Dr. Marques Sophie
Office 519
Linear algebra II
SpringSemester 2016
marques@cims.nyu.edu
Problem Set # 11
Justify all your answers completely (Or with a proof or with a counter example)
unless mentioned differently. No step should be a mystery or bring a question. The grader
cannot be expected to work his way through a sprawling mess of identities presented without
a coherent narrative through line. If he can’t make sense of it in finite time you could lose
serious points. Coherent, readable exposition of your work is half the job in mathematics.
You will loose serious points if your exposition is messy, incomplete, uses mathematical
symbols not adapted...
Exercise 1:
Find the Jordan canonical form of the following matrices over C and deduce the corresponding canonical form for those matrices over R and give for both the base change
matrices that permit to get those forms:


4 −1 −1
4 
1. A =  −2 0
2 −2 2


2 1 0 0
 −1 4 0 0 

2. B = 
 −1 4 1 −1 
−1 3 2 −1


3 2 −2 −1
 1 0 −1 0 

3. C = 
 3 3 −2 −1 
4 0 −2 −1
Solution:
1. The characteristic polynomial is
4 − x −1 −1 4 = −x3 + 6x2 − 16x + 16 = −(x − 2)(x2 − 4x + 8).
det(A − xI) = −2 −x
2
−2 2 − x
Hence, the eigenvalues are
λ1 = 2,
λ2 = 2 − 2i,
1
λ3 = 2 + 2i.
Denote the corresponding eigenvectors v1 , v2 , v3 . Since the eigenvalues are distinct,
the Jordan canonical form is simply given by the diagonal matrix


2
0
0
0 
[A]X = 0 2 − 2i
0
0
2 + 2i
where X = {v1 , v2 , v3 } is an ordered basis of eigenvectors. Solving for ker(A−λi I)
for i = 1, 2, 3, we have
ker(A − λ1 I) = Span{(1, 1, 1)}
ker(A − λ2 I) = Span{((3 − i), (3 + 4i), 5)}
ker(A − λ3 I) = Span{((3 + i), (3 − 4i), 5)}.
We can now assign
v1 = (1, 1, 1),
v2 = ((3 − i), (3 + 4i), 5),
v3 = ((3 + i), (3 − 4i), 5).
Since the matrix is diagonalizable, the dimension of each eigenspace is 1. Let
f = x + iy ∈ Eλ2 (TC ). Using the notation from class where J is the conjugate
operator, it follows that J(f ) ∈ Eλ3 (TC ). If
W = Cf ⊕ CJ(f ),
we have shown in class that Eλ1 ⊕ Eλ2 = W and the set Y = {x, y} is a basis for
the T -invariant subspace
(W )R (T := LA ). Using polar coordinates for λ2 = 2−2i
√ −iπ/4
iθ
so
gives λ2 = re = 8e
√
cos θ sin θ
cos(− π4 ) sin(− π4 )
[T |(W )R ]Y = r
= 8
.
− sin θ cos θ
− sin(− π4 ) cos(− π4 )
where Y = {x, y}. Since v3 = J(v2 ), let f = v2 so Y = {(3, 3, 5), (−1, 4, 0)}.
Notice that Z = {v1 } ∪ Y forms an ordered basis for R3 . The corresponding
canonical form for A over R is then

 

2 √ 0
0
2 0 0
√
π
π 


[T ]Z = 0
√8 cos(− 4π) √ 8 sin(− 4π ) = 0 2 −2 .
0 2 2
0 − 8 sin(− 4 )
8 cos(− 4 )
with change of basis matrix


