# Linear Algebra/Jordan Canonical Form/Solutions

## Solutions[edit | edit source]

- 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.

- Answer

- The characteristic polynomial is and the minimal polynomial is the same.
- The characteristic polynomial is . The minimal polynomial is .
- The characteristic polynomial is and the minimal polynomial is the same.
- The characteristic polynomial is The minimal polynomial is the same.
- The characteristic polynomial is . The minimal polynomial is .
- The characteristic polynomial is and the minimal polynomial is the same.
- The characteristic polynomial is and the minimal polynomial is the same.
- The characteristic polynomial is and the minimal polynomial is .
- 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.

- 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.
- 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

- 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.
- 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

- Problem 4

Find the change of basis matrices for each example.

- 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.

- Here is the arrow diagram.
The matrix to move from the lower left to the upper left is this.
- We want this matrix and its inverse.
- The concatenation of these bases for the
generalized null spaces will do for the basis for the
entire space.

*This exercise is recommended for all readers.*

- Problem 5

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

- 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.

- 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 .

- 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
- 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
- The characteristic polynomial is
.
For the eigenvalue , calculation of the
powers of yields this.
- The characteristic polynomial of this
matrix is .
This matrix has only a single eigenvalue, .
By finding the powers of we have
- 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
- 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

*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.

- Answer

- The characteristic polynomial is .
For we have
- The characteristic polynomial is
.
For ,

*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.

- Find the formula for the characteristic polynomial of a matrix.
- Show that
trace is invariant under similarity, and so we can sensibly
speak of the "trace of a map". (
*Hint:*see the prior item.) - Is trace invariant under matrix equivalence?
- Show that the trace of a map is the sum of its eigenvalues (counting multiplicities).
- Show that the trace of a nilpotent map is zero. Does the converse hold?

- Answer

- The characteristic polynomial is this.
- 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 .
- No, there are matrices and that are
equivalent (for some nonsingular and )
but that have different traces.
An easy example is this.
- Put the matrix in Jordan form. By the first item, the trace is unchanged.
- The first part is easy; use the third item.
The converse does not hold: this matrix

- 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 .