From A First Course in Linear Algebra
Version 1.05
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/
The previous section introduced eigenvalues and eigenvectors, and concentrated
on their existence and determination. This section will be more about theorems,
and the various properties eigenvalues and eigenvectors enjoy. Like a good
relay,
we will lead-off with one of our better theorems and save the very best for the
anchor leg.
Theorem EDELI
Eigenvectors with Distinct Eigenvalues are Linearly Independent
Suppose that is an
square matrix and
is a set of eigenvectors
with eigenvalues
such that
whenever . Then
is a linearly
independent set.
Proof If , then the set is linearly independent since eigenvectors are nonzero (Definition EEM), so assume for the remainder that .
We will prove this result by contradiction (Technique CD). Suppose to the contrary that is a linearly dependent set. Define to be an integer such that is linearly independent and is linearly dependent. We have to ask if there is even such an integer? Since eigenvectors are nonzero, the set is linearly independent. Think of adding in vectors to this set, one at a time, , , ,…Since we are assuming that is linearly dependent, eventually this set will convert from being linearly independent to being linearly dependent. In other words, it is the addition of the vector that converts the set from linear independence to linear dependence. So there is such a , and futhermore .
Since is linearly dependent there are scalars, , some non-zero, so that
Then,
This is a relation of linear dependence on the linearly independent set , so the scalars must all be zero. That is, for . However, we have the hypothesis that the eigenvalues are distinct, so for . Thus for .
This reduces the original relation of linear dependence on to the simpler equation . By Theorem SMEZV we conclude that or . Eigenvectors are never the zero vector (Definition EEM), so . So all of the scalars , are zero, contradicting their introduction as the scalars creating a nontrivial relation of linear dependence on the set . With a contradiction in hand, we conclude that must be linearly independent.
There is a simple connection between the eigenvalues of a matrix and whether or not the matrix is nonsingular.
Theorem SMZE
Singular Matrices have Zero Eigenvalues
Suppose is a square
matrix. Then is
singular if and only if
is an eigenvalue of .
Proof We have the following equivalences:
With an equivalence about singular matrices we can update our list of equivalences about nonsingular matrices.
Theorem NME8
Nonsingular Matrix Equivalences, Round 8
Suppose that is a
square matrix of size .
The following are equivalent.
Proof The equivalence of the first and last statements is the contrapositive of Theorem SMZE, so we are able to improve on Theorem NME7.
Certain changes to a matrix change its eigenvalues in a predictable way.
Theorem ESMM
Eigenvalues of a Scalar Multiple of a Matrix
Suppose is a square
matrix and is an
eigenvalue of .
Then is an
eigenvalue of .
Proof Let be one eigenvector of for . Then
So is an eigenvector of for the eigenvalue .
Unfortunately, there are not parallel theorems about the sum or product of arbitrary matrices. But we can prove a similar result for powers of a matrix.
Theorem EOMP
Eigenvalues Of Matrix Powers
Suppose is a
square matrix, is
an eigenvalue of ,
and is an integer.
Then is an
eigenvalue of .
Proof Let be one eigenvector of for . Suppose has size . Then we proceed by induction on (Technique I). First, for ,
so is an eigenvalue of in this special case. If we assume the theorem is true for , then we find
So is an eigenvector of for , and induction tells us the theorem is true for all .
While we cannot prove that the sum of two arbitrary matrices behaves in any reasonable way with regard to eigenvalues, we can work with the sum of dissimilar powers of the same matrix. We have already seen two connections between eigenvalues and polynomials, in the proof of Theorem EMHE and the characteristic polynomial (Definition CP). Our next theorem strengthens this connection.
Theorem EPM
Eigenvalues of the Polynomial of a Matrix
Suppose is a square
matrix and is an
eigenvalue of . Let
be a polynomial in
the variable . Then
is an eigenvalue
of the matrix .
Proof Let be one eigenvector of for , and write . Then
So is an eigenvector of for the eigenvalue .
Example BDE
Building desired eigenvalues
In Example ESMS4 the
symmetric matrix
is shown to have the three eigenvalues . Suppose we wanted a matrix that has the three eigenvalues . We can employ Theorem EPM by finding a polynomial that converts to , to , and to . Such a polynomial is called an interpolating polynomial, and in this example we can use
We will not discuss how to concoct this polynomial, but a text on numerical analysis should provide the details. In our case, simply verify that , and .
Now compute
Theorem EPM tells us that if transforms the eigenvalues in the desired manner, then will have the desired eigenvalues. You can check this by computing the eigenvalues of directly. Furthermore, notice that the multiplicities are the same, and the eigenspaces of and are identical.
Inverses and transposes also behave predictably with regard to their eigenvalues.
Theorem EIM
Eigenvalues of the Inverse of a Matrix
Suppose is a square
nonsingular matrix and
is an eigenvalue of .
