, such that the equation c₁f₁(x) + c₂f₂(x)+ ... + c_kf_k(x) = 0 is true for all x.  If a set of functions is NOT dependent, then we say it is independent.
     
     Using this definition, {x, x²} is an independent set of functions.  To show that, for any particular constants c₁, c₂  consider the equation c₁x + c₂x² = 0.  We can factor this equation to obtain x( c₁ + c₂x) = 0.  Now consider the possibilities.  First, if c₁ and c₂  both = 0, then the equation is true for every x. If c₂ = 0 but not c₁, then the only solution is x = 0.  If c₁ = 0 but not c₂, then the only solution is again x = 0.  Finally,  if neither c₁  nor c₂ is equal to 0, then there are two solutions: x = 0 and x = -c₁ / c₂ . Reviewing these possibilities, we observe that the equation holds for ALL x only if both coefficients are 0. So, {x, x²} is not a dependent set, and so must be an independent set of functions
     
     On the other hand, the set  {cos²x, sin²x, 2-cos²x}  is a dependent set of functions.  To justify this, we use the trigonometric identity cos²x + sin²x = 1, which is true for all x.  Therefore
             1cos²x + 2sin²x -1(2-cos²x) = 1cos²x + 2sin²x -2 + cos²x = 2cos²x + 2sin²x- 2 = 0
     holds for all x.  Since the coefficients in this equation are 1, 2, and -1, and these are not all 0, this shows that the set  {cos²x, sin²x, 2-cos²x}  is a dependent set of functions.