Linear Algebra/Jordan Canonical Form/Solutions

Solutions edit

Problem 1

Do the check for Example 2.3.

Answer

We are required to check that

 

That calculation is easy.

Problem 2

Each matrix is in Jordan form. State its characteristic polynomial and its minimal polynomial.

  1.  
  2.  
  3.  
  4.  
  5.  
  6.  
  7.  
  8.  
  9.  
Answer
  1. The characteristic polynomial is   and the minimal polynomial is the same.
  2. The characteristic polynomial is  . The minimal polynomial is  .
  3. The characteristic polynomial is   and the minimal polynomial is the same.
  4. The characteristic polynomial is   The minimal polynomial is the same.
  5. The characteristic polynomial is  . The minimal polynomial is  .
  6. The characteristic polynomial is   and the minimal polynomial is the same.
  7. The characteristic polynomial is   and the minimal polynomial is the same.
  8. The characteristic polynomial is   and the minimal polynomial is  .
  9. The characteristic polynomial is   and the minimal polynomial is the same.
This exercise is recommended for all readers.
Problem 3

Find the Jordan form from the given data.

  1. The matrix   is   with the single eigenvalue  . The nullities of the powers are:   has nullity two,   has nullity three,   has nullity four, and   has nullity five.
  2. The matrix   is   with two eigenvalues. For the eigenvalue   the nullities are:   has nullity two, and   has nullity four. For the eigenvalue   the nullities are:   has nullity one.
Answer
  1. The transformation   is nilpotent (that is,   is the entire space) and it acts on a string basis via two strings,   and  . Consequently,   can be represented in this canonical form.
     
    and therefore   is similar to this this canonical form matrix.
     
  2. The restriction of the transformation   is nilpotent on the subspace  , and the action on a string basis is given as  . The restriction of the transformation   is nilpotent on the subspace  , having the action on a string basis of   and  . Consequently the Jordan form is this
     
    (note that the blocks are arranged with the least eigenvalue first).
Problem 4

Find the change of basis matrices for each example.

  1. Example 2.13
  2. Example 2.14
  3. Example 2.15
Answer

For each, because many choices of basis are possible, many other answers are possible. Of course, the calculation to check if an answer gives that   is in Jordan form is the arbiter of what's correct.

  1. Here is the arrow diagram.
     
    The matrix to move from the lower left to the upper left is this.
     
    The matrix   to move from the upper right to the lower right is the inverse of  .
  2. We want this matrix and its inverse.
     
  3. The concatenation of these bases for the generalized null spaces will do for the basis for the entire space.
     
    The change of basis matrices are this one and its inverse.
     
This exercise is recommended for all readers.
Problem 5

Find the Jordan form and a Jordan basis for each matrix.

  1.  
  2.  
  3.  
  4.  
  5.  
  6.  
  7.  
Answer

The general procedure is to factor the characteristic polynomial   to get the eigenvalues  ,  , etc. Then, for each   we find a string basis for the action of the transformation   when restricted to  , by computing the powers of the matrix   and finding the associated null spaces, until these null spaces settle down (do not change), at which point we have the generalized null space. The dimensions of those null spaces (the nullities) tell us the action of   on a string basis for the generalized null space, and so we can write the pattern of subdiagonal ones to have  . From this matrix, the Jordan block   associated with   is immediate  . Finally, after we have done this for each eigenvalue, we put them together into the canonical form.

  1. The characteristic polynomial of this matrix is  , so it has only the single eigenvalue  .

     

    (Thus, this transformation is nilpotent:   is the entire space). From the nullities we know that  's action on a string basis is  . This is the canonical form matrix for the action of   on  

     

    and this is the Jordan form of the matrix.

     

    Note that if a matrix is nilpotent then its canonical form equals its Jordan form.

    We can find such a string basis using the techniques of the prior section.

     

    The first basis vector has been taken so that it is in the null space of   but is not in the null space of  . The second basis vector is the image of the first under  .

  2. The characteristic polynomial of this matrix is  , so it is a single-eigenvalue matrix. (That is, the generalized null space of   is the entire space.) We have
     
    and so the action of   on an associated string basis is  . Thus,
     
    the Jordan form of T is
     
    and choosing vectors from the above null spaces gives this string basis (many other choices are possible).
     
  3. The characteristic polynomial   has two roots and they are the eigenvalues   and  . We handle the two eigenvalues separately. For  , the calculation of the powers of   yields
     
    and the null space of   is the same. Thus this set is the generalized null space  . The nullities show that the action of the restriction of   to the generalized null space on a string basis is  . A similar calculation for   gives these null spaces.
     
    (The null space of   is the same, as it must be because the power of the term associated with   in the characteristic polynomial is two, and so the restriction of   to the generalized null space   is nilpotent of index at most two— it takes at most two applications of   for the null space to settle down.) The pattern of how the nullities rise tells us that the action of   on an associated string basis for   is  . Putting the information for the two eigenvalues together gives the Jordan form of the transformation  .
     