Then is an eigenvalue
of the matrix .
Proof Notice that since is assumed nonsingular, exists by Theorem NI, but more importantly, does not involve division by zero since Theorem SMZE prohibits this possibility.
Let be one eigenvector of for . Suppose has size . Then
So is an eigenvector of for the eigenvalue .
The theorems above have a similar style to them, a style you should consider using when confronted with a need to prove a theorem about eigenvalues and eigenvectors. So far we have been able to reserve the characteristic polynomial for strictly computational purposes. However, the next theorem, whose statement resembles the preceding theorems, has an easier proof if we employ the characteristic polynomial and results about determinants.
Theorem ETM
Eigenvalues of the Transpose of a Matrix
Suppose is a square
matrix and is an
eigenvalue of . Then
is an eigenvalue
of the matrix .
Proof Let be one eigenvector of for . Suppose has size . Then
So and have the same characteristic polynomial, and by Theorem EMRCP, their eigenvalues are identical and have equal algebraic multiplicities. Notice that what we have proved here is a bit stronger than the stated conclusion in the theorem.
If a matrix has only real entries, then the computation of the characteristic polynomial (Definition CP) will result in a polynomial with coefficients that are real numbers. Complex numbers could result as roots of this polynomial, but they are roots of quadratic factors with real coefficients, and as such, come in conjugate pairs. The next theorem proves this, and a bit more, without mentioning the characteristic polynomial.
Theorem ERMCP
Eigenvalues of Real Matrices come in Conjugate Pairs
Suppose is a square matrix
with real entries and
is an eigenvector of
for the eigenvalue .
Then is an
eigenvector of for
the eigenvalue .
Proof
So is an eigenvector of for the eigenvalue .
This phenomenon is amply illustrated in Example CEMS6, where the four complex eigenvalues come in two pairs, and the two basis vectors of the eigenspaces are complex conjugates of each other. Theorem ERMCP can be a time-saver for computing eigenvalues and eigenvectors of real matrices with complex eigenvalues, since the conjugate eigenvalue and eigenspace can be inferred from the theorem rather than computed.
A polynomial of degree will have exactly roots. From this fact about polynomial equations we can say more about the algebraic multiplicities of eigenvalues.
Theorem DCP
Degree of the Characteristic Polynomial
Suppose that is a square
matrix of size . Then the
characteristic polynomial of ,
, has
degree .
Proof We will prove a more general result by induction (Technique I). Then the theorem will be true as a special case. We will carefully state this result as a proposition indexed by , .
: Suppose that is an matrix whose entries are complex numbers or linear polynomials in the variable of the form , where is a complex number. Suppose further that there are exactly entries that contain and that no row or column contains more than one such entry. Then, when , is a polynomial in of degree , with leading coefficient , and when , is a polynomial in of degree or less.
Base Case: Suppose is a matrix. Then its determinant is equal to the lone entry (Definition DM). When , the entry is of the form , a polynomial in of degree with leading coefficient . When , then and the entry is simply a complex number, a polynomial of degree . So is true.
Induction Step: Assume is true, and that is an matrix with entries of the form . There are two cases to consider.
Suppose . Then every row and every column will contain an entry of the form . Suppose that for the first row, this entry is in column . Compute the determinant of by an expansion about this first row (Definition DM). The term associated with entry of this row will be of the form
The submatrix is an matrix with terms of the form , no more than one per row or column. By the induction hypothesis, will be a polynomial in of degree with coefficient . So this entire term is then a polynomial of degree with leading coefficient .
The remaining terms (which constitute the sum that is the determinant of ) are products of complex numbers from the first row with cofactors built from submatrices that lack the first row of and lack some column of , other than column . As such, these submatrices are matrices with entries of the form , no more than one per row or column. Applying the induction hypothesis, we see that these terms are polynomials in of degree or less. Adding the single term from the entry in column with all these others, we see that is a polynomial in of degree and leading coefficient .
The second case occurs when . Now there is a row of that does not contain an entry of the form . We consider the determinant of by expanding about this row (Theorem DER), whose entries are all complex numbers. The cofactors employed are built from submatrices that are matrices with either or entries of the form , no more than one per row or column. In either case, , and we can apply the induction hypothesis to see that the determinants computed for the cofactors are all polynomials of degree or less. Summing these contributions to the determinant of yields a polynomial in of degree or less, as desired.
Definition CP tells us that the characteristic polynomial of an matrix is the determinant of a matrix having exactly entries of the form , no more than one per row or column. As such we can apply to see that the characteristic polynomial has degree .
Theorem NEM
Number of Eigenvalues of a Matrix
Suppose that is a square
matrix of size with
distinct eigenvalues .
