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Unformatted text preview: stating the dimension of V ⊥ . 1
1
1
0
1
1
1 = , 2 = , 3 = .
v
0 v
1 v
0
1
0
0 (b) Give a basis for V and state the dimension of V . Justify, why the vectors in the basis
provided as your answer indeed form a basis of V .
(c) Find the matrices P and P ⊥ whose application result in the orthogonal projections on
V and on V ⊥ , respectively. (d) For what value of a parameter p, p being a real number, the projection of a vector 2
3p
0 w=
1 , onto space V is the shortest? What is the length of such shortest
0
vector?
Solution: (a) V ⊥ is equal to the left null space of A = [1 2 3 ], that is, all vectors
vvv
y
satisfying AT = 0. To solve it, note that reduction of AT is as follows
y 10
0
1
100
1
1001
100
1
1 −1 3 → 0 1 0 −1 +r
−r
AT = 1 1 1 0 1 → 0 1 1 −1 → 0 1
0 1 0 −1 −2
r
0 0 −1
0 ×(−1)
001
0
1 1 0 0 −1
r y
100
1 1
y
Therefore, system AT = 0 is equivalent to 0 1 0 −1 2 = 0, so y4 is a free variy y3 001
0
y4
able (nonpivot column) and y1 = −y4 , y2 = y4 , y3 = 0. This gives the solution of AT = 0
y...
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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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