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Unformatted text preview: 1 1 as = y4 y 0 .
1
1 −1 1
. A basis B for V ⊥ is a vector ,
a
Answer: V ⊥ is the line spanned by a vector = a
0
1
that is, B = { }. The dimension of V ⊥ is 1.
a (b) The dimension of V is “the dimension of the large space, R4 , minus the dimension of
its perpendicular complement V ⊥ .” Therefore, the dimension of V is = 4 − 1 = 3.
A basis B for V can be formed either by columns of A, or by rows of any reduced form
of AT . Thus, it be given, B = {1 , 2 , 3 }
as v v
v can 0
0
1 0 1 0
, .
or as B = , 0 1 0 0
−1
1
Both versions of B span space V (the ﬁrst by deﬁnition, second by the property of matrix
reduction). They must be independent, since the dimension of V is 3. (This can be deduced
either as above, or by counting number of pivots in the reduced version of the matrix AT .)
(c) The formula for the matrix representing the orthogonal projection onto the column
space of a matrix A is A(AT A)−1 AT . We will use it ﬁrst to ﬁnd P ⊥ , for which the matrix A
is given by a single column . So, AT = (−1, 1, 0, 1) · (−1, 1, 0, 1) = 1 + 1 + 0 + 1 =...
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This document was uploaded on 02/19/2014 for the course MATH 441 at WVU.
 Spring '08
 STAFF
 Math, Linear Algebra, Algebra

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