Then
Proof By the definition of the algebraic multiplicity (Definition AME), we can factor the characteristic polynomial as
where is a nonzero constant. (We could prove that , but we do not need that specificity right now. See Exercise PEE.T30) The left-hand side is a polynomial of degree by Theorem DCP and the right-hand side is a polynomial of degree . So the equality of the polynomials’ degrees gives the equality .
Theorem ME
Multiplicities of an Eigenvalue
Suppose that is a
square matrix of size
and
is an eigenvalue. Then
Proof Since is an eigenvalue of , there is an eigenvector of for , . Then , so , since we can extend into a basis of (Theorem ELIS).
To show that is the most involved portion of this proof. To this end, let and let be a basis for the eigenspace of , . Construct another vectors, , so that
is a basis of . This can be done by repeated applications of Theorem ELIS. Finally, define a matrix by
where is an matrix whose columns are . The columns of are linearly independent by design, so is nonsingular (Theorem NMLIC) and therefore invertible (Theorem NI). Then,
So
Preparations in place, we compute the characteristic polynomial of ,
What can we learn then about the matrix ?
Now imagine computing the characteristic polynomial of by computing the characteristic polynomial of using the form just obtained. The first columns of are all zero, save for a on the diagonal. So if we compute the determinant by expanding about the first column, successively, we will get successive factors of . More precisely, let be the square matrix of size that is formed from the last rows and last columns of . Then
This says that is a factor of the characteristic polynomial at least times, so the algebraic multiplicity of as an eigenvalue of is greater than or equal to (Definition AME). In other words,
as desired.
Theorem NEM says that the sum of the algebraic multiplicities for all the eigenvalues of is equal to . Since the algebraic multiplicity is a positive quantity, no single algebraic multiplicity can exceed without the sum of all of the algebraic multiplicities doing the same.
Theorem MNEM
Maximum Number of Eigenvalues of a Matrix
Suppose that is a
square matrix of size .
Then cannot have
more than distinct
eigenvalues.
Proof Suppose that has distinct eigenvalues, . Then
Recall that a matrix is Hermitian (or self-adjoint) if (Definition HM). In the case where is a matrix whose entries are all real numbers, being Hermitian is identical to being symmetric (Definition SYM). Keep this in mind as you read the next two theorems. Their hypotheses could be changed to “suppose is a real symmetric matrix.”
Theorem HMRE
Hermitian Matrices have Real Eigenvalues
Suppose that is a
Hermitian matrix and
is an eigenvalue of .
Then .
Proof Let be one eigenvector of for the eigenvalue . Then by Theorem PIP we know . So
If a complex number is equal to its conjugate, then it has a complex part equal to zero, and therefore is a real number.
Notice the appealing symmetry to the justificcations given for the steps of this proof. In the center is the ability to pitch a Hermitian matrix from one side of the inner product to the other.
Look back and compare Example ESMS4 and Example CEMS6. In Example CEMS6 the matrix has only real entries, yet the characteristic polynomial has roots that are complex numbers, and so the matrix has complex eigenvalues. However, in Example ESMS4, the matrix has only real entries, but is also symmetric, and hence Hermitian. So by Theorem HMRE, we were guaranteed eigenvalues that are real numbers.
In many physical problems, a matrix of interest will be real and symmetric, or Hermitian. Then if the eigenvalues are to represent physical quantities of interest, Theorem HMRE guarantees that these values will not be complex numbers.
The eigenvectors of a Hermitian matrix also enjoy a pleasing property that we will exploit later.
Theorem HMOE
Hermitian Matrices have Orthogonal Eigenvectors
Suppose that is a
Hermitian matrix and
and are two eigenvectors
of for different
eigenvalues. Then
and are orthogonal
vectors.
Proof Let be an eigenvector of for and let be an eigenvector of for a different eigenvalue . So we have . Then
This equality says that and are orthogonal vectors (Definition OV).
Notice again how the key step in this proof is the fundamental property of a Hermitian matrix (Theorem HMIP) — the ability to swap across the two argumanrs of the inner product. We’ll build on these results and continue to see some more interesting properties in Section OD.
T10 Suppose that
is a square matrix. Prove that the constant term of the characteristic polynomial of
is equal to the
determinant of .
Contributed by Robert Beezer Solution [1201]
T20 Suppose that
is a square matrix. Prove that a single vector may not be an eigenvector of
for
two different eigenvalues.
Contributed by Robert Beezer Solution [1202]
T30 Theorem DCP tells us that the characteristic polynomial of a square matrix of
size has
degree .
By suitably augmenting the proof of Theorem DCP prove that the coefficient of
in the characteristic
polynomial is .
Contributed by Robert Beezer
T10 Contributed by Robert Beezer Statement [1200]
Suppose that the characteristic polynomial of
is
Then
T20 Contributed by Robert Beezer Statement [1200]
Suppose that the vector
is an eigenvector of for
the two eigenvalues
and ,
where .
Then ,
so
which is a contradiction.