## Transcribed Text

Problem 1. Suppose V is a linear space over R, with basis
e1 e2 e3
. U, W is two subspace
of V. For each of the following problem, Find a basis of U ∩ V
(1) U = {ae1 + be2 + ce3|a + b + c = 0} , V = {pe1 + se2 + ce3|3p + 2s + c = 0}
(2) U = {ae1 + be2 + ce3|2a + b + 3c = 0} , V = {be1 + ae2 + ce3|4b + 2a + 6c = 0} (Caution:
Do some substitution to make sure the scalar in front of each vector of basis are the same)
Problem 2. Suppose V is a linear space over R, with basis
e1 e2 e3 e4
. U, W is two
subspace of V. For each of the following problem, Find a basis of U + V
(1) U = {ae1+be2+ce3+de4|a+b+c = 0; a+b = 0} , V = {ae1+be2+ce3|2a+2b+c = 0; c = 0}
Problem 3. Suppose V is a linear space over R, with basis
e1 e2 e3 e4
,T : V −→ V is
a linear transformation, For each of the following case, determine whether the subspace W is an
invariant subspace of T. I’ll show an example over here.
Example:
T
e1 e2 e3 e4
=
e1 e2 e3 e4
2
4
5
−9
0
9
5
−12
0
8
−1
−5
1
0
1
0
W = {ae1 + be2 + ce3 + de4|a + b + c + d = 0}
We just want to show that TW ⊂ W. We see, we like the matrix form, rewrite
W = {ae1 + be2 + ce3 + de4|
1 1 1 1
a
b
c
d
= 0}
Suppose v =
e1 e2 e3 e4
a
b
c
d
, then
(2) U = {ae1 + be2 + ce3|a + b + 3c = 0} , V = {ae1 + be2 + ce3|4b + 2a +
Tv = T
e1 e2 e3 e4
a
b
c
d
=
e1 e2 e3 e4
2
4
5
−9
0
9
5
−12
0
8
−1
−5
1
0
1
0
a
b
c
d
We want to see the final vector is in W or not, we discover the coordinate of this final vector is
2
4
5
−11
−2
9
5
−12
0
6
−1
−5
1
0
−1
0
a
b
c
d
To see whether this vector is in W or not, we just test left multiply by
1 1 1 1
, and we
see
1 1 1 1
2
4
5
−9
0
9
5
−12
0
8
−1
−5
1
0
1
0
a
b
c
d
=
2 2 2 2
a
b
c
d
= 2
1 1 1 1
a
b
c
d
= 2 ∗ 0
= 0
The last equal is true because the vector is already taken in W, and the coordinate of vector in
W satisfies this equality.
This means Tv ∈ W. so indeed, W is an invariant subspace. Determine whether the following
are invariant subspaces of T.
(1)
T
e1 e2 e3 e4
=
e1 e2 e3 e4
5
6
3
−9
2
7
0
−12
0
5
0
9
−2
3
0
0
W = {ae1 + be2 + ce3 + de4|3a + 2b + c
(2)
T
e1 e2 e3 e4
=
e1 e2 e3 e4
9
0
−10
−9
3
−1
−3
−12
11
−1
−2
9
10
−1
0
0
W = {ae1 + be2 + ce3 + de4|a + 10b + c = 0}
4
Problem 4. Suppose V is 4-dimensional space over R, with chosen basis
e1 e2 e3 e4
, T
is a linear transformation T : V −→ V , and have the matrix representation:
T
e1 e2 e3 e4
=
e1 e2 e3 e4
1
−2
0
1
1
−2
0
1
0
0
1
0
1
−2
0
1
(1) Find a basis of Im T.
(2) We learned in class that image and kernel are all invariant subspace for linear transformation. With your basis for invariant subspace Im T you found above, find the matrix
representation of restriction T|Im T
T|Im T : Im T −→ Im T
(3) Find a basis of Im T2 What is the dimension of Im T2?
(4) With your basis for invariant subspace Im T2
, find the matrix representation of restriction
of T|Im T2
T|Im T2 : Im T2 −→ Im T2
(5) What is the dimension of Im T3? Does the map T|Im T2 : Im T2 −→ Im T2 an isomorphism?
(6) Draw the power-rank curve for T, (means the graph with x axis the power of T, and y axis
with
Problem 5. Suppose T : V −→ V is a linear map, and dim V = 5 dim Im(T) = 3, T
3 = 0.
Without using the result in the class. Analise the following.
(1) What is the dimension of Ker(T
2
) ∩ Im(T)?(Does everything in Image of T killed by T
2?)
(2) What is the dimension of Ker(T)? Why?(Using dim(ker) + dim(im) = dim(domain))
(3) What is the dimension of ker(T)∩Im(T) at most be? (it at most be the dimension of kernel
of T)
(4) In the above, we know the dimension of ker(T) ∩ Im(T) could be 0,1,2. Now we assume if
the dimension of ker(T) ∩ Im(T) is 1, then what is the dimension of Im(T
2
)?
(5) With assumption in (4), if ker(T) ∩ Im(T) is 1, then what is the dimension of ker(T) ∩
Im(T
2
) at most be? why? (ker(T) ∩ Im(T
2
) is a subspace of ker(T) ∩ Im(T))
(6) Continue assumption in (4), if ker(T)∩Im(T
2
) at most 1, what is the dimension of Im(T
3
)
at least be? why? but we already assumed T
3 = 0, show that then the assumption in (4) is
wrong.
(7) Analise what will happen if ker(T) ∩ Im(T) = 0?(Hint, then restriction on Im T would be
isomorphism. Isomorphism all the way.)
(8) With above analysis, we know the only case is dim(Ker(T) ∩ Im(T)) = 2, then find
dim(Im(T
2
))
Problem 6. Suppose T : V −→ V is a linear map, and dim V = 6, dim Im(T
3
) = 3, dim Im(T
5
) =
2Using the non-increasing and non-accelerating property of the rank curve we learned in class, answer the following question.
(1) Draw the power-rank curve, fits other power-rank, what is the rank-stable number?
(2) Look at your curve, quick answer the following question. (Sometimes the formula dim(ker)+
dim(Im) = dim(Domain) is super useful!)
1 What is the dimension of ImT ∩ KerT2?
2 What is the dimension of ImT2 ∩ KerT3?
3 What is the dimension of ImT ∩ KerT3?
4 For what number n does the restriction of linear transformation T on ImT n
is an isomorphism?
Problem 7. Suppose A is a 10 × 10 square matrix, we know rank(A3
) = 4, A6 = 0, Draw the
power-rank curve, fits other ranks, and for each rank say reasons why you fits there. Then find
rank(A), rank(A2
), rank(A4
), rank(A5
)
9
Problem 8. Suppose T,R is two commute linear transformation of linear space V, that is, T R =
RT, then show that kerR and ImR are invariant subspaces for T.

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