Linear Algebra/Matrix Multiplication/Solutions

A=(285)

SolutionsEdit

This exercise is recommended for all readers.
Problem 1

Compute, or state "not defined".

  1.  
  2.  
  3.  
  4.  
Answer
  1.  
  2.  
  3. Not defined.
  4.  
This exercise is recommended for all readers.
Problem 2

Where

 

compute or state "not defined".

  1.  
  2.  
  3.  
  4.  
Answer
  1.  
  2.  
  3.  
  4.  
Problem 3

Which products are defined?

  1.   times  
  2.   times  
  3.   times  
  4.   times  
Answer
  1. Yes.
  2. Yes.
  3. No.
  4. No.
This exercise is recommended for all readers.
Problem 4

Give the size of the product or state "not defined".

  1. a   matrix times a   matrix
  2. a   matrix times a   matrix
  3. a   matrix times a   matrix
  4. a   matrix times a   matrix
Answer
  1.  
  2.  
  3. Not defined.
  4.  
This exercise is recommended for all readers.
Problem 5

Find the system of equations resulting from starting with

 

and making this change of variable (i.e., substitution).

 
Answer

We have

 

which, after expanding and regrouping about the  's yields this.

 

The starting system, and the system used for the substitutions, can be expressed in matrix language.

 

With this, the substitution is  .

Problem 6

As Definition 2.3 points out, the matrix product operation generalizes the dot product. Is the dot product of a   row vector and a   column vector the same as their matrix-multiplicative product?

Answer

Technically, no. The dot product operation yields a scalar while the matrix product yields a   matrix. However, we usually will ignore the distinction.

This exercise is recommended for all readers.
Problem 7

Represent the derivative map on   with respect to   where   is the natural basis  . Show that the product of this matrix with itself is defined; what the map does it represent?

Answer

The action of   on   is  ,  ,  , ... and so this is its   matrix representation.

 

The product of this matrix with itself is defined because the matrix is square.

 


The map so represented is the composition

 

which is the second derivative operation.

Problem 8

Show that composition of linear transformations on   is commutative. Is this true for any one-dimensional space?

Answer

It is true for all one-dimensional spaces. Let   and   be transformations of a one-dimensional space. We must show that   for all vectors. Fix a basis   for the space and then the transformations are represented by   matrices.

 

Therefore, the compositions can be represented as   and  .

 

These two matrices are equal and so the compositions have the same effect on each vector in the space.

Problem 9

Why is matrix multiplication not defined as entry-wise multiplication? That would be easier, and commutative too.

Answer

It would not represent linear map composition; Theorem 2.6 would fail.

This exercise is recommended for all readers.
Problem 10
  1. Prove that   and   for positive integers  .
  2. Prove that   for any positive integer   and scalar  .
Answer

Each follows easily from the associated map fact. For instance,   applications of the transformation  , following   applications, is simply   applications.

This exercise is recommended for all readers.
Problem 11
  1. How does matrix multiplication interact with scalar multiplication: is  ? Is  ?
  2. How does matrix multiplication interact with linear combinations: is  ? Is  ?
Answer

Although these can be done by going through the indices, they are best understood in terms of the represented maps. That is, fix spaces and bases so that the matrices represent linear maps  .

  1. Yes; we have both   and   (the second equality holds because of the linearity of  ).
  2. Both answers are yes. First,   and   both send   to  ; the calculation is as in the prior item (using the linearity of   for the first one). For the other,   and   both send   to  .
Problem 12

We can ask how the matrix product operation interacts with the transpose operation.

  1. Show that  .
  2. A square matrix is symmetric if each   entry equals the   entry, that is, if the matrix equals its own transpose. Show that the matrices   and   are symmetric.
Answer

We have not seen a map interpretation of the transpose operation, so we will verify these by considering the entries.

  1. The   entry of   is the   entry of  , which is the dot product of the  -th row of   and the  -th column of  . The   entry of   is the dot product of the  -th row of   and the  -th column of  , which is the dot product of the  -th column of   and the  -th row of  . Dot product is commutative and so these two are equal.
  2. By the prior item each equals its transpose, e.g.,  .
This exercise is recommended for all readers.
Problem 13

Rotation of vectors in   about an axis is a linear map. Show that linear maps do not commute by showing geometrically that rotations do not commute.

Answer

Consider   rotating all vectors   radians counterclockwise about the   and   axes (counterclockwise in the sense that a person whose head is at   or   and whose feet are at the origin sees, when looking toward the origin, the rotation as counterclockwise).

   

Rotating   first and then   is different than rotating   first and then  . In particular,   so  , while   so  , and hence the maps do not commute.

Problem 14

In the proof of Theorem 2.12 some maps are used. What are the domains and codomains?

Answer

It doesn't matter (as long as the spaces have the appropriate dimensions).

For associativity, suppose that   is  , that   is  , and that   is  . We can take any   dimensional space, any   dimensional space, any   dimensional space, and any   dimensional space— for instance,  ,  ,  , and   will do. We can take any bases  ,  ,  , and  , for those spaces. Then, with respect to   the matrix   represents a linear map  , with respect to   the matrix   represents a  , and with respect to   the matrix   represents an  . We can use those maps in the proof.

The second half is done similarly, except that   and   are added and so we must take them to represent maps with the same domain and codomain.

Problem 15

How does matrix rank interact with matrix multiplication?

  1. Can the product of rank   matrices have rank less than  ? Greater?
  2. Show that the rank of the product of two matrices is less than or equal to the minimum of the rank of each factor.
Answer
  1. The product of rank   matrices can have rank less than or equal to   but not greater than  . To see that the rank can fall, consider the maps   projecting onto the axes. Each is rank one but their composition  , which is the zero map, is rank zero. That can be translated over to matrices representing those maps in this way.
     