1 3 −1
S = 1 3 4 
1 5 0
so A = S[T ]Z S −1 .
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2. The characteristic polynomial is
2 − x
1
0
0 −1 4 − x
0
0 det(B − xI) = 4
1−x
−1 −1
−1
3
2
−1 − x
= (2 − x)(4 − x)(x2 − 1 + 2) + (x2 − 1 + 2)
= (x − i)(x + i)(x − 3)2 .
The eigenvalues (including multiplicity) given by the roots are
λ1 = 3,
λ2 = 3,
λ3 = i,
λ4 = −i.
Solving for x in (B − λi I)x = 0 for i = 1, 3, 4 and (B − λ1 I)2 x = 0 gives
ker(B − λ1 I) = {(1, 1, 1, 1)}
ker(B − λ1 I)2 = {(1, 0, 0, 0), (0, 1, 1, 1)}
ker(B − λ3 I) = {(0, 0, 1, 1 − i)}
ker(B − λ4 I) = {(0, 0, 1, 1 + i)}.
Define vectors that are in each of the above kernels
v1
v2
v3
v4
= (1, 1, 1, 1),
= (−1, 0, 0, 0),
= (0, 0, 1, 1 − i),
= (0, 0, 1, 1 + i),
respectively. Hence, X = {v1 , v2 , v3 , v4 } is an ordered basis for the vector space
that puts B in Jordan canonical form


3 1 0 0
0 3 0 0 

J =
0 0 i 0  .
0 0 0 −i
with change of basis matrix S that has ordered columns given by the vectors in
X.
To find the corresponding form over R, let f = v3 . If f = x + iy, then x =
(0, 0, 1, 1) and y = (0, 0, 0, −1). Since λ4 = λ3 , we have that Y = {x, y} is an
ordered basis for Mλ3 ,λ4 = Eλ3 ⊕ Eλ4 . Writing λ3 in polar coordinates gives us
π
λ3 = i = ei 2 = reiθ so r = 1 and θ = π2 . Letting T = LB ,
cos π2 sin π2
0 1
[T |Eλ3 ,λ4 ]Y =
=
.
− sin π2 cos π2
−1 0
3
Let Z = {v1 , v2 , x, y} be an ordered basis for the vector space over R. The matrix
with columns equal to the ordered vectors in Z gives us the desired change of
basis matrix to get the corresponding canonical form over R. Denote this matrix


1 −1 0 0
1 0 0 0 

P =
1 0 1 0  .
1 0 1 −1
The corresponding canonical form

3
0
[T ]Z = 
0
0
is then
1 0
3 0
0 0
0 −1

0
0
 = P −1 BP.
1
0
3. The characteristic polynomial is given by
3 − x 2
−2
−1 1
−x
−1
0 det(C − xI) = 3 −2 − x
−1 3
4
0
−2
−1 − x
= (3 − x)(−x3 − 3x2 − 3 − 3x) − 2x2 − 2 + 6x2 + 4x + 6 + 4x2 + 2x + 6
= x4 + 2x2 + 1.
The roots of the polynomial gives us the eigenvalues (including multiplicity)
λ1 = i,
λ2 = i,
λ3 = −i,
λ4 = −i.
Examining the kernels ker(C − λi I), we have generalized eigenvectors
1+i
1+i
, 0,
, 1)
2
2
2 − i 1 + i 3 − 2i
,
,
, 0)
v2 = (
2
2
2
1−i
1−i
v3 = (
, 0,
, 1)
2
2
2 + i 1 − i 3 + 2i
,
,
, 0)
v4 = (
2
2
2
∈ ker(C − λ1 I)
v1 = (
∈ ker(C − λ1 I)2 ∼ ker(C − λ1 I)
∈ ker(C − λ3 I)
∈ ker(C − λ3 I)2 ∼ ker(C − λ3 I)
with (C − λ1 )(v2 ) = v1 and (C − λ3 )(v4 ) = v3 . Hence, the Jordan canonical form
is given by taking the basis X = {v1 , v2 , v3 , v4 } for V to get


i 1 0 0
0 i 0 0 

[T ]X = 
0 0 −i 1  .
0 0 0 −i
4
where T = LC .
For the form over R, first notice that λ1 = λ2 = λ3 = λ4 . Putting this into polar
π
coordinates gives λ1 = i = reiθ = ei 2 so r = 1 and θ = π2 . Let f1 = x + iy =
v1 = Jv3 and f2 = u + iw = v2 = Jv4 . Then Y = {x, y, u, w} gives us a basis
for Mλ1 ,λ3 . But V = Mλ1 ⊕ Mλ3 so Y is a basis for the vector space. Hence, the
corresponding canonical form is

