## SolutionsEdit

- Problem 1

For each, find the characteristic polynomial and the eigenvalues.

- Answer

- This
- ; ,
- ; ,
- ;
- ;

*This exercise is recommended for all readers.*

- Problem 2

For each matrix, find the characteristic equation, and the eigenvalues and associated eigenvectors.

- Answer

- The characteristic equation is . Its roots, the eigenvalues, are and . For the eigenvectors we consider this equation.
- The characteristic equation is

- Problem 3

Find the characteristic equation, and the eigenvalues and associated eigenvectors for this matrix. *Hint.* The eigenvalues are complex.

- Answer

The characteristic equation

has the complex roots and . This system

For Gauss' method gives this reduction.

(For the calculation in the lower right get a common denominator

to see that it gives a equation.) These are the resulting eigenspace and eigenvector.

For the system

leads to this.

- Problem 4

Find the characteristic polynomial, the eigenvalues, and the associated eigenvectors of this matrix.

- Answer

The characteristic equation is

and so the eigenvalues are (this is a repeated root of the equation) and . For the rest, consider this system.

When then the solution set is this eigenspace.

When then the solution set is this eigenspace.

So these are eigenvectors associated with and .

*This exercise is recommended for all readers.*

- Problem 5

For each matrix, find the characteristic equation, and the eigenvalues and associated eigenvectors.

- Answer

- The characteristic equation is
- The characteristic equation is

*This exercise is recommended for all readers.*

- Problem 6

Let be

Find its eigenvalues and the associated eigenvectors.

- Answer

With respect to the natural basis the matrix representation is this.

Thus the characteristic equation

is . To find the associated eigenvectors, consider this system.

Plugging in gives

- Problem 7

Find the eigenvalues and eigenvectors of this map .

- Answer

, ,

*This exercise is recommended for all readers.*

- Problem 8

Find the eigenvalues and associated eigenvectors of the differentiation operator .

- Answer

Fix the natural basis . The map's action is , , , and and its representation is easy to compute.

We find the eigenvalues with this computation.

Thus the map has the single eigenvalue . To find the associated eigenvectors, we solve

to get this eigenspace.

- Problem 9
- Prove that

the eigenvalues of a triangular matrix (upper or lower triangular) are the entries on the diagonal.

- Answer

The determinant of the triangular matrix is the product down the diagonal, and so it factors into the product of the terms .

*This exercise is recommended for all readers.*

- Problem 10

Find the formula for the characteristic polynomial of a matrix.

- Answer

Just expand the determinant of .

- Problem 11

Prove that the characteristic polynomial of a transformation is well-defined.

- Answer

Any two representations of that transformation are similar, and similar matrices have the same characteristic polynomial.

*This exercise is recommended for all readers.*

- Problem 12

- Can any non- vector in any nontrivial vector space be a eigenvector? That is, given a from a nontrivial , is there a transformation and a scalar such that ?
- Given a scalar , can any non- vector in any nontrivial vector space be an eigenvector associated with the eigenvalue ?

- Answer

- Yes, use and the identity map.
- Yes, use the transformation that multiplies by .

*This exercise is recommended for all readers.*

- Problem 13

Suppose that and . Prove that the eigenvectors of associated with are the non- vectors in the kernel of the map represented (with respect to the same bases) by .

- Answer

If then under the map .

- Problem 14

Prove that if are all integers and then

has integral eigenvalues, namely and .

- Answer

The characteristic equation

simplifies to . Checking that the values and satisfy the equation (under the condition) is routine.

*This exercise is recommended for all readers.*

- Problem 15

Prove that if is nonsingular and has eigenvalues then has eigenvalues . Is the converse true?

- Answer

Consider an eigenspace . Any is the image of some (namely, ). Thus, on (which is a nontrivial subspace) the action of is , and so is an eigenvalue of .

*This exercise is recommended for all readers.*

- Problem 16

Suppose that is and are scalars.

- Prove that if has the eigenvalue with an associated eigenvector then is an eigenvector of associated with eigenvalue .
- Prove that if is diagonalizable then so is .

- Answer

- We have .
- Suppose that is diagonal. Then is also diagonal.

*This exercise is recommended for all readers.*

- Problem 17

Show that is an eigenvalue of if and only if the map represented by is not an isomorphism.

- Answer

The scalar is an eigenvalue if and only if the transformation is singular. A transformation is singular if and only if it is not an isomorphism (that is, a transformation is an isomorphism if and only if it is nonsingular).

- Problem 18

- Show that if is an eigenvalue of then is an eigenvalue of .
- What is wrong with this proof generalizing that? "If is an eigenvalue of and is an eigenvalue for , then is an eigenvalue for , for, if and then "?

- Answer

- Where the eigenvalue is associated with the eigenvector then . (The full details can be put in by doing induction on .)
- The eigenvector associated wih might not be an eigenvector associated with .

- Problem 19

Do matrix-equivalent matrices have the same eigenvalues?

- Answer

No. These are two same-sized, equal rank, matrices with different eigenvalues.

- Problem 20

Show that a square matrix with real entries and an odd number of rows has at least one real eigenvalue.

- Answer

The characteristic polynomial has an odd power and so has at least one real root.

- Problem 21

Diagonalize.

- Answer

The characteristic polynomial has distinct roots , , and . Thus the matrix can be diagonalized into this form.

- Problem 22

Suppose that is a nonsingular matrix. Show that the **similarity transformation** map sending is an isomorphism.

- Answer

We must show that it is one-to-one and onto, and that it respects the operations of matrix addition and scalar multiplication.

To show that it is one-to-one, suppose that , that is, suppose that , and note that multiplying both sides on the left by and on the right by gives that . To show that it is onto, consider and observe that .

The map preserves matrix addition since follows from properties of matrix multiplication and addition that we have seen. Scalar multiplication is similar: .

- ? Problem 23

Show that if is an square matrix and each row (column) sums to then is a characteristic root of . (Morrison 1967)

- Answer

*This is how the answer was given in the cited source.*

If the argument of the characteristic function of is set equal to , adding the first rows (columns) to the th row (column) yields a determinant whose th row (column) is zero. Thus is a characteristic root of .

## ReferencesEdit

- Morrison, Clarence C. (proposer) (1967), "Quickie",
*Mathematics Magazine***40**(4): 232. - Strang, Gilbert (1980),
*Linear Algebra and its Applications*(Second ed.), Harcourt Brace Jovanovich.