Name:
18.700 Practice Problems for Exam 4
1.(10 points) Let T be an operator on such that T 2 = T .
(a) Find all the possible eigenvalues of T .
If T v = λv, then
λ2 v = T 2 v = T v = λv,
which implies λ2 = λ, i.e., λ is 0 or 1.
(b) Prove T can be diagonalizable.
There are various way to see this. One way is G(0, T ) = Null(T n ) = Null(T ) as
T n = T . Similarly, G(1, T ) = Null(T − I)n = Null(T − I).
A different proof will be, v = (v − T v) + T v. Since T (v − T v) = 0, (v − T v) ∈
E(0, T ); and T (T v) = T v, T v ∈ E(1, T ). So
V = E(0, T ) ⊕ E(1, T ).
2. (20 points) Let V be a inner product space.
(a) Let T be a self-adjoint operator. If λ is the eignevalue with the largest absolute
value. Prove for any vector v, kT vk ≤ |λkvk.
By the spectral theorem, we know there is a orthonormal basis, e1 , ..., en , such
that T (ei ) = λi ei . Therefore, if we write v = a1 v1 + · · · + an vn .
kT (v)k = kλ1 a1 v1 + · · · + λn an vn k
p
=
|a1 λ1 |2 + · · · + |an λn |2
p
≤ |λ| · |a1 |2 + · · · + |an |2
= |λ|kvk.
(b) Prove for any operator T on V , there exists a constant C > 0, such that
kT vk ≤ Ckvk.
There are many ways to prove this. One is following: We fix an orthonormal
basis. Denote M(T ) by (Aij ) under this basis. If we write w = (b1 , ..., bn ) = T v,
where v = (a1 , ..., an ), then
|bi |2 = |Ai1 a1 + · · · Ain an |2 ≤ (|Ai1 |2 + · · · |Ain |2 )(|a1 |2 + · · · + |an |2 ).
qP
2
So if we choose C =
i,j |Aij |
kT (v)k2 = kwk2
= |b1 |2 + · · · + |bn |2
X
≤ (
|Aij |2 ) · |a1 |2 + · · · + |an |2
i,j
2
= C kvk2 .
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3. (20 points) Let A, B be two operators on complex linear space V . Assume AB =
BA.
(a) Prove any generalized eigenspace of A is invariant under B.
Since AB = BA, we see Ak B = BAk . Therefore, if v ∈ G(λi , A) = Null(A−λi I)n ,
then
(A − λi I)n (Bv) = B(A − λi I)n (v) = 0,
which implies Bv ∈ G(λi , A), i.e. G(λi , A) is invariant under B. And this is true for
any λi .
(b) Use the above fact to prove that
V = ⊕λi ,µj G(λi , µj , V )
where λi and µj are distinct pairs of eigenvalues of A and B and
G(λi , µj , V ) = {v ∈ V |(A−λi I)p v = (B−µj I)q v = 0 for some positive integers p, q.}
Using the above result, fixing λi , we can apply the generalized eigenspace decomposition theorem for B|G(λi ,A) , i.e. the restriction of B on G(λi , A). Then we
know
G(λi , A) = ⊕µj G(µj , B|G(λi ,A) ).
Then we have
V = ⊕λi G(λi , A) = ⊕λi ,µj G(λi , µj , V ).
4.(20 points) Let N : R3 → R3 be a nilpotent operator represented by the matrix


1
1
3
2
2
6
−1 −1 −3
in the standard ordered basis. Find a Jordan basis B of R3 such that the matrix of
N relative to the basis B is in the Jordan form.
Since N 6= 0 and


0 0 0
N 2 = 0 0 0 ,
0 0 0
dim(Null(N )) = 2 and dim(Range(N )) = 1 (this can be seen either using the theorem
we proved in the class, or Range(N ) ⊂ Null(N )).
So we know that the Jordan basis will be of the form (N v1 , v1 , v2 ) for any v1 such
that N v1 6= 0 and v2 such that N v2 = 0 but v2 6= Range(N ).
5. (20 points) Find the Jordan canonical

2 1
0 2

0 0
0 0
form of the matrix

0 0
0 0
.
3 0
−1 3
2
We have the matrix with the block form
A 0
2 1
3 0
where A =
,B =
.
0 B
0 2
−1 3
So the for the first block A, v1 = e1 , v2 = e2 already gives a Jordan basis. Forthe

0
0
0 0

second block B, N = B − 3I =
. So the Jordan basis is given by e3 = 
1
−1 0
0
 
0
0

and N e3 = 
 0 . So the Jordan basis is given by
−1
(e1 , e2 , N e3 , e3 ).
6. (10 points) If T ∈ L(V ) is normal over a complex vector space V , prove that
Null(T k ) = Null(T ) and Range(T k ) = Range(T ) for all k > 0.
We apply the spectral theorem and know that there is orthonormal basis e1 , ..., en
such that T ei = λi ei . Thus Null(T ) = Span(ei | λi = 0) and Range(T ) = Span(ei | λi 6=
0).
For T k , we can use the same orthonormal basis, only the eigenvalues become
k
λi from λi . But then we see Null(T ) = Null(T k ) and Range(T ) = Range(T k ), as
whether λi is 0 if and only if the same thing for λki .
7. (10 points) If T is a normal operator on a complex vector space. Then the minimal
polynomial of T has the no repeated roots.
T is diagonalizable by the spectral theorem. Then V = ⊕m
i=1 E(λi , T ) for distinct
λi . Now consider the polynomial p(z) = (z − λ1 ) · · · (z − λm ), which has no repeated
roots by the definition.
On the other hand, for any i, write p(z) = qi (z)(z − λi ), then
p(T )|E(λi ,T ) = q(T )(T − λi I)|E(λi ,T ) = 0.
This is for any E(λi , T ), so p(T ) = 0 on V .
8. (10 points) Prove that S is an isometry if and only if all the singular values of S
are equal 1.
If S is isometry, then S ∗ S = I whose positive square root is I, so all the singular
values are all 1. Conversely, if S has all its singular values to be 1, then S ∗ S is a
positive self-adjoint operator whose eigenvalues are all 1, which means S ∗ S = I, i.e.,
S is isometry.
9. (10 points) Suppose T ∈ L(V ) and n = dim V . Prove that
RangeT n = RangeT n+1 = RangeT n+2 = · · ·
3
This is like what we did for Null space. We know
V ⊃ RangeT ⊃ RangeT 2 ⊃ RangeT 3 ⊃ · · ·
If RangeT k = RangeT k+1 for some non-negative integer k, then
RangeT k+2 = T (T k+1 (V )) = T (T k (V )) = RangeT k+1 ,
and keep going, we know RangeT k = RangeT k+m for any non-negative integer m.
Since we can not have
n > dim(RangeT ) > dim RangeT 2 · · · > dim RangeT n+1 ,
there will be some k ≤ n such that Range(T k ) = Range(T k+1 ).
10. (10 points) Suppose T, S ∈ L(V ) and ST is nilpotent. Prove that T S is nilpotent.
Since ST is nilpotent, then (ST )n = 0 for some n. Then
(T S)n+1 = T ST S · · · T S = T (ST )n S = 0.
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