Consider the linear transform on ℜ3 corresponding
to contraction in the x-axis by a factor of 1/2 followed by
rotation of 30°(=π/6) counter-clockwise about the z-axis.
Find the determinant of the standard matrix for this transform.
[2] 1/2
Construct the matrix representing this transformation. [4]
√3/4
-1/2
0
1/4
√3/2
0
0
0
1
(the order of matrix multiplication is important!)
Consider the class of rotations about the origin in
ℜ2, by an angle of θ. For every value of
θ between 0 and 2π, find all eigenvalues and eigenvectors
of the rotation. This question may be solved geometrically, without
explicitly constructing the transform matrices. [8]
Geometric reasoning is probably a better approach here than algebraic
manipulation using the characteristic equation (det(λI-A)=0):
When θ = 0 or 2π, all x in ℜ2 are eigenvectors,
with λ = 1.
When θ = π, all x in ℜ2 are eigenvectors,
with λ = -1.
For all other θ, there are no eigenvectors.
Consider the following 4x6 matrix A:
1
-2
2
-1
0
2
-3
6
-1
0
2
2
2
-4
1
0
2
0
1
2
3
0
2
-2
Find the rank of the matrix. [4] 4
Find a basis for the rowspace of the matrix. [2] {R1, R2, R3, R4}
(i.e., all the rows; the matrix is full-rank.)
Find a basis for the columnspace of the matrix. [2] {C1, C2, C3, C4}
What dimension is the nullspace? [1] 2
Find all vectors x for which Ax = 0. [2] {C5, C6} form a basis for
the nullspace, so all linear combinations of those two columns.
Using either the Vandermonde or Newton methods, find the
lowest-degree polynomial that interpolates through the points
(-1, 2), (0, -1), (1, 0), (2, 1). Check your answer. [6]
-1 -(1/3) x +2 x2 -(2/3) x3
Consider the unit circle in ℜ2, i.e.,
all vectors with unit length.
Using the standard operations on ℜ2,
is the unit circle a vector space?
(If so, find the zero element (additive identity) and
negatives (additive inverses).
If not, it's sufficient to demonstrate one axiom that is false.) [4] No: (0,0) is not in the set.
Let's redefine vector addition on the unit circle as
(cos(θ1), sin(θ1)) +
(cos(θ2), sin(θ2)) =
(cos(θ1 + θ2),
sin(θ1 + θ2)).
Also, redefine scalar multiplication on the unit circle as
k(cos(θ), sin(θ)) = (cos(kθ), sin(kθ)).
With these new operations, is the unit circle in ℜ2
a vector space? [5] Yes:
e.g., zero element is (cos(0),sin(0)) = (1,0).
Negative is -(cos(θ), sin(θ)) =
(cos(2π-θ), sin(2π-θ)).