 

cos θ sin θ
1
0
0 1 1 0
− sin θ cos θ


0
1 
 = −1 0 0 1
[T ]Y = 
 0


0
cos θ sin θ
0 0 0 1
0
0
− sin θ cos θ
0 0 −1 0
with base change matrix
1
2
1
2
1 − 12

0 0 1 1 
2
2 
S=
 1 1 3 −1  .
2
2
2
1 0 0 0
Exercise 2: Prove the following important property of the exponential map
det(eA ) = eT r(A) f or all A ∈ M (n, F)
holds when F = C. Then explain how complexification can be used to show this result
remains true when F = R.
Solution: If F = R an R-basis {ei } for V gives a C-basis {e˜j = ej + i0} in VC and it is
trivial to check that
1. (eA )C = e(AC )
2. det(AC ) = det(A).
3. T r(AC ) = T r(A).
Thus it suffices to consider the case in which F = C, A = AC , V = VC . Then A can be
put into Jordan canonical form.
If sp(A) = {λi }, V = ⊕i Mλi . The generalized eigenspace decomposition, and pi =
projection onto Mλi . The generalized
eigenspace decomposition, and pi =projection
Pr
onto Mλi . The operator D =
i=1 λi Ti is diagonalized by any basis X that runs
through Mλ1 , Mλ2 , · · · in succession and if this basis is suitably chosen A − D = N is
upper triangular, with [D, N ] = 0. Now


J1
0


r
·

 X
=
·
mi λi
T r(A) = T r(D + N ) = T r 




i=1
·
0
Jr
5
where

∗
λi
·


Ji = 







·
·
0

λi
with (mi = dim(Mλi ) the algebraic multiplicity of λi in pT (x)). Hence,


E1
0


·


A


·
e =



·
0
Er
where

eλi
∗
·


Ei = 


·
·
eλi
0
So that
A
det(e ) =
r
Y
i=1
λi m i
(e )
=
r
Y
emi λi = e






P
i
mi λi
= eT r(A)
i=1
Exercise 3: Find A ∈ M (2, C) such that LA (x) = A · x is diagonalizable, but not
orthogonally diagonalizable with respect to the standard inner product on C2 .
Solution: Let {e1 , e2 } the standard ON basis in C2 ; f1 = e1 , f2 = e1 + e2 and define
linear operator T : C2 → C2 such that T (f1 ) = f1 , T (f2 ) = 2f2 . Obviously, T is
diagonalized
over the bases {f1 , f2 }. Its matrix with the standard basis X = {e1 , e2 } is
1 1
A=
since T (e1 ) = e1 , T (e2 ) = e1 + 2e2 . The eigenspaces Eλ (T ), λ = 1 and
0 2
λ = 2, are uniquely determined and are not orthogonal. Therefore no ON basis can
diagonalize T .
Exercise 4: Let V be an inner product space T : V → V diagonalizable. Prove that T
is orthogonally diagonalizable if and only if the eigenspace Eλ (T ) in V = ⊕λ∈Sp(T ) Eλ (T )
are orthogonal.
Solution: The Eλ (T ) are defined without reference to any inner product on V . If the
Eλ (T ) are orthogonal, we may take ON basis in each Eλ (T ); combining then we get
an ON basis for V consisting of T -eigenvectors (because T (ei ) = λei , for any vector
ei ∈ Eλ (T ).) That proves (⇐).
On the other hand, if there is an ON basis {ei } such that T (ei ) = λei , 1 ≤ i ≤ n, we
6
may lump together the ei associated with a single eigenvalue λ.
If sp(T ) = {µj : 1 ≤ j ≤ r} and we let Ij = {i ∈ {1, n} : λi = µj }. Then [1, n] = ∪rj=1 Ij
(disjoint union) and Vj = (⊕i∈Ij Cei ) is included the eigenspace Eµj (T ). Obviously,
these eigenspaces are orthogonal (since ei ⊥ ei , if i 6= i0 in [1, n]) and V = ⊕rj=1 Eµj (T )
since T is diagonalizable by hypotheses. But V = ⊕rj=1 Vj (orthogonal direct sum ) and
Vj ⊆ Eµj (T ) then Vj = Eµj (T ) if V is finite dimensional. Since Vi ⊥ Vj if j 6= j 0 , we
must then have Eµj (T ) ⊥ Eµ0j (T ) for j = j 0 and the eigenspace Eµj (T ) are orthogonal,
as claimed.
Exercise 5:
Decide whether
0
1. A =
1

