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MATH 204
Test 4 Solutions December 5, 2006 Name
Put your answers in the space provided. Show your reasoning. Calculators may be used. Calculator answers without proper justification may not receive full credit. The maximum score on the test is 30 points.
- 6 points Define the linear transformation T from M 3 × 2 to M 2 × 2 by T (A) =
[ 0 1 0 0 0 1
] A
[ 1 0 1 0
] .
T (
a b c d e f
) =
[ 0 1 0 0 0 1
]
a b c d e f
[ 1 0 1 0
]
[ c d e f
] [ 1 0 1 0
]
[ c + d 0 e + f 0
]
1a. If A =
, compute T (A). T (A) =
[ 12 0 24 0
]
1b. Find a basis for the kernel of T.
A general matrix in the kernel of T is of the form
a b c −c e −e
, so a basis for the kernel of^ T^ is
,
,
,
1c. Find a basis for the range of T.
A general matrix in the range of T is of the form
[ u 0 v 0
] , so a basis for the range of T is
{[ 1 0 0 0
] ,
[ 0 0 1 0
]}
1d. Give the dimensions of the kernel, range and domain of T.
The dimension of the kernel is 4 ; the range is 2 ; and the domain is 6
- 4 points Consider the matrix A =
. Without using your calculator find (in factored form) the
characteristic polynomial of A.
det(A − λI) =
∣∣ ∣∣ ∣∣ ∣∣ ∣
1 − λ 0 2 0 0 1 − λ 0 1 0 0 6 − λ 0 1 1 2 1 − λ
∣∣ ∣∣ ∣∣ ∣∣ ∣
= (6 − λ)
∣∣ ∣∣ ∣∣ ∣
1 − λ 0 0 0 1 − λ 1 1 1 1 − λ
∣∣ ∣∣ ∣∣ ∣
= (6 − λ)(1 − λ)
∣∣ ∣∣ ∣
1 − λ 1 1 1 − λ
∣∣ ∣∣ ∣ = (6^ −^ λ)(1^ −^ λ)(1^ −^2 λ^ +^ λ
(^2) − 1) = (6 − λ)(1 − λ)(λ − 2)λ
- 3 point The matrix
has eigenvalues 1 and 5. Find bases for the corresponding eigenspaces.
λ = 1
A − I =
⇒
x 1 = x 1 x 2 = − 2 x 1 −x 3 x 3 = x 3
or
x 1 x 2 x 3
= x 1
+ x 3
A basis for λ = 1’s eigenspace is
,
A−I =
⇒
x 1 = − 12 x 2 − 12 x 3 x 2 = x 2 x 3 = x 3
or
x 1 x 2 x 3
= x 1
+x 3
A basis for λ = 1’s eigenspace is
,
λ = 5
A − 5 I =
⇒
x 1 = 0 x 2 = x 3 x 3 = x 3
or
x 1 x 2 x 3
= x 3
A basis for λ = 5’s eigenspace is
- 3 points Let T : R^2 → R^2 be the linear transformation which projects vectors in R^2 perpendicularly onto the
line spanned by the vector [1, 3]T^.
4a. Explain why 1 is an eigenvalue and why [1, 3]T^ is one of its eigenvectors.
If x is a multiple of [1, 3]T^ , then T (x) = x. I.e. x is an eigenvector with eigenvalue 1
4b. Explain why [3, −1]T^ is also an eigenvector and give its eigenvalue. (Hint: Draw a picture)
If y is a multiple of [3, −1]T^ , then T (y) = 0. I.e. y is an eigenvector with eigenvalue 0
- 4 points Without using your calculator find the eigenvalues and eigenvectors of the matrix A =
Write your answers without fractions. There will only be two linearly independent eigenvectors.
det(A − λI) =
∣∣ ∣∣ ∣∣ ∣
−λ 1 1 − 2 5 − λ − 2 0 1 1 − λ
∣∣ ∣∣ ∣∣ ∣
∣∣ ∣∣ ∣∣ ∣
−λ 0 λ − 2 5 − λ − 2 0 1 1 − λ
∣∣ ∣∣ ∣∣ ∣
∣∣ ∣∣ ∣∣ ∣
−λ 0 0 − 2 5 − λ − 4 0 1 1 − λ
∣∣ ∣∣ ∣∣ ∣
= λ
∣∣ ∣∣ ∣
5 − λ − 4 1 1 − λ
∣∣ ∣∣ ∣ =^ −λ^
( λ^2 − 6 λ + 5 + 4
) = −λ(λ − 3)^2 eigenvalues are 0 and 3
λ = 0
A− 0 I =
⇒
⇒
x 1 = − 72 x 3 x 2 = −x 3 x 3 = x 3
eigenvector is
λ = 3
A− 3 I =
⇒
⇒
x 1 = x 3 x 2 = 2 x 3 x 3 = x 3
eigenvector is
