Download MATH 304 Test 2 Solutions: Orthogonal Projections, Eigenvectors, and Line Fitting and more Exams Linear Algebra in PDF only on Docsity!
MATH 304
Test 2 Solutions February 16, 2007 Name
Put your answers in the space provided. Show your reasoning. The maximum score on the test is 30 points. Calculators may be used unless specifically restricted.
- 4 points Let A =
and let v =
. Let W be the column space of A. Write v as the sum
of two vectors, one in W and the other in W ⊥.
First observe that the columns of A are orthogonal. We first find the projection of v onto W and then subtract this projection from v to find its projection onto W ⊥.
projW v =
+
=
+
=
projW ⊥ v =
So v =
+
- 3 points Suppose that x is an eigenvector of A corresponding to the eigenvalue λ. Show that x is an eigenvector of 5 I − A. What is the corresponding eigenvalue?
(5I − A)x = 5x − Ax = 5x − λx = (5 − λ)x
So x is an eigenvector for 5 I − A and the eigenvalue is 5 − λ
- 4 points Find the equation of the line, y = b+mx, which best fits the three points (1, 1), (2, −3) and (6, −7).
The points yield three equations
b + m = 1 b + 2m = − 3 b + 6m = − 7
or Ax =
[ b m
]
= b
AT^ A =
[ 1 1 1 1 2 6
]
=
[ 3 9 9 41
] AT^ b =
[ 1 1 1 1 2 6
]
=
[ − 9 − 47
]
[ 3 9 − 9 9 41 − 47
] ⇒
[ 1 3 − 3 0 14 − 20
] ⇒
[ 7 21 − 21 0 7 − 10
] ⇒
[ 7 0 9 0 7 − 10
]
So we have b − 9 / 7 , m = − 10 / 7 and the equation of the line is y = −
x +
- Let W = span
,
,
and let v =
4a. 2 points Find the projection of v onto W. Circle your answer.
4b. 2 points Determine an orthogonal basis for W ⊥. Circle your answer twice,
projW v =
The vector
−
=
is a non-zero vector which spans W ⊥^ and so
an orthogonal basis for W ⊥^ is
or
- Let A =
and B =
A and B have the same column spaces and let us call it W.
5a. 2 points Find the vector in W which is closest to v = [2, 1 , 2 , 0]T^.
We use the columns of B since they are orthogonal. We project v onto W.
projW v =
- 4 points Find x, y, z, s, t so that the matrix U =
x 23 z 2 3
1 3 y 2 3 s^ t
is a symmetric orthogonal matrix. Circle
your answer.
By symmetry: y = s and z = 23.
By unit columns: 1 = x^2 +
( 2 3
) 2
( 2 3
) 2 so x = ± 13 1 =
( 2 3
) 2
( 1 3
) 2
By orthogonal columns: 23 x + 29 + 23 s = 0 so s = −x − 13. Since s > 0 , x = 13 and y = s = − (^23)
1 3
2 3 +^
2 3
( − (^23)
)
- 23 t = 29 − 49 + 23 t = 0 so t = (^13)
The symmetric orthogonal matrix is then
1 3
2 3
2 3 2 3
1 3 −^
2 3 2 3 −^
2 3
1 3
- 3 points Consider the following three vectors in R^4 : a 1 = [1, − 1 , 1 , −1]T^ , a 2 = [1, 1 , 0 , 0]T^ ,
a 3 = [1, − 1 , − 1 , 0]T^. Use the Gram-Schmidt process to construct an orthogonal basis with no fractions for Span(a 1 , a 2 , a 3 ).
a 1 and a 2 are already orthogonal.
t =
−^
−^
=
−^
=
Our orthogonal basis is