    We can take elements of the nullspaces to get an appropriate basis.
     
  4. The characteristic polynomial is  . For the eigenvalue  , calculation of the powers of   yields this.
     
    The null space of   is the same, and so this is the generalized null space  . Thus the action of the restriction of   to   on an associated string basis is  . For  , computing the powers of   yields
     
    and so the action of   on a string basis for   is  . Therefore the Jordan form is
     
    and a suitable basis is this.
     
  5. The characteristic polynomial of this matrix is  . This matrix has only a single eigenvalue,  . By finding the powers of   we have
     
    and so the action of   on an associated string basis is  . The Jordan form is this
     
    and one choice of basis is this.
     
  6. The characteristic polynomial   has only a single root, so the matrix has only a single eigenvalue  . Finding the powers of   and calculating the null spaces
     
    shows that the action of the nilpotent map   on a string basis is   and  . Therefore the Jordan form is
     
    and an appropriate basis (a string basis associated with  ) is this.
     
  7. The characteristic polynomial is a bit large for by-hand calculation, but just manageable  . This is a single-eigenvalue map, so the transformation   is nilpotent. The null spaces
     
    and the nullities show that the action of   on a string basis is   and  . The Jordan form is
     
    and finding a suitable string basis is routine.
     
This exercise is recommended for all readers.
Problem 6

Find all possible Jordan forms of a transformation with characteristic polynomial  .

Answer

There are two eigenvalues,   and  . The restriction of   to   could have either of these actions on an associated string basis.

 

The restriction of   to   could have either of these actions on an associated string basis.

 

In combination, that makes four possible Jordan forms, the two first actions, the second and first, the first and second, and the two second actions.

 
Problem 7

Find all possible Jordan forms of a transformation with characteristic polynomial  .

Answer

The restriction of   to   can have only the action  . The restriction of   to   could have any of these three actions on an associated string basis.

 

Taken together there are three possible Jordan forms, the one arising from the first action by   (along with the only action from  ), the one arising from the second action, and the one arising from the third action.

 
This exercise is recommended for all readers.
Problem 8

Find all possible Jordan forms of a transformation with characteristic polynomial   and minimal polynomial  .

Answer

The action of   on a string basis for   must be  . Because of the power of   in the minimal polynomial, a string basis for   has length two and so the action of   on   must be of this form.

 

Therefore there is only one Jordan form that is possible.

 
Problem 9

Find all possible Jordan forms of a transformation with characteristic polynomial   and minimal polynomial  .

Answer

There are two possible Jordan forms. The action of   on a string basis for   must be  . There are two actions for   on a string basis for   that are possible with this characteristic polynomial and minimal polynomial.

 

The resulting Jordan form matrics are these.

 
This exercise is recommended for all readers.
Problem 10
Diagonalize these.
  1.  
  2.  
Answer
  1. The characteristic polynomial is  . For   we have
     
    (of course, the null space of   is the same). For  ,
     
    (and the null space of   is the same). We can take this basis
     
    to get the diagonalization.
     
  2. The characteristic polynomial is  . For  ,
     
    and the null space of   is the same. For  
     
    and the null space of   is the same. We can take this basis
     
    to get a diagonalization.
     
This exercise is recommended for all readers.
Problem 11

Find the Jordan matrix representing the differentiation operator on  .

Answer

The transformation   is nilpotent. Its action on   is  . Its Jordan form is its canonical form as a nilpotent matrix.

 
This exercise is recommended for all readers.
Problem 12

Decide if these two are similar.

 
Answer

Yes. Each has the characteristic polynomial  . Calculations of the powers of   and   gives these two.

 

(Of course, for each the null space of the square is the entire space.) The way that the nullities rise shows that each is similar to this Jordan form matrix

 

and they are therefore similar to each other.

Problem 13

Find the Jordan form of this matrix.

 

Also give a Jordan basis.

Answer

Its characteristic polynomial is   which has complex roots  . Because the roots are distinct, the matrix is diagonalizable and its Jordan form is that diagonal matrix.

 

To find an associated basis we compute the null spaces.

 

For instance,

 

and so we get a description of the null space of   by solving this linear system.

 

(To change the relation   so that the leading variable   is expressed in terms of the free variable  , we can multiply both sides by  .)

As a result, one such basis is this.

 
Problem 14

How many similarity classes are there for   matrices whose only eigenvalues are   and  ?

Answer

We can count the possible classes by counting the possible canonical representatives, that is, the possible Jordan form matrices. The characteristic polynomial must be either   or  . In the   case there are two possible actions of   on a string basis for  .

 

There are two associated Jordan form matrices.

 

Similarly there are two Jordan form matrices that could arise out of  .

 

So in total there are four possible Jordan forms.

This exercise is recommended for all readers.
Problem 15

Prove that a matrix is diagonalizable if and only if its minimal polynomial has only linear factors.

