Hi, joannehuang-ga:
I've numbered the problems 0 to 3 and interspersed the solutions
among the parts of the problem statements. In problem 2 I've used
both operator expressions and matrix representations to cover the
same material, thinking that the combination of both approaches may
be helpful to your understanding.
regards, mathtalk-ga
0. Let dim(V)=n=2k+1, and let T:V -> V be a linear tranformation
satisfing T^n=0 but not T^(n-1)=0.
Prove that there is no linear operator S for which S^2=T.
Proof: First note that we must assume k > 0, for if n = 1, then
there exists a trivial counterexample T = 0. (Recall T^0 = I by
convention.)
Given k > 0, the assumptions of the problem amount to saying that
f(x) = x^n is the minimal polynomial for T, since f(T) = 0 but no
proper divisor of f(x) satisfies that condition.
If there were S such that S^2 = T, then the minimal polynomial for
S would have to divide x^(2n), since S^(2n) = T^n = 0. Now the
characteristic polynomial of S has degree n, so the degree of the
minimal polynomial is at most n. Hence S^n = 0 because the minimal
polynomial for S must be x^m with m <= n.
Since S^n = 0, also S^(n+1) = S^2(k+1) = T^(k+1) = 0. But:
k+1 < 2k+1 = n (because k > 0)
would contradict that x^n is the minimal polynomial of T. Therefore
no such S can exist. QED
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
1. Let V = P_2(R), polynomials of degree at most 2, be a real inner
produce space with <f(x),g(x)> = INTEGRAL f(x)g(x) dx OVER [0,1].
Let U be the subspace of V defined by U = { f(x) in V : f(1) = 0 }.
(a) Construct a basis B_1 for U and prove that is indeed a basis.
Answer: Consider the standard basis {1,x,x^2} for V. By inspection
each of these basis functions is 1 at x=1, so:
B_1 = {x - 1,x^2 - 1}
contains two functions that are 0 at x=1, so B_1 is a subset of U.
We claim B_1 is a basis for subspace U.
Linear independence of B_1: Suppose there exist scalars a,b s.t.
a(x - 1) + b(x^2 - 1) = 0 in P_2(R)
That is bx^2 + ax - (a+b) = 0 identically as polynomials. Comparing
coefficients of x and x^2 on both sides shows us that a = b = 0. So
the linear independence of B_1 is proven.
Spanning of U by B_1: Let f(x) be any polynomial of degree at most
two, and suppose f(x) belongs to U. That is:
f(x) = ax^2 + bx + c for some real coefficients a,b,c
f(1) = a + b + c = 0
c = -(a + b)
Thus f(x) = ax^2 + bx - (a + b) = a(x^2 - 1) + b(x - 1) shows us
that B_1 is a spanning set of U.
Since a basis is by definition a linearly independent spanning set,
we have shown B_1 is a basis for U.
(b) Use the Gram-Schmidt method on the basis B_1 to construct an
orthonormal basis B_2 for U.
Answer: The first step of Gram-Schmidt is to "normalize" the first
element of our basis. Here the inner product is the integral, so:
<x - 1,x - 1> = INTEGRAL (x-1)^2 dx OVER [0,1]
= INTEGRAL u^2 du OVER [0,1] by substitution u = 1-x
= 1/3
so the unit length multiple of x - 1 is sqrt(3)(x - 1).
The second step of Gram-Schmidt is to find the residue of the next
basis element after subtracting its projection onto the first element
and then normalizing this residue to have unit length. A well-known
trick for avoiding "square roots" in the computation is by taking the
residue like this:
(x^2 - 1) - (<x^2 - 1,x - 1>/<x - 1,x - 1>) (x - 1)
We already found the denominator of that scalar coefficient above, so
it remains only to find its numerator:
<x^2 - 1,x - 1> = INTEGRAL (x^2 - 1)(x - 1) dx OVER [0,1]
= INTEGRAL (x + 1)(x - 1)^2 dx OVER [0,1]
= INTEGRAL (2 - u) u^2 du OVER [0,1] as before
= (2/3) - (1/4) = 5/12
Therefore we have as a residue:
(x^2 - 1) - (5/4) (x - 1) = x^2 - (5/4)x + (1/4)
To normalize this polynomial we require its length (squared):
<x^2 - (5/4)x + 1/4,x^2 - (5/4)x + 1/4>
= INTEGRAL (1/16)(4x^2 - 5x + 1)^2 dx OVER [0,1]
= (1/16) INTEGRAL (4x - 1)^2 (x - 1)^2 dx OVER [0,1]
= (1/16) INTEGRAL (3 - 4u)^2 u^2 du OVER [0,1]
= (1/16) INTEGRAL 9u^2 - 24u^3 + 16u^4 du OVER [0,1]
= (1/16) [ 3 - 6 + (16/5) ] = (1/16)(1/5)
After dividing by the appropriate square root, the normalized
residue becomes:
sqrt(5) (4x^2 - 5x + 1)
To summarize the output of Gram-Schmidt gives us this:
B_2 = { sqrt(3) (x - 1), sqrt(5) (4x^2 - 5x + 1) }
(c) Determine an f(x) in U such that:
INTEGRAL |1 + x - f(x)|^2 dx
is as small as possible.
