Download Rank Nullity Worksheet TRUE or FALSE? Justify your answer. 1

Rank Nullity Worksheet
TRUE or FALSE? Justify your answer.
1. There exists a 4 × 5 matrix of rank 3 and such that the dimension of the space spanned by
its columns is 4.
Solution note: False. The dimension of the image is the rank of A.
2. There exists a surjective linear transformation T : R5 → R4 given by multiplication by a rank
3 matrix.
Solution note: False. Surjective means the image is all of R4 , which has dimension
4. The dimension of the image is the rank of the matrix, which is 3.
3. If A is 6 × 7 and rref (A) has 6 pivots, then the map given by multiplication by A is injective.
Solution note: True. The rank of A is 6 and the source has dimension 6. By ranknullity, the kernel has dimension 0. This means the map is injective.
4. If A is a 4 × 5 matrix, then it is possible for rank(A) to be 3 and dim(ker(A)) to be 3.
Solution note: No. dim ker + dim im = dim source = 5. So we would have 3 + 3 = 5,
which is impossible.
5. There exists a 4 × 5 matrix A of rank 3 such that dim(ker(A)) is 2.
Solution note: True. It is possible and in fact always true by Rank Nullity.
6. If A is a 4 × 5 matrix and B is a 5 × 3 matrix, then rank(A) ≤ rank(B).
Solution note: False! B could be the zero matrix, which has rank 0. But A will have
rank more than 0 if it has even one non-zero entry.
7. If matrix A has columns C1 , . . . , C4 which admit the relation C1 + C2 − C3 − 3C4 = 0, then
[1 1 − 1 − 3]T is a solution to A~x = 0.
Solution note: True. A relation on the columns is the same as a solution to the
corresponding homogenerous system of linear equations.
8. If A is a 4 × 5 matrix and B is a 5 × 3 matrix, then rank(A) ≤ rank(AB).
Solution note: False. If B is the zero matrix, so is AB. So its rank is zero, regardless
of A.
9. There does not exist a linear transformation T : R3 → R3 such that ker(T ) and im(T ) are
both lines in R3 .
Solution note: True. By rank-nullity, in that case, we would have 3 = 1 + 1.
10. If the linear system A~x = ~b has at least 5 solutions for some choice of ~b, then it must have at
least 5 solutions for any other choice of ~b.
Solution note: If it has at least 5 solutions, it must have infinitely many.
1
~
it has no solutions for some values of b. A counterexample is A =
0
0
A~x =
has infinitely many solutions, which is at least 5. But A~x =
0
solutions (is inconsistent).
But
maybe
0
. Then
0
0
has no
1
11. If A is an n × n matrix such that rank(A2 ) < n, then rank(A) < n as well.
Solution note: True. Suppose A has rank n. Since A is n × n, this means it is
invertible. So also A2 is invertible. So its rank would be n.
12. If A is a 4 × 5 matrix and B is a 5 × 3 matrix, then rank(AB) ≤ rank(B).
Solution note: True. Think about the composition linear transformation
B
A
R3 −→ R5 → R4 .
The image of AB is the same as the image of imB under A. The dimension of a
vector space can only go down under a transformation, so the dimension of imAB is
at most dim imB. So rank AB ≤ rank B.
13. If A is a 4 × 5 matrix and B is a 5 × 3 matrix, then rank(AB) ≤ rank(A).
Solution note: True. Think about the composition linear transformation
B
A
R3 −→ R5 → R4 .
The image of AB is contained in the image of A, so dimension imAB ≤ dim imA.
14. For each natural number n ≥ 1, there is a square matrix A such that An is the zero matrix.
n 0 0
0 0
Solution note: True. For any n,
=
.
0 0
0 0
15. For each natural number n, there is a square matrix A such that An is the identity matrix.
Solution note: TRUE. Think about 2 × 2 matrices as giving linear transformations
of R2 to itself. The question can be rephrased as asking whether or not there is
a transformation that, when composed with itself n times, gives us the identity
transformation. Rotation through an angle of 2π
n will work. Explicitly, the matrix
2π
2π
cos n −sin n
A=
is a square matrix such that An is the identity matrix.
2π
sin 2π
cos
n
n
16. If A is a 5 × 5 matrix such that A20 = I5 , then the rank of A is 5.
Solution note: TRUE. If A20 = I5 , then A19 A = AA19 = I5 . This means that A is
invertible, so has full rank 5.
17. The linear transformation
R2
→
R2
1
1
1
3
sending
to
and
to
is given by left
1
2
−1
−2
multiplication by a diagonal matrix.
Solution note: FALSE. It is easy to compute the matrix
of this map. We just need
1
1 1
to see where it sends ~e1 and ~e2 . Since ~e1 = 2 (
+
), we see that T (~e1 ) =
1
−1
1
1
3
2
2
1
1 1
e2 =
2 (T ( 1 + T ( −1 )) = 2 ( 2 + −2 ) = 0 . So the first column is 0 . Since ~
1
1
1
1
3
−1
1 1
e2 ) = 21 (T (
− T(
)) = 12 (
−
)=
.
2 ( 1 − −1 ), we see that T (~
1
−1
2
−2
2
2 −1
, which is NOT DIAGONAL.
So the matrix is
0 2
18. The linear transformation P7 → R sending a polynomial f to f 0 (0) has a six-dimensional
kernel.
Solution note: False. The map is surjective (since f (x) = cx is sent to c ∈ R for any
c). By rank nullity, the kernel is 7 dimensional, since the image is 1 dimensional.
19. There is exactly one linear transformation
R2
→
R2
3
1
1
0
.
to
and
to
sending
−2
−1
2
2
Solution note: TRUE. It is easy to compute the matrix of this map, so map is
completely determined
Weneed
T (~e2).
T (~
e1 ) and
information.
to find
by the given
3.5
1/2
1
3
0
0
1
.
=
) + 21 T (
+
) = T(
)=
+ 21
Compute T (~e1 ) = T (
−1
1
−1
−2
2
2
−1
0
1
to
, we know that ~e2 is sent
This is the first column. Also, since T sends
2
2
1/2
1/2
to
. So the second column of the matrix is
. Thus the map T is left
1
1
3.5 .5
multiplication by
. This is the only map with the stated properties.
−1 1
20. If we know the output of a linear transformation T : R3 → R3 on some basis {~v1 , ~v2 , ~v3 } for
R3 , then we can find the matrix of T .
Solution note: TRUE. To find the matrix, we would need to find where ~e1 , ~e2 , and
~e3 go. But each of these is a linear combination of {~v1 , ~v2 , ~v3 }, and T is linear.
So, for example, we can find ~e1 by writing ~e1 = a~v1 + b~v2 + c~v3 . Then T (~e1 ) =
aT (~v1 ) + bT (~v2 ) + cT (~v3 ). This is a column vector inR3 and the first column of the
matrix of T . Similarly, we can find T (~e2 ) and T (~e3 ) which are the second and third
columns of the matrix.
21. The matrix of the projection of R3 onto to some plane through the origin in R3 has rank 2.
Solution note: TRUE. The rank of the matrix is the dimension of the image. Clearly,
when we project onto a plane, the image of that mappign is exactly the plane, which
is dimension 2.