For the purposes of this section, we fix a linear operator T on an n-dimensional vector space V such that the characteristic polynomial of T splits. Let λ1, λ2, …, λk
By Theorem 7.7 (p. 484), each generalized eigenspace Kλi
is the Jordan canonical form of T. In this matrix, each O is a zero matrix of appropriate size.
In this section, we compute the matrices Ai
To aid in formulating the uniqueness theorem for J, we adopt the following convention: The basis βi
To illustrate the discussion above, suppose that, for some i, the ordered basis βi
To help us visualize each of the matrices Ai
The array consists of ni
Counting from left to right, the jth column consists of the pj
Denote the end vectors of the cycles by v1, v2, …, vni
Notice that the dot diagram of Ti
Now let rj
In Example 1, with ni=4, p1=p2=3, p3=2
Here r1=4, r2=3
We now devise a method for computing the dot diagram of Ti
To determine the dot diagram of Ti
For any positive integer r, the vectors in βi
Clearly, N((T−λiI)r)⊆Kλi
Let a and b denote the number of vectors in S1
In the case that r=1
The dimension of Eλi
Exercise.
We are now able to devise a method for describing the dot diagram in terms of the ranks of operators.
Let rj
r1=dim(V)−rank(T−λiI).
rj=rank((T−λiI)j−1)−rank((T−λiI)j)
By Theorem 7.9, for j=1, 2, …, p−1
Hence
and for j>1
Theorem 7.10 shows that the dot diagram of Ti
For any eigenvalue λi
We apply these results to find the Jordan canonical forms of two matrices and a linear operator.
Let
We find the Jordan canonical form of A and a Jordan canonical basis for the linear operator T=LA
Thus A has two distinct eigenvalues, λ1=2
Suppose that β1
and
(Actually, the computation of r2
So
Since λ2=3
Setting β=β1∪β2
and so J is the Jordan canonical form of A.
We now find a Jordan canonical basis for T=LA
From this diagram we see that v1∈N((T−2I)2)
It is easily seen that
is a basis for N((T−2I)2)=Kλ1
Then
Now simply choose v2
Thus we have associated the Jordan canonical basis
with the dot diagram in the following manner.
By Theorem 7.6 (p. 483), the linear independence of β1
Since λ2=3
Thus
is a Jordan canonical basis for LA
Notice that if
then J=Q−1AQ
Let
We find the Jordan canonical form J of A, a Jordan canonical basis for LA
The characteristic polynomial of A is det(A−tI)=(t−2)2(t−4)2
We begin by computing the dot diagram of T1
hence the dot diagram of T1
Therefore
where β1
Next we compute the dot diagram of T2
Thus
where β2
Choose v to be any solution to the system of linear equations
for example,
Thus
Therefore
is a Jordan canonical basis for LA
Finally, we define Q to be the matrix whose columns are the vectors of β
Then J=Q−1AQ
Let V be the vector space of polynomial functions in two real variables x and y of degree at most 2. Then V is a vector space over R and α={1, x, y, x2, y2, xy}
For example, if f(x, y)=x+2x2−3xy+y
We find the Jordan canonical form and a Jordan canonical basis for T.
Let A=[T]α
and hence the characteristic polynomial of T is
Thus λ=0
and since
r2=rank(A)−rank(A2)=3−1=2
Because there are a total of six dots in the dot diagram and r1=3
We conclude that the Jordan canonical form of T is
We now find a Jordan canonical basis for T. Since the first column of the dot diagram of T consists of three dots, we must find a polynomial f1(x, y)
and
Likewise, since the second column of the dot diagram consists of two dots, we must find a polynomial f2(x, y)
Since our choice must be linearly independent of the polynomials already chosen for the first cycle, the only choice in α
Finally, the third column of the dot diagram consists of a single polynomial that lies in the null space of T. The only remaining polynomial in α
Thus β={2, 2x, x2, y, xy, y2}
In the three preceding examples, we relied on our ingenuity and the context of the problem to find Jordan canonical bases. The reader can do the same in the exercises. We are successful in these cases because the dimensions of the generalized eigenspaces under consideration are small. We do not attempt, however, to develop a general algorithm for computing Jordan canonical bases, although one could be devised by following the steps in the proof of the existence of such a basis (Theorem 7.7 p. 484).
