Click the arrow to see answer.
for any integer. In the case of so . Thus the eigenvalues of are just where is an eigenvalue of .
Additional Exercises for Chapter 8: Eigenvalues and Eigenvectors
Review Exercises
If is an invertible matrix, compare the eigenvalues of and . More generally, for an
arbitrary integer, compare the eigenvalues of and .
If is an matrix and is a nonzero constant, compare the eigenvalues of and
.
Click the arrow to see answer.
Say Then and so the eigenvalues of are just where is an eigenvalue of .
Let be invertible matrices which commute. That is, . Suppose is an eigenvector of .
Show that then must also be an eigenvector for .
Click the arrow to see answer.
. Here it is assumed that .
Suppose is an matrix and it satisfies for some a positive integer larger than 1.
Show that if is an eigenvalue of then equals either 0 or .
Click the arrow to see answer.
Let be the eigenvector. Then and so Hence if then and so
Show that if and , then whenever are scalars,
Does this imply that is an eigenvector? Explain.
Click the arrow to see answer.
The formula follows from properties of matrix multiplications. However, this vector might not be an eigenvector because it might equal and eigenvectors cannot equal .
Is it possible for a nonzero matrix to have only as an eigenvalue?
Click the arrow to see answer.
Yes. works.
If is the matrix of a linear transformation which rotates all vectors in through
explain why cannot have any real eigenvalues. Is there an angle such that rotation
through this angle would have a real eigenvalue? What eigenvalues would be
obtainable in this way?
Let be the matrix of the linear transformation which rotates all vectors in through
an angle of . For which values of does have a real eigenvalue?
Click the arrow to see answer.
When you think of this geometrically, it is clear that the only two values of are 0 and or these added to integer multiples of
Let be the linear transformation which reflects vectors about the axis. Find a
matrix for and then find its eigenvalues and eigenvectors.
Click the arrow to see answer.
The matrix of is . The eigenvectors and eigenvalues are:
Let be the linear transformation which rotates all vectors in counterclockwise
through an angle of Find a matrix of and then find eigenvalues and eigenvectors.
Click the arrow to see answer.
The matrix of is . The eigenvectors and eigenvalues are:
Let be the linear transformation which reflects all vectors in through the plane.
Find a matrix for and then obtain its eigenvalues and eigenvectors.
Click the arrow to see answer.
The matrix of is The eigenvectors and eigenvalues are:
Find the eigenvalues and eigenvectors of the matrix
One eigenvalue is Diagonalize if possible.
Click the arrow to see answer.
The eigenvalues are . The eigenvectors corresponding to the eigenvalues are: Therefore this matrix is not diagonalizable.
Find the eigenvalues and eigenvectors of the matrix
One eigenvalue is Diagonalize if possible.
Click the arrow to see answer.
The eigenvectors and eigenvalues are: The matrix needed to diagonalize the above matrix is and the diagonal matrix is
Find the eigenvalues and eigenvectors of the matrix
One eigenvalue is Diagonalize if possible.
Click the arrow to see answer.
The eigenvectors and eigenvalues are: The matrix needed to diagonalize the above matrix is and the diagonal matrix is
Find the eigenvalues and eigenvectors of the matrix
One eigenvalue is Diagonalize if possible.
Click the arrow to see answer.
This one has some complex eigenvalues.
Challenge Exercises
Suppose is an matrix consisting entirely of real entries but is a complex eigenvalue
having the eigenvector, Here and are real vectors. Show that then is also an
eigenvalue with the eigenvector, .
Given a polynomial and a square matrix , the matrix is called the evaluation of at
. Let . Show that for all polynomials .
If is diagonalizable and is a polynomial such that for all eigenvalues of , show that
(here, the final O is the zero matrix the same size as ).
The characteristic polynomial of (see Definition def:char_poly_complex) certainly satisfies the requirement
that for all eigenvalues of . In solving this problem you have proved a special case of
the Cayley-Hamilton theorem, see Theorem th:Cayley_Hamilton. In fact, if is the characteristic
polynomial of , then whether or not is diagonalizable.
Assume the matrix is similar to an upper triangular matrix. If , show that is equal
to the zero matrix.
Show that is similar to for all matrices .
Let . If treat the cases and separately.
If , reduce to the case using Exercise prob:5_5_12 prob:5_5_12d.
Suppose is an matrix and let be an eigenvector such that . Also suppose the characteristic
polynomial of is
Explain why
Use this to prove that the Cayley-Hamilton theorem holds for any diagonalizable
matrix . (The Cayley-Hamilton theorem says that satisfies its characteristic
equation, i.e.,
. (For a proof of the general case, see Theorem th:Cayley_Hamilton)
Suppose the characteristic polynomial of an matrix is . Find where is an
integer.
Click the arrow to see answer.
The eigenvalues are distinct because they are the roots of . Hence if is a given vector with then so .
Octave Exercises
Use Octave to check your work on Problems prb:8.9 to prb:8.12. The first steps of prb:8.9 are in the code
below. See if you can finish the rest of the problem.
To use Octave, go to the Sage Math Cell Webpage, copy the code below into the cell, select OCTAVE as the language, and press EVALUATE.
% Eigenvalues and eigenvectors
A=[-6 -92 12; 0 0 0; -2 -31 4];
poly(A)
% We should interpret our result as
% z^3+2z^2+0+0 = z^2(z+2), we interpret the super-small number as a zero.
% It was caused by rounding errors in the poly() algorithm
% Then we compute a basis for each eigenspace.
%To check our work, we can write:
[P,D]=eig(A)
% If we try to diagonalize P, we are warned that P is singular.
inv(P)*A*P
% This is because columns 2 and 3 of P are the same.
% The eigenvalue 0 has algebraic multiplicity 2, but geometric multiplicity 1
% This matrix A is NOT diagonalizable
Use Octave to check your work on Problems prb:8.20 to prb:8.31. The first steps of prb:8.31 are in the code
below. See if you can finish the rest of the problem.
To use Octave, go to the Sage Math Cell Webpage, copy the code below into the cell, select OCTAVE as the language, and press EVALUATE.
% Eigenvalues and eigenvectors
A=[14 -12 5; -6 2 -1; -69 51 -21];
poly(A)
% We can use synthetic division to determine
% z^3+5z^2+16z+30 = (z+3)(z^2+2z+10), and use the
% quadratic formula to determine the two complex eigenvalues.
% Then we compute a basis for each eigenspace.
%To check our work, we can write:
[P,D]=eig(A)
%Notice that complex numbers are written as ordered pairs in Octave.
% Now to diagonalize A, we observe
inv(P)*A*P
% This is D, if we interpret the super-small numbers as zeros.
% They were caused by rounding errors in the eig() algorithm
Bibliography
These problems come from Chapter 7 of Ken Kuttler’s A First Course in Linear Algebra. (CC-BY)
Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, pp. 359–401.