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Math 2270-2 Midterm 2, February 27, 2009
Solutions
Problem 1 (20 points). Consider the linear transformation from R^3 into R^3 given by the matrix
A =
Find:
a) A basis of the kernel of this linear transformation; b) The nullity of this linear transformation.
Solution: Multiplying the first row by 4 and subtracting from the second row, and multiplying it by 7 and subtracting from the third row we get (^)
Dividing the second row by −6 and the third by −12 we get
Hence by subtracting the second row from the third, we get:
Multiplying the second row by 3 and subtracting it from the first one we get (^)
What is the reduced row echelon form of A. Let
x y z
1
be a vector in the kernel. Then x, y and z satisfy the equations
x +^12 z = 0 y +^12 z = 0
Hence, we have
x y z
−^12 z −^12 z z
= −^1
z
and the kernel is spanned by the vector
In particular, nullity of A is equal to 1.
Problem 2 (20 points). Consider the linear transformation from the first problem. Find: a) A basis of the image of this linear transformation; b) The rank of this linear transformation.
Solution: Since the reduced row echelon form of A is equal to
we know that the rank of A is 2 and a basis of the image of A is given by the first two columns of A, i.e., by the vectors
(^) and
by dividing the second vector by 3 we get a slightly simpler basis
(^) and
What is the reduced row echelon form of the augmented matrix. This implies that the solutions are c 1 = − 2 t, c 2 = t and c 3 = t for arbitrary t. It follows that u, v and w are linearly dependent. Moreover, u and v are linearly independent (since c 3 = 0 implies t = 0). Hence w is redundant (since w = 2u − v).
Problem 4 (20 points). Consider the linear space P 2 of all polynomials in t of degree ≤ 2. Show that T (f ) = f ′′^ − 2 f ′^ is a linear transformation from P 2 into P 2. Find the matrix of this linear transformation in the basis { 1 , t, t^2 }.
a) Find the image and the kernel of this linear transformation. b) Is this linear transformation an isomorphism?
(Explain your answer!).
Solution:
We have T (1) = 0, T (t) = −2 and T (t^2 ) = 2 − 4 t. Hence the matrix is equal to
A =
We have
T (a + bt + ct^2 ) = 2c − 2(b + 2ct) = 2(c − b) − 4 ct.
Hence, the kernel consists of all constant polynomials. The image con- sists of all polynomials of degree ≤ 1. Since the kernel is nonzero, this map is not an isomorphism.
Problem 5 (20 points). Consider the linear transformation T from R^2 into R^2 given by the matrix
A =
[
]
Find the matrix B of this linear transformation in the basis
[
]
[
]
a) by direct calculation; b) using the change of basis matrix S.
Solution: a) We have [ 1 − 1 1 1
] [
]
[
]
[
]
[
]
and (^) [ 1 − 1 1 1
] [
]
[
]
[
]
[
]
Hence, we have
B =
[
]
b) On the other hand,
S =
[
]
and
S−^1 =
[
]
Finally, we have
B = S−^1 AS
=
[
] [
] [
]
[
] [
]
[
]
