1. Suppose L is a linear transformation from
to
.
a. Can L be surjective? If yes then provide an example and explain why it is surjective. If not
then prove that no linear transformation from
to
could be surjective.
b. Can L be injective? If yes the provide an example and explain why the example is injective.
If not then prove that no linear transformation from
to
could be injective.
Hint:
Think of L(X) =
X
or
L(X) =
2. Suppose L is a linear transformation from
X
to
.
a. Can L be surjective? If yes the provide an example and explain why it is surjective. If not
then prove that no linear transformation from
to
could be surjective.
b. Can L be injective? If yes the provide an example and explain why the example is injective.
If not then prove that no linear transformation from
to
could be surjective.
Hint:
Think of L(X) =
3. Let T :
X
->
or L(X) =
X
be defined by
a. Prove that T is a linear transformation.
Hint: You must check that
and
Equivalently you can verify that
b. Calculate the image of T ( that is, calculate a basis for the image of T)
hint: This can be done directly by noting that if L: V-> W is a linear transformation and B is
a spanning set for V then L(B) is a spanning set for the image of L. Here
so B =
{1,x} and L(B) = { L(1), L(x)} =
is a spanning set for the image of L. We can check directly that these are independent so this
must be a basis for the image.
c. Calculate the Kernel of L ( that is, calculate a basis for the kernel of T)
hint: For any linear transformation L, defined on a vector space V we have dim(Kernel(L)) +
dim(Image(L)) = dim(V)
Here we have that dim(V) =2 = dim(Image(L)) so dim(Kernel(L)) = 0 which means that
Kernel(L) = O, which ( by convention) has the empty set as its basis. The empty set is often
denoted denoted by
, or {} , but it is fine to simply denote it by "the empty set".
Another way to do (c) is to show directly that Kernel(L) = by taking
calculating
and
Recall that If L(f) = 0 then
= 0 in
but since
are
independent this says a = 0 and a+b = 0 so a=b=0 which says that f = 0. This says that the
kernel of f is O.
hint: Computational Approach One can take a direct computational approach by using the
coordinate mappings between
and
and
and
to translate the context from L:
->
to L* :
->
where L* (v) =
from
to
L*(
(v)
to
where
is the inverse of the coordinate mapping
and
is the coordinate mapping h ->
with C the basis
.
L* is given by the matrix
from
= [L*(
),
)] =
=
Since
Under
has rank 2 its columns are independent so they are a basis for the column space.
the column space of
corresponds to the image of L so
is a basis for the image.
Since the rank of
is 2 ist null space is O. Since the null space of
corresponds to
the kernel of L under the coordinate mapping we see that the kernel of L is O
4. What are the possible dimensions of subspaces of
matrices over the reals.
, the vector space of 2 by 2
For each possible dimension give an example of a subspace of
You must explain why it has that dimension.
Hint: If V is any vector space and
of that dimension.
is a basis for V then every subset of
B is independent and is therefore a basis for its linear span. Thus if
then the linear
span of S is a subspace of dimension the number of elements of S. There are subsets of B with
0,1 ..., k elements so V has subspaces of those dimensions.
V cannot have a subspace of any dimension greater than k since any k+1 elements in a vector
space of dimension k are dependent. This can be seen by using coordinates to translate the
problem to
were the k+1 vectors can thought of as the columns of a k by k+1 matrix.
5. Let A =
, <A|I> =
An REF of <A|I> is
a. Calculate the consistency matrix G of A.
Hint: G =
b. Use G to determine which of AX=
,
AX =
is solvable. Explain.
Hint: AX=B is consistent if and only GB=O
G
= 0, G
=
c. If at least one of the equations in (b) is solvable then choose a solvable one and compute its
complete solution.
AX =
is equivalent to H(AX) = H
or
X=
x = 2y -8, 10y = 30
y=3, x=-2
X=
is the complete solution
6. Suppose V is a vector space of dimension 5 and L: V ->
whose image is spanned by
S=
a linear transformation
. What is the dimension of the kernel of L?
Hint: The image of V is is the column space of
which has REF =
and hence has
rank 2. Thus the dimension of the column space is 2 so the dimension of the image of L is 2.
Since dim(V) = dim(kernel(L)) + dim(Image(L))
we see that kernel of L has dimension = 3
7. Suppose V and W are vector spaces and L: V -> W is a linear transformation from V to
W.
Suppose
is a subset of V and L(S) =
.
Suppose L(S) is a linearly dependent set. Prove that S is linearly dependent.
Hint : Suppose
0 = L(0) =
with not all
then
=
Now deduce a contradiction, using independence of L(S).
8. What are all values of a for which determinant (
Hint:
) = 10?
9. Let Q be the matrix
a.
(x is a variable)
Fill in the missing entry: Adjoint (Q ) =
Hint : the 1,2 entry of the adjoint is the 2,1 cofactor =
determinant (
)
This is also the determinant of
Alternately, since Q *Adjoint(Q) = determinant(Q)*Identity, the (1,2) entry has to be 0 so
[2,x,-1]
=
b. What is the determinant of Q?
Hint: one can calculate directly that it is
Alternately , Since Q *Adjoint(Q) = determinant(Q)*Identity, any member of the diagnal of
Q *Adjoint(Q) is the determinant.
The (1,1) entry of the product is [ 2,x,-1]
= -4 + 3x -3 = 3x-7
Note also the original "___" could be calculated once we know the determinant since for
instance the (1,2) entry has to be
0 so
[2,]
=
c. What is the inverse of Q?
Hint: The inverse of Q is
adjoint(Q)
This can also be calculated by row reduction of <A|I>
10.
a. If
A=
, B=
and C = A- xB (x a variable) then
determinant (C) = _______
Hint:
b. If
A=
determinant (C) = _______
Hint: - (1-x)*(5-x)*(3-x)
,
B=
and C = A- xB (x a variable) then
11.
Suppose V is a vector space and
Let f =
a. If
, g=
is a basis for V
, h=
and W is the linear span of S. Then find a basis for W.
Hint : Use the coordinate map construct an isomorphism between V and
Under the isomorhism f ->
corresponds to
, g->
the column space of the matrix
, h->
and the linear span of S
which has REF =
Thus {f,g} is a basis for W.
b. Suppose L: V -> V is a linear transformation from V to V such that
Find a basis for the kernel of L
Hint : Under the isomorphism between V and
defined by the coordinate mapping the
linear transformation L corresponds to L*:
->
which is given by the matrix
=
which has an REF =
A basis for the nullspace of
+
is then
,
which corresponds to {
}
which is thus a basis for the kernel of L
12. Prove that any subset of an indpendent set is independent
Hint:
Let S be an independent and T a subset of S. Suppose a linear combination of the
elements of T is O.
Add to this combination the sum of terms, each consisting of
members of S that are not in T.
where
ranges over all
The result is a linear combination of all of the members of S which is O. Since S is independent
all of the coefficients is 0. But this includes all of the coefficients of T so those are all 0. This
implies that T is independent.
13. If S is a linearly independent set and W is the linear span of S then, in terms of S, what is
the dimension of W?
Hint: S is a linearly independent spanning set for W. Thus it is a basis for W. Thus the
dimension of W is the number of elements in S.
14. Give an example of set of vectors with 5 non-zero vectors in
dimension 2
Hint : Consider
15. If L is a linear transformation from V to W prove that L(O) = O
Hint: If v is any vector in V then L(O) =
>
=O
whose linear span has