Answer

Jordan form is unique. A diagonal matrix is in Jordan form. Thus the Jordan form of a diagonalizable matrix is its diagonalization. If the minimal polynomial has factors to some power higher than one then the Jordan form has subdiagonal  's, and so is not diagonal.

Problem 16

Give an example of a linear transformation on a vector space that has no non-trivial invariant subspaces.

Answer

One example is the transformation of   that sends   to  .

Problem 17

Show that a subspace is   invariant if and only if it is   invariant.

Answer

Apply Lemma 2.7 twice; the subspace is   invariant if and only if it is   invariant, which in turn holds if and only if it is   invariant.

Problem 18

Prove or disprove: two   matrices are similar if and only if they have the same characteristic and minimal polynomials.

Answer

False; these two   matrices each have   and  .

 
Problem 19

The trace of a square matrix is the sum of its diagonal entries.

  1. Find the formula for the characteristic polynomial of a   matrix.
  2. Show that trace is invariant under similarity, and so we can sensibly speak of the "trace of a map". (Hint: see the prior item.)
  3. Is trace invariant under matrix equivalence?
  4. Show that the trace of a map is the sum of its eigenvalues (counting multiplicities).
  5. Show that the trace of a nilpotent map is zero. Does the converse hold?
Answer
  1. The characteristic polynomial is this.
     
    Note that the determinant appears as the constant term.
  2. Recall that the characteristic polynomial   is invariant under similarity. Use the permutation expansion formula to show that the trace is the negative of the coefficient of  .
  3. No, there are matrices   and   that are equivalent   (for some nonsingular   and  ) but that have different traces. An easy example is this.
     
    Even easier examples using   matrices are possible.
  4. Put the matrix in Jordan form. By the first item, the trace is unchanged.
  5. The first part is easy; use the third item. The converse does not hold: this matrix
     
    has a trace of zero but is not nilpotent.
Problem 20

To use Definition 2.6 to check whether a subspace is   invariant, we seemingly have to check all of the infinitely many vectors in a (nontrivial) subspace to see if they satisfy the condition. Prove that a subspace is   invariant if and only if its subbasis has the property that for all of its elements,   is in the subspace.

Answer

Suppose that   is a basis for a subspace   of some vector space. Implication one way is clear; if   is   invariant then in particular, if   then  . For the other implication, let   and note that   is in   as any subspace is closed under linear combinations.

This exercise is recommended for all readers.
Problem 21

Is   invariance preserved under intersection? Under union? Complementation? Sums of subspaces?

Answer

Yes, the intersection of   invariant subspaces is   invariant. Assume that   and   are   invariant. If   then   by the invariance of   and   by the invariance of  .

Of course, the union of two subspaces need not be a subspace (remember that the  - and  -axes are subspaces of the plane   but the union of the two axes fails to be closed under vector addition, for instance it does not contain  .) However, the union of invariant subsets is an invariant subset; if   then   or   so   or  .

No, the complement of an invariant subspace need not be invariant. Consider the subspace

 

of   under the zero transformation.

Yes, the sum of two invariant subspaces is invariant. The check is easy.

Problem 22

Give a way to order the Jordan blocks if some of the eigenvalues are complex numbers. That is, suggest a reasonable ordering for the complex numbers.

Answer

One such ordering is the dictionary ordering. Order by the real component first, then by the coefficient of  . For instance,   but  .

Problem 23

Let   be the vector space over the reals of degree   polynomials. Show that if   then   is an invariant subspace of   under the differentiation operator. In  , does any of  , ...,   have an invariant complement?

Answer

The first half is easy— the derivative of any real polynomial is a real polynomial of lower degree. The answer to the second half is "no"; any complement of   must include a polynomial of degree  , and the derivative of that polynomial is in  .

Problem 24

In  , the vector space (over the reals) of degree   polynomials,

 

and

 

are the even and the odd polynomials;   is even while   is odd. Show that they are subspaces. Are they complementary? Are they invariant under the differentiation transformation?

Answer

For the first half, show that each is a subspace and then observe that any polynomial can be uniquely written as the sum of even-powered and odd-powered terms (the zero polynomial is both). The answer to the second half is "no":   is even while   is odd.

Problem 25

Lemma 2.8 says that if   and   are invariant complements then   has a representation in the given block form (with respect to the same ending as starting basis, of course). Does the implication reverse?

Answer

Yes. If   has the given block form, take   to be the first   vectors of  , where   is the   upper left submatrix. Take   to be the remaining   vectors in  . Let   and   be the spans of   and  . Clearly   and   are complementary. To see   is invariant (  works the same way), represent any   with respect to  , note the last   components are zeroes, and multiply by the given block matrix. The final   components of the result are zeroes, so that result is again in  .

Problem 26

A matrix   is the square root of another   if  . Show that any nonsingular matrix has a square root.

Answer

Put the matrix in Jordan form. By non-singularity, there are no zero eigenvalues on the diagonal. Ape this example:

 

to construct a square root. Show that it holds up under similarity: if   then  .