1
2. A =  0
0
A, B are unitarily equivalent.
1
0 1/2
and B =
0
1/2 0



1 0
1 0 0
2 2  and B =  0 2 0 .
0 3
0 0 3
Solution:
1. If B = U AU ∗ then in particular B is similar to A (conjugate by some 2 × 2
matrix with det(S) 6= 0). But sp(A) = roots of λ2 − 1 = {−1, 1}, sp(B) =
roots of λ2 − 1/4 = {−1/2, 1/2}. But sp(SAS −1 ) = sp(A) for any matrix.
2. Now sp(A) = sp(B) = {1, 2, 3} (the diagonal entries of an upper triangular matrix
are its eigenvalues).
Solving (A − λI)x = 0 for λ = 1, 2, 3, e compute the eigenspaces for A.
 
0
Eλ=1 = C  0  = Ce3 , Eλ=2 = C(e1 + e2 ), Eλ=3 = C · (e1 + 2e2 + e3 )
1
Obviously, B is orthogonally diagonalizable, diagonalized by X = {e1 , e2 , e3 } (ON
basis). If B is unitarily equivalent to A, with A = U BU ∗ for some U ∈ U (2),
then U ∗ AU = B then A is diagonalized by the ON basis Y = U (X ), vectors in Y
are eigenvectors, but the eigenvectors are obviously not orthogonal. Hence B is
not unitarily equivalent to A.
Exercise 6:
If T is a unitary operator on a finite dimensional inner product space. Prove that it
has a unitary square root U (such that U 2 = T ).
Solution: T being normal is orthogonally diagonalizable say with ON bases of eigenvectors. T (fi ) = λi fi with each λi ∈ sp(T ) (reapeats allowed). These eigenvalues must
7
have |λ| = 1. In fact if U v = λv, (v 6= 0) then
||v||2 = (v, v) = (U v, U v) = (λv, λv) = λλ̄(v, v) = |λ|2 ||v||2
hence |λ|2 = 1, it has polar form λ = eiθ = cos(θ) + i(sin(θ)), for some real λ; then
µ = eiθ/2 satisfies µ2 = λ and |µ| = 1.
Define U , so
 iθ /2

e 1
0


·



·
[U ]Y = 




·
iθn /2
0
e
when Y is the ON basis of eigenvectors for T . Obviously U ∗ U = I since

 −iθ /2
0
e 1


·


∗


·
U =



·
−iθn /2
0
e
Exercise 7:
Let U : V → V be a linear operator on a finite dimensional inner product space. If
||U (x)|| = ||x||, for all x ∈ V (U is an isometry), is U necessarily a unitary operator?
Solution: If U is a bijection, the polarization identities imply that (U x, U y)V = (x, y)V ,
making U unitary. Since dim(V ) < ∞, U is surjective if and only if it is one-to-one if
and only if ker(U ) = {x ∈ V : U x = 0} is trivial. But U x = 0 if and only ||U x||2 = ||x||2
if and only if x = 0 in V . The answer is yes.
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