Linear Algebra/Changing Representations of Vectors/Solutions

Solutions

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This exercise is recommended for all readers.
Problem 1

In  , where

 

find the change of basis matrices from   to   and from   to  . Multiply the two.

Answer

For the matrix to change bases from   to   we need that   and that  . Of course, the representation of a vector in   with respect to the standard basis is easy.

 

Concatenating those two together to make the columns of the change of basis matrix gives this.

 

The change of basis matrix in the other direction can be gotten by calculating   and   (this job is routine) or it can be found by taking the inverse of the above matrix. Because of the formula for the inverse of a   matrix, this is easy.

 
This exercise is recommended for all readers.
Problem 2

Find the change of basis matrix for  .

  1.  ,  
  2.  ,  
  3.  ,  
  4.  ,  
Answer

In each case, the columns   and   are concatenated to make the change of basis matrix  .

  1.  
  2.  
  3.  
  4.  
Problem 3

For the bases in Problem 2, find the change of basis matrix in the other direction, from   to  .

Answer

One way to go is to find   and  , and then concatenate them into the columns of the desired change of basis matrix. Another way is to find the inverse of the matrices that answer Problem 2.

  1.  
  2.  
  3.  
  4.  
This exercise is recommended for all readers.
Problem 4

Find the change of basis matrix for each  .

  1.  
  2.  
  3.  
Answer

The columns vector representations  , and  , and   make the change of basis matrix  .

  1.  
  2.  
  3.  

E.g., for the first column of the first matrix,  .

This exercise is recommended for all readers.
Problem 5

Decide if each changes bases on  . To what basis is   changed?

  1.  
  2.  
  3.  
  4.  
Answer

A matrix changes bases if and only if it is nonsingular.

  1. This matrix is nonsingular and so changes bases. Finding to what basis   is changed means finding   such that
     
    and by the definition of how a matrix represents a linear map, we have this.
     
    Where
     
    we can either solve the system
     
    or else just spot the answer (thinking of the proof of Lemma 1.4).
     
  2. Yes, this matrix is nonsingular and so changes bases. To calculate  , we proceed as above with
     
    to solve
     
    and get this.
     
  3. No, this matrix does not change bases because it is nonsingular.
  4. Yes, this matrix changes bases because it is nonsingular. The calculation of the changed-to basis is as above.
     
Problem 6

Find bases such that this matrix represents the identity map with respect to those bases.

 
Answer

This question has many different solutions. One way to proceed is to make up any basis   for any space, and then compute the appropriate   (necessarily for the same space, of course). Another, easier, way to proceed is to fix the codomain as   and the codomain basis as  . This way (recall that the representation of any vector with respect to the standard basis is just the vector itself), we have this.

 
Problem 7

Conside the vector space of real-valued functions with basis  . Show that   is also a basis for this space. Find the change of basis matrix in each direction.

Answer

Checking that   is a basis is routine. Call the natural basis  . To compute the change of basis matrix   we must find   and  , that is, we need   such that these equations hold.

 

Obviously this is the answer.

 

For the change of basis matrix in the other direction we could look for   and   by solving these.

 

An easier method is to find the inverse of the matrix found above.

 
Problem 8

Where does this matrix

 

send the standard basis for  ? Any other bases? Hint. Consider the inverse.

Answer

We start by taking the inverse of the matrix, that is, by deciding what is the inverse to the map of interest.

 

This is more tractable than the representation the other way because this matrix is the concatenation of these two column vectors

 

and representations with respect to   are transparent.

 

This pictures the action of the map that transforms   to   (it is, again, the inverse of the map that is the answer to this question). The line lies at an angle   to the   axis.

 

This map reflects vectors over that line. Since reflections are self-inverse, the answer to the question is: the original map reflects about the line through the origin with angle of elevation  . (Of course, it does this to any basis.)

This exercise is recommended for all readers.
Problem 9

What is the change of basis matrix with respect to  ?

Answer

The appropriately-sized identity matrix.

Problem 10

Prove that a matrix changes bases if and only if it is invertible.

Answer

Each is true if and only if the matrix is nonsingular.

