Download MATH 425: Homework 2 Solution - Orthogonal Complements and Gram-Schmidt Process and more Assignments Linear Algebra in PDF only on Docsity!
MATH 425 Written Homework 2 Solution Radford 02/07/
- (20 points) We have noted that orthogonal complements are subspaces. Thus S⊥^ is a subspace of R^4.
(a) Let v =
x y z w
∈ R^4. Then v ∈ S⊥^ if and only if = 0 = by Lemma
4.3.1. Thus v ∈ S⊥^ if and only if
1 x + 2y + 1z + 0w = 0 4 x + 1y + 2z + 3w = 0.
Row reduction yields ( 1 2 1 0 4 1 2 3
) −→ · · · −→
( 1 0 3 / 7 6 / 7 0 1 2 / 7 − 3 / 7
) .
Therefore v ∈ S⊥^ if and only if
x y z w
= z
which means {
,
} = {v 1 , v 2 } is a basis for S⊥. ( 10 )
(b) q 1 =
7 v 1 || 7 v 1 ||
.
w 2 = v 2 − q 1
−
−
Thus q 2 =
31 w 2 || 31 w 2 ||
. An answer is {q 1 , q 2 }. ( 10 )
- (20 points) From the calculation
( x y
) ,
( z w
) > =
( x y
)t ( 1 2 2 7
) ( z w
)
( x y
) ( z + 2w 2 z + 7w
) = xz+2xw+2yz+7yw
we have
<
( x y
) ,
( z w
) > = xz + 2xw + 2yz + 7yw (1)
for all
( x y
) ,
( z w
) ∈ R^2. We apply the Gram-Schmidt process to the standard basis
{e 1 , e 2 } for R^2.
By (1) note that = <
( 1 0
) ,
( 1 0
) > = 1. Therefore q 1 = e 1. ( 10 ) By (1)
again
w 2 = e 2 − q 1 =
( 0 1
) − <
( 0 1
) ,
( 1 0
) >
( 1 0
)
( 0 1
) − 2
( 1 0
)
( − 2 1
) .
Using (1) again we see = <
( − 2 1
) ,
( − 2 1
) > = 3 and thus q 2 =
( − 2 1
) .
One answer is {
( 1 0
) ,
( − 2 1
) }. ( 10 )
- (20 points) Since V = S⊕T is an orthogonal sum, by definition any v ∈ V can be written v = s + t for some s ∈ S, t ∈ T and for all s ∈ S and t ∈ T it follows that = 0.
Thus y =
x +
(b) A =
and thus AtA =
=
.
By Theorem 4.6.1 the polynomial f (x) = a 0 +a 1 x+a 2 x^2 is determined by the linear system
AtA
a 0 a 1 a 2
= Aty; that is
a 0 a 1 a 2
=
.
There are various ways of solving this system; using row reduction or by finding the inverse of AtA. The latter can be done easily enough by computing the classical adjoint. In any
case
a 0 a 1 a 2
=
. Thus f (x) =
x −
x^2. ( 10 )
