Stat 511 Lecture 5 "Handout"
The following are equivalent regarding c ∈ ℜk
1) ∃ a ∈ℜn such that a′Xβ = c′β ∀β
2) c ∈ C ( X′ ) ( c′ is a linear combination of the rows of X )
3) Xβ1 = Xβ 2 ⇒ c′β1 = c′β 2
Condition 3) is a condition about the lack of ambiguity of the value of c′β . Condition 1) is
c′β = a′Xβ = a′EY = Ea′Y
so that a′Y can be used to estimate c′β in an unbiased fashion.
Proof: First suppose that 2) holds. c ∈ C ( X′ ) ⇒ ∃ a such that c′ = aX . So for this a ,
c′β = a′Xβ ∀β and 1) holds.
Next suppose that 1) holds, i.e. that ∃ a ∈ ℜn such that a′Xβ = c′β ∀β . Suppose Xβ1 = Xβ 2 .
For this a ,
c′β1 = a′Xβ1 = a′Xβ 2 = c′β 2
and 3) holds.
Finally, suppose that 3) holds. It is "obviously" equivalent to write
X ( β1 − β 2 ) = 0 