Answer: The f(x) in U which minimizes this "norm squared" is just
the projection of "vector" 1 + x onto the subspace U of V. Since
we have just computed an orthonormal basis B_2 for U, a natural way
to do this is to simply "project" 1 + x onto each of the two basis
elements in B_2 = {g(x),h(x)}:
f(x) = <1 + x, g(x)> g(x) + <1 + x, h(x)> h(x)
where g(x) = sqrt(3) (x - 1), h(x) = sqrt(5) (4x^2 - 5x + 1).
Once the appropriate integrals are evaluated, we find:
f(x) = -2(x - 1) - (5/3)(4x^2 - 5x + 1)
= -(20/3)x^2 + (19/3)x + (1/3)
As a check on this answer I verified that f(x) - (1 + x) is indeed
orthogonal to both basis elements x - 1 and x^2 - 1 of B_1.
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Note: In this problem the expression T*T represents T' ° T, or the
composition of T' (the adjoint of T) with T (first do T, then T').
2. Let <,> be the usual (Euclidean) inner product on R^3, and let
T be a linear transformation on R^3 defined by:
T(x,y,z) = (y,2z,0)
(a) Compute sqrt( T*T ).
Answer: Let's begin by determining the adjoint T', which is the
unique operator on R^3 such that:
<T(x,y,z),(a,b,c)> = <(x,y,z),T'(a,b,c)>
for all vectors (x,y,z), (a,b,c) in R^3. By comparison of terms on
both sides, it is found that:
<T(x,y,z),(a,b,c)> = ay + 2bz = <(x,y,z),(0,a,2b)>
T'(a,b,c) = (0,a,2b)
Therefore the compostion T*T is like this:
T*T(x,y,z) = T'(T(x,y,z)) = T'(y,2z,0) = (0,y,4z)
The same result can be had by working with a matrix representation
of T, say with respect to the standard ordered basis of R^3. In
that case the adjoint T' corresponds to the matrix transpose:
/ 0 1 0 \ / 0 0 0 \
T |--> | 0 0 2 | and T' |--> | 1 0 0 |
\ 0 0 0 / \ 0 2 0 /
so that T*T = T' ° T corresponds to the product of their matrix
representations:
/ 0 0 0 \
T*T |--> | 0 1 0 |
\ 0 0 4 /
In any case here the computation of sqrt( T*T ) can be done by
inspection:
sqrt( T*T )(x,y,z) = (0,y,2z)
so that sqrt( T*T ) "squared" gives T*T as above. Note that the
matrix representation of T*T is not just symmetric but already
diagonal if done with respect to the standard basis:
/ 0 0 0 \ / 0 0 0 \
T*T |--> | 0 1 0 | , so sqrt( T*T ) |--> | 0 1 0 |
\ 0 0 4 / \ 0 0 2 /
and sqrt( T*T ) amounts to taking the principle square root of each
diagonal entry of the matrix representation of T*T in this case.
Note that the matrix representations of both T*T and sqrt( T*T ) are
nonnegative diagonal matrices. This corresponds to the property of
these operators being positive semi-definite and symmetric.
(b) Compute orthonormal bases for each of:
V_1 = range(sqrt( T*T )),
V_2 = range(sqrt( T*T ))^perp = orthogonal complement of V_1,
V_3 = range(T), and
V_4 = range(T)^perp = orthogonal complement of V_3.
Answers: Note that the range of sqrt( T*T ), like the range of T*T,
consists of vectors whose first component is zero. So we can just
use a subset of the standard basis of R^3 omitting the first vector:
B_1 = orthonormal basis for V_1 = {(0,1,0),(0,0,1)}
and for the orthogonal complement:
B_2 = orthonormal basis for V_2 = {(1,0,0)}
The range of T on the other hand consists of vectors whose third
component is zero. So:
B_3 = orthonormal basis for V_3 = {(1,0,0),(0,1,0)}
and for the orthogonal complement:
B_4 = orthonormal basis for V_4 = {(0,0,1)}
(c) Compute an isometry S for which T = S ° sqrt( T*T ).