Problem 11

Finish the proof of Lemma 1.4.

Answer

What remains to be shown is that left multiplication by a reduction matrix represents a change from another basis to  .

Application of a row-multiplication matrix   translates a representation with respect to the basis   to one with respect to  , as here.

 

Applying a row-swap matrix   translates a representation with respect to   to one with respect to  . Finally, applying a row-combination matrix   changes a representation with respect to   to one with respect to  .

 
 

(As in the part of the proof in the body of this subsection, the various conditions on the row operations, e.g., that the scalar   is nonzero, assure that these are all bases.)

This exercise is recommended for all readers.
Problem 12

Let   be a   nonsingular matrix. What basis of   does   change to the standard basis?

Answer

Taking   as a change of basis matrix  , its columns are

 

and, because representations with respect to the standard basis are transparent, we have this.

 

That is, the basis is the one composed of the columns of  .

This exercise is recommended for all readers.
Problem 13
  1. In   with basis   we have this represenatation.
     
    Find a basis   giving this different representation for the same polynomial.
     
  2. State and prove that any nonzero vector representation can be changed to any other.

Hint. The proof of Lemma 1.4 is constructive— it not only says the bases change, it shows how they change.

Answer
  1. We can change the starting vector representation to the ending one through a sequence of row operations. The proof tells us what how the bases change. We start by swapping the first and second rows of the representation with respect to   to get a representation with respect to a new basis  .
     
    We next add   times the third row of the vector representation to the fourth row.
     
    (The third element of   is the third element of   minus   times the fourth element of  .) Now we can finish by doubling the third row.
     
  2. Here are three different approaches to stating such a result. The first is the assertion: where   is a vector space with basis   and   is nonzero, for any nonzero column vector   (whose number of components equals the dimension of  ) there is a change of basis matrix   such that  . The second possible statement: for any ( -dimensional) vector space   and any nonzero vector  , where   are nonzero, there are bases   such that   and  . The third is: for any nonzero   member of any vector space (of dimension  ) and any nonzero column vector (with   components) there is a basis such that   is represented with respect to that basis by that column vector. The first and second statements follow easily from the third. The first follows because the third statement gives a basis   such that   and then   is the desired  . The second follows from the third because it is just a doubled application of it. A way to prove the third is as in the answer to the first part of this question. Here is a sketch. Represent   with respect to any basis   with a column vector  . This column vector must have a nonzero component because   is a nonzero vector. Use that component in a sequence of row operations to convert   to  . (This sketch could be filled out as an induction argument on the dimension of  .)
Problem 14

Let   be vector spaces, and let   be bases for   and   be bases for  . Where   is linear, find a formula relating   to  .

Answer

This is the topic of the next subsection.

This exercise is recommended for all readers.
Problem 15

Show that the columns of an   change of basis matrix form a basis for  . Do all bases appear in that way: can the vectors from any   basis make the columns of a change of basis matrix?

Answer

A change of basis matrix is nonsingular and thus has rank equal to the number of its columns. Therefore its set of columns is a linearly independent subset of size   in   and it is thus a basis. The answer to the second half is also "yes"; all implications in the prior sentence reverse (that is, all of the "if ... then ..." parts of the prior sentence convert to "if and only if" parts).

This exercise is recommended for all readers.
Problem 16

Find a matrix having this effect.

 

That is, find a   that left-multiplies the starting vector to yield the ending vector. Is there a matrix having these two effects?

  1.  
  2.  

Give a necessary and sufficient condition for there to be a matrix such that   and  .

Answer

In response to the first half of the question, there are infinitely many such matrices. One of them represents with respect to   the transformation of   with this action.

 

The problem of specifying two distinct input/output pairs is a bit trickier. The fact that matrices have a linear action precludes some possibilities.

  1. Yes, there is such a matrix. These conditions
     
    can be solved
     
    to give this matrix.
     
  2. No, because
     
    no linear action can produce this effect.
  3. A sufficient condition is that   be linearly independent, but that's not a necessary condition. A necessary and sufficient condition is that any linear dependences among the starting vectors appear also among the ending vectors. That is,
     
    The proof of this condition is routine.