The following result may be thought of as a corollary to Theorem 7.10.
Let A and B be n×n
If A and B have the same Jordan canonical form J, then A and B are each similar to J and hence are similar to each other.
Conversely, suppose that A and B are similar. Then A and B have the same eigenvalues. Let JA
We determine which of the matrices
are similar. Observe that A, B, and C have the same characteristic polynomial −(t−1)(t−2)2
Since JA=JC
The reader should observe that any diagonal matrix is a Jordan canonical form. Thus a linear operator T on a finite-dimensional vector space V is diagonalizable if and only if its Jordan canonical form is a diagonal matrix. Hence T is diagonalizable if and only if the Jordan canonical basis for T consists of eigenvectors of T. Similar statements can be made about matrices. Thus, of the matrices A, B, and C in Example 5, A and C are not diagonalizable because their Jordan canonical forms are not diagonal matrices.
Label the following statements as true or false. Assume that the characteristic polynomial of the matrix or linear operator splits.
The Jordan canonical form of a diagonal matrix is the matrix itself.
Let T be a linear operator on a finite-dimensional vector space V that has a Jordan canonical form J. If β
Linear operators having the same characteristic polynomial are similar.
Matrices having the same Jordan canonical form are similar.
Every matrix is similar to its Jordan canonical form.
Every linear operator with the characteristic polynomial (−1)n(t−λ)n
Every linear operator on a finite-dimensional vector space has a unique Jordan canonical basis.
The dot diagrams of a linear operator on a finite-dimensional vector space are unique.
Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial of T splits. Suppose that λ1=2, λ2=4
λ1=2 |
λ2=4 |
λ3=−3 |
---|---|---|
Find the Jordan canonical form J of T.
Let T be a linear operator on a finite-dimensional vector space V with Jordan canonical form
Find the characteristic polynomial of T.
Find the dot diagram corresponding to each eigenvalue of T.
For which eigenvalues λi
For each eigenvalue λi
Compute the following numbers for each i, where Ui
rank(Ui)
rank(U2i)
nullity(Ui)
nullity(U2i)
For each of the matrices A that follow, find a Jordan canonical form J and an invertible matrix Q such that J=Q−1AQ
A=(−33−2−76−31−12)
A=(01−1−44−2−211)
A=(0−1−1−3−1−2756)
A=(0−312−21−12−21−12−2−314)
For each linear operator T, find a Jordan canonical form J of T and a Jordan canonical basis β
(a) V is the real vector space of functions spanned by the set of real-valued functions {et, tet, t2et, e2t}
(b) T is the linear operator on P3(R)
(c) T is the linear operator on P3(R)
(d) T is the linear operator on M2×2(R)
(e) T is the linear operator on M2×2(R)
(f) V is the vector space of polynomial functions in two real variables x and y of degree at most 2, as defined in Example 4, and T is the linear operator on V defined by
Let A be an n×n
Let A be an n×n
(a) Prove that [TW]γ′=([TW]γ)t
(b) Let J be the Jordan canonical form of A. Use (a) to prove that J and Jt
(c) Use (b) to prove that A and At
Let T be a linear operator on a finite-dimensional vector space, and suppose that the characteristic polynomial of T splits. Let β
(a) Prove that for any nonzero scalar c, {cx: x∈β}
(b) Suppose that γ
(c) Apply (b) to obtain a Jordan canonical basis for LA
Suppose that a dot diagram has k columns and m rows with pj
(a) m=p1
(b) pj=max{i: ri≥j}
(c) r1≥r2≥⋯≥rm.
(d) Deduce that the number of dots in each column of a dot diagram is completely determined by the number of dots in the rows.