Answer: From a comparison of the operator definitions:
T(x,y,z) = (y,2z,0)
sqrt( T*T )(x,y,z) = (0,y,2z)
we see that one possible choice for S would be coordinate rotation:
S(x,y,z) = (y,z,x)
which also happens to be orientation preserving. Verify:
S ° sqrt( T*T )(x,y,z) = S(0,y,2z) = (y,2z,0) = T(x,y,z)
In this case the inverse of S, rotating the coordinates back in the
opposite direction, is also S', the adjoint of S. In terms of their
matrix representations wrt to the standard order basis of R^3:
/ 0 1 0 \ / 0 0 1 \
S |--> | 0 0 1 | and S' |--> | 1 0 0 |
\ 1 0 0 / \ 0 1 0 /
(d) Compute the singular value decomposition of T.
Answer: Since sqrt( T*T ) is already diagonalized, one simple answer
might be to write, using what we have already discussed above:
T = S ° sqrt( T*T ) ° I
However it is customary when writing the singular value decompostion
to put diagonal entries (singular values) in decreasing order, i.e.:
/ 2 0 0 \ / 0 0 0 \
| 0 1 0 | rather than | 0 1 0 |
\ 0 0 0 / \ 0 0 2 /
The distinction is difficult to make when one is using only the
operator notation to discuss the decomposition, but in terms of the
matrix representations we would multiply on the left and right by:
/ 0 0 1 \
U = | 0 1 0 |
\ 1 0 0 /
which corresponds to the (orientation reversing) isometry that swaps
first and third coordinates. U is symmetric and its own inverse.
The conventional ordering of singular values would then be given by:
/ 0 1 0 \ / 0 1 0 \ / 2 0 0 \
| 0 0 2 | = | 0 0 1 | U | 0 1 0 | U
\ 0 0 0 / \ 1 0 0 / \ 0 0 0 /
/ 0 1 0 \ / 2 0 0 \
= | 1 0 0 | | 0 1 0 | U
\ 0 0 1 / \ 0 0 0 /
In other words the standard basis matrix representation for T can be
written as V D U, where U,V are orthogonal matrices:
/ 0 1 0 \
V = | 1 0 0 |
\ 0 0 1 /
and diagonal matrix D puts the singular values in descending order.
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
/ 0 0 1 \
3. Let A = | 1 0 0 | .
\ 0 1 0 /
(a) Determine the Jordan form for A over the reals R.
Answer: For details about the Jordan form over the reals, see here:
http://www.numbertheory.org/courses/MP274/realjord.pdf
In this case the permutation matrix A obviously satisfies A^3 = I
because the permutation of rows induced by A is a cycle of order 3.
One can also show that x^3 - 1 is A's characteristic polynomial:
det(xI - A) = x^3 - 1
Hence A has one real eigenvalue (characteristic root) x = 1, and a
pair of complex conjugate roots from the remaining factor, namely:
x^3 - 1 = (x - 1)(x^2 + x + 1)
x^2 + x + 1 = 0
x = -(1/2) + i sqrt(3)/2, -(1/2) - i sqrt(3)/2
A corresponding real Jordan form for A is then:
/ 1 0 0 \
| 0 a b | where a = -(1/2) and b = sqrt(3)/2.
\ 0 -b a /
(b) Determine the Jordan form for A over the complex numbers C.
As the analysis above shows, A has three distinct eigenvalues over
the complex numbers, so the Jordan form of A in that context is:
/ 1 0 0 \
| 0 z 0 | where z = -(1/2) + i sqrt(3)/2,
\ 0 0 z'/
and z' = -(1/2) - i sqrt(3)/2 is the complex conjugate of z.
Other Links of Interest:
If you want to do symbolic integration, Wolfram supplies an online
calculator for indefinite integrals here:
http://www.integrals.com/
Search Strategy
Keywords: "real Jordan form"
://www.google.com/search?hl=en&lr=&ie=UTF-8&oe=UTF-8&q=%22real+Jordan+form%22&btnG=Google+Search |
Clarification of Answer by
mathtalk-ga
on
06 May 2003 21:50 PDT
Hi, joannehuang-ga:
I will use row vectors (x,y,z,w) to denote elements of R^4, even
though the representation of T by matrix A with respect to a
standard basis is by convention a multiplication by A on the left
of a vector Au = v. To reconcile this inconsistency (since u,v
are then column vectors rather than row vectors), I recall the use
of ' to denote transposition (thus turning rows into columns).