    To prove that the product of rank   matrices cannot have rank greater than  , we can apply the map result that the image of a linearly dependent set is linearly dependent. That is, if   and   both have rank   then a set in the range   of size larger than   is the image under   of a set in   of size larger than   and so is linearly dependent (since the rank of   is  ). Now, the image of a linearly dependent set is dependent, so any set of size larger than   in the range is dependent. (By the way, observe that the rank of   was not mentioned. See the next part.)
  2. Fix spaces and bases and consider the associated linear maps   and  . Recall that the dimension of the image of a map (the map's rank) is less than or equal to the dimension of the domain, and consider the arrow diagram.
     
    First, the image of   must have dimension less than or equal to the dimension of  , by the prior sentence. On the other hand,   is a subset of the domain of  , and thus its image has dimension less than or equal the dimension of the domain of  . Combining those two, the rank of a composition is less than or equal to the minimum of the two ranks. The matrix fact follows immediately.
Problem 16

Is "commutes with" an equivalence relation among   matrices?

Answer

The "commutes with" relation is reflexive and symmetric. However, it is not transitive: for instance, with

 

  commutes with   and   commutes with  , but   does not commute with  .

This exercise is recommended for all readers.
Problem 17

(This will be used in the Matrix Inverses exercises.) Here is another property of matrix multiplication that might be puzzling at first sight.

  1. Prove that the composition of the projections   onto the   and   axes is the zero map despite that neither one is itself the zero map.
  2. Prove that the composition of the derivatives   is the zero map despite that neither is the zero map.
  3. Give a matrix equation representing the first fact.
  4. Give a matrix equation representing the second.

When two things multiply to give zero despite that neither is zero, each is said to be a zero divisor.

Answer
  1. Either of these.
     
  2. The composition is the fifth derivative map   on the space of fourth-degree polynomials.
  3. With respect to the natural bases,
     
    and their product (in either order) is the zero matrix.
  4. Where  ,
     
    and their product (in either order) is the zero matrix.
Problem 18

Show that, for square matrices,   need not equal  .

Answer

Note that  , so a reasonable try is to look at matrices that do not commute so that   and   don't cancel: with

 

we have the desired inequality.

 
This exercise is recommended for all readers.
Problem 19

Represent the identity transformation   with respect to   for any basis  . This is the identity matrix  . Show that this matrix plays the role in matrix multiplication that the number   plays in real number multiplication:   (for all matrices   for which the product is defined).

Answer

Because the identity map acts on the basis   as  , ...,  , the representation is this.

 

The second part of the question is obvious from Theorem 2.6.

Problem 20

In real number algebra, quadratic equations have at most two solutions. That is not so with matrix algebra. Show that the   matrix equation   has more than two solutions, where   is the identity matrix (this matrix has ones in its   and   entries and zeroes elsewhere; see Problem 19).

Answer

Here are four solutions.

 
Problem 21
  1. Prove that for any   matrix   there are scalars   that are not all   such that the combination   is the zero matrix (where   is the   identity matrix, with  's in its   and   entries and zeroes elsewhere; see Problem 19).
  2. Let   be a polynomial  . If   is a square matrix we define   to be the matrix   (where   is the appropriately-sized identity matrix). Prove that for any square matrix there is a polynomial such that   is the zero matrix.
  3. The minimal polynomial   of a square matrix is the polynomial of least degree, and with leading coefficient  , such that   is the zero matrix. Find the minimal polynomial of this matrix.
     
    (This is the representation with respect to  , the standard basis, of a rotation through   radians counterclockwise.)
Answer
  1. The vector space   has dimension four. The set   has five elements and thus is linearly dependent.
  2. Where   is  , generalizing the argument from the prior item shows that there is such a polynomial of degree   or less, since   is a  -member subset of the  -dimensional space  .
  3. First compute the powers
     
    (observe that rotating by   three times results in a rotation by  , which is indeed what   represents). Then set   equal to the zero matrix
     
     
    to get this linear system.
     
    Apply Gaussian reduction.
     
    Setting  ,  , and   to zero makes   and   also come out to be zero so no degree one or degree zero polynomial will do. Setting   and   to zero (and   to one) gives a linear system
     
    that can be solved with   and  . Conclusion: the polynomial   is minimal for the matrix  .
Problem 22

The infinite-dimensional space   of all finite-degree polynomials gives a memorable example of the non-commutativity of linear maps. Let   be the usual derivative and let   be the shift map.

 

Show that the two maps don't commute  ; in fact, not only is   not the zero map, it is the identity map.

Answer

The check is routine:

 

while

 

so that under the map   we have  .

Problem 23

Recall the notation for the sum of the sequence of numbers  .

 

In this notation, the   entry of the product of   and   is this.

 

Using this notation,

  1. reprove that matrix multiplication is associative;
  2. reprove Theorem 2.6.
Answer
  1. Tracing through the remark at the end of the subsection gives that the   entry of   is this  
     
    (the first equality comes from using the distributive law to multiply through the  's, the second equality is the associative law for real numbers, the third is the commutative law for reals, and the fourth equality follows on using the distributive law to factor the  's out), which is the   entry of  .
  2. The  -th component of   is
     
    and so the  -th component of   is this
     
     
    (the first equality holds by using the distributive law to multiply the  's through, the second equality represents the use of associativity of reals, the third follows by commutativity of reals, and the fourth comes from using the distributive law to factor the  's out).