Let T be a linear operator whose characteristic polynomial splits, and let λ
(a) Prove that dim(Kλ)
(b) Deduce that Eλ=Kλ
The following definitions are used in Exercises 11–19.
A linear operator T on a vector space V is called nilpotent if Tp=T0
Let T be a linear operator on a finite-dimensional vector space V, and let β
Prove that any square upper triangular matrix with each diagonal entry equal to zero is nilpotent.
Let T be a nilpotent operator on an n-dimensional vector space V, and suppose that p is the smallest positive integer for which Tp=T0
(a) N(Ti)⊆N(Ti+1)
(b) There is a sequence of ordered bases β1, β2, …, βp
(c) Let β−βp
(d) The characteristic polynomial of T is (−1)ntn
Prove the converse of Exercise 13(d): If T is a linear operator on an n-dimensional vector space V and (−1)ntn
Give an example of a linear operator T on a finite-dimensional vector space over the field of real numbers such that T is not nilpotent, but zero is the only eigenvalue of T. Characterize all such operators. Visit goo.gl/
Let T be a nilpotent linear operator on a finite-dimensional vector space V. Recall from Exercise 13 that λ=0
Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial of T splits, and let λ1, λ2, …, λk
(a) Prove that S is a diagonalizable linear operator on V.
(b) Let U=T−S
Let T be a linear operator on a finite-dimensional vector space V over C, and let J be the Jordan canonical form of T. Let D be the diagonal matrix whose diagonal entries are the diagonal entries of J, and let M=J−D
(a) M is nilpotent.
(b) MD=DM.
(c) If p is the smallest positive integer for which Mp=O
and, for any positive integer r≥p
Let F=C
be the m×m
(a) Nm=O
(b) For any integer r≥m
(c) limr→∞Jr
|λ|<1
λ=1
(Note that limr→∞λr
(d) Prove Theorem 5.12 on page 284.
The following definition is used in Exercises 20 and 21.
For any A∈Mn×n(C)
Let A, B∈Mn×n(C)
(a) ||A||m≥0
(b) ||A||m=0
(c) ||cA||m=|c|⋅||A||m
(d) ||A+B||m≤||A||m+||B||m
(e) ||AB||m≤n||A||m||B||m.
Let A∈Mn×n(C)
(a) ||Ak||m≤1
(b) There exists a positive number c such that ||Jk||m≤c
(c) Each Jordan block of J corresponding to the eigenvalue λ=1 is a 1×1 matrix.
(d) limk→∞Ak exists if and only if 1 is the only eigenvalue of A with absolute value 1.
(e) Theorem 5.19(a), using (c) and Theorem 5.18.
The next exercise requires knowledge of absolutely convergent series as well as the definition of eA for a matrix A. (See page 310.)
Use Exercise 20(d) to prove that eA exists for every A∈Mn×n(C).
Let x′=Ax be a system of n linear differential equations, where x is an n-tuple of differentiable functions x1(t), x2(t), …, xn(t) of the real variable t, and A is an n×n coefficient matrix as in Exercise 16 of Section 5.2. In contrast to that exercise, however, do not assume that A is diagonalizable, but assume that the characteristic polynomial of A splits. Let λ1, λ2, …, λk be the distinct eigenvalues of A.
(a) Prove that if u is the end vector of a cycle of generalized eigenvectors of LA of length p and u corresponds to the eigenvalue λi, then for any polynomial f(t) of degree less than p, the function
is a solution to the system x′=Ax.
(b) Prove that the general solution to x′=Ax is a sum of the functions of the form given in (a), where the vectors u are the end vectors of the distinct cycles that constitute a fixed Jordan canonical basis for LA.
Use Exercise 23 to find the general solution to each of the following systems of linear equations, where x, y, and z are real-valued differentiable functions of the real variable t.
x′=2x+yy′=2y−zz′=3z
x′=2x+yy′=2y+zz′=2z