Because the matrices A are all essentially in Jordan canonical form,
a basis of the eigenvectors and generalized eigenvectors can be
chosen to be "standard" basis vectors. The question seems to be a
little more than simply exhibiting a basis of (generalized) eigen-
vectors, as it asks for "all the eigenvectors" (resp. all the
generalized eigenvectors) associated with an eigenvalue. However
for convenience we will define these "standard" basis vectors:
e_1 = (1,0,0,0)'
e_2 = (0,1,0,0)'
e_3 = (0,0,1,0)'
e_4 = (0,0,0,1)'
(a)
/ 2 1 0 0 \
| 0 2 0 0 |
| 0 0 2 0 |
\ 0 0 0 2 /
The minimal polynomial is (x - 2)^2; the characteristic polynomial
is (x - 2)^4. The only eigenvalue is 2. The (true) eigenvectors
associated with eigenvalue 2 are the non-zero linear combinations
of e_1, e_3, and e_4, i.e. (a,0,b,c)' where not all a,b,c are zero.
The generalized eigenvalues are the non-zero vectors u such that:
(A - 2I)^k u = 0 for some k > 1
Here e_2 is a particular generalized eigenvector, but more generally
any vector in R^4 outside the span of { e_1, e_3, e_4 } is in this
case a generalized eigenvector, i.e. any vector (a,b,c,d)' where b
is not zero.
(b)
/ 2 1 0 0 \
| 0 2 0 0 |
| 0 0 2 1 |
\ 0 0 0 2 /
The minimal polynomial is again (x - 2)^2; the characteristic
polynomial (x - 2)^4. Again the only eigenvalue is 2.
The eigenvectors are the non-zero vectors spanned by { e_1, e_3 },
i.e. vectors (a,0,b,0)' where not both a,b are zero. Generalized
eigenvectors are non-zero vectors outside the span of { e_1, e_3 },
i.e. vectors (a,b,c,d)' where not both b,d are zero.
(c)
/ 2 1 0 0 \
| 0 2 1 0 |
| 0 0 2 1 |
\ 0 0 0 2 /
The minimal polynomial and characteristic polynomial are both
(x - 2)^4. Again the only eigenvalue is 2.
The only eigenvectors are non-zero multiples of e_1, i.e. vectors
(a,0,0,0)' where a is non-zero. All the other non-zero vectors are
generalized eigenvectors, i.e. (a,b,c,d)' where not all b,c,d are
zero.
(d)
/ 2 1 0 0 \
| 0 2 0 0 |
| 0 0 3 0 |
\ 0 0 0 4 /
The minimal polynomial and characteristic polynomial are both
(x - 2)^2 (x - 3) (x - 4). There are three eigenvalues 2, 3, 4.
The eigenvalue 2 has as eigenvectors all non-zero multiples of
e_1, i.e. vectors (a,0,0,0)' where a is not zero. It has for
generalized eigenvectors the rest of the non-zero vectors spanned
by { e_1, e_2 }, i.e. vectors (a,b,0,0)' where b is not zero.
The eigenvalue 3 has as eigenvectors all non-zero multiples of
e_3, i.e. vectors (0,0,a,0)' where a is not zero. There are no
additional generalized eigenvectors associated with eigenvalue 3.
The eigenvalue 4 has as eigenvectors all non-zero multiples of
e_4, i.e. vectors (0,0,0,a)' where a is not zero. There are no
additional generalized eigenvectors associated with eigenvalue 4.
(e)
/ 2 0 0 0 \
| 0 2 0 0 |
| 0 0 3 0 |
\ 0 0 0 4 /
The minimal polynomial is (x - 2)(x - 3)(x - 4). The characteristic
polynomial is (x - 2)^2 (x - 3)(x - 4). Again there are three
eigenvalues 2,3,4.
The eigenvalue 2 has a two-dimensional "space" of eigenvectors, the
non-zero linear combinations of e_1 and e_2, i.e. vectors (a,b,0,0)'
where not both a,b are zero. There are no additional generalized
eigenvectors associated with eigenvalue 2.
The eigenvalue 3 has just a one-dimensional "space" of eigenvectors,
the non-zero multiples of e_3, i.e. vectors (0,0,a,0)' where a is
not zero. There are no additional generalized eigenvectors for the
eigenvalue 3.
The eigenvalue 4 has just a one-dimensional "space" of eigenvectors,
the non-zero multiples of e_4, i.e. vectors (0,0,0,a)' where a is
not zero. There are no additional generalized eigenvectors for the
eigenvalue 4.
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