Linear Algebra/Eigenvalues and Eigenvectors/Solutions

< Linear Algebra‎ | Eigenvalues and Eigenvectors

SolutionsEdit

Problem 1

For each, find the characteristic polynomial and the eigenvalues.

  1.  
  2.  
  3.  
  4.  
  5.  
Answer
  1. This
     
    simplifies to the characteristic equation  . Because the equation factors into   there is only one eigenvalue  .
  2.  ;  ,  
  3.  ;  ,  
  4.  ;  
  5.  ;  
This exercise is recommended for all readers.
Problem 2

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

  1.  
  2.  
Answer
  1. The characteristic equation is  . Its roots, the eigenvalues, are   and  . For the eigenvectors we consider this equation.
     
    For the eigenvector associated with  , we consider the resulting linear system.
     
    The eigenspace is the set of vectors whose second component is twice the first component.
     
    (Here, the parameter is   only because that is the variable that is free in the above system.) Hence, this is an eigenvector associated with the eigenvalue  .
     
    Finding an eigenvector associated with   is similar. This system
     
    leads to the set of vectors whose first component is zero.
     
    And so this is an eigenvector associated with  .
     
  2. The characteristic equation is
     
    and so the eigenvalues are   and  . To find eigenvectors, consider this system.
     
    For   we get
     
    leading to this eigenspace and eigenvector.
     
    For   the system is
     
    leading to this.
     
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.

  1.  
  2.  
Answer
  1. The characteristic equation is
     
    and so the eigenvalues are   and also the repeated eigenvalue  . To find eigenvectors, consider this system.
     
    For   we get
     
    leading to this eigenspace and eigenvector.
     
    For   the system is
     
    leading to this.
     
  2. The characteristic equation is
     
    and the eigenvalues are   and (by using the quadratic equation)   and  . To find eigenvectors, consider this system.
     
    Substituting   gives the system
     
    leading to this eigenspace and eigenvector.
     
    Substituting   gives the system
     
     
    (the middle coefficient in the third equation equals the number  ; find a common denominator of   and then rationalize the denominator by multiplying the top and bottom of the frsction by  )
     
    which leads to this eigenspace and eigenvector.
     
    Finally, substituting   gives the system
     
     
    which gives this eigenspace and eigenvector.
     
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
  1. 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  ?
  2. Given a scalar  , can any non-  vector in any nontrivial vector space be an eigenvector associated with the eigenvalue  ?
Answer
  1. Yes, use   and the identity map.
  2. 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.

  1. Prove that if   has the eigenvalue   with an associated eigenvector   then   is an eigenvector of   associated with eigenvalue  .
  2. Prove that if   is diagonalizable then so is  .
Answer
  1. We have  .
  2. 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
  1. Show that if   is an eigenvalue of   then   is an eigenvalue of  .
  2. 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  "?
(Strang 1980)
Answer
  1. Where the eigenvalue   is associated with the eigenvector   then  . (The full details can be put in by doing induction on  .)
  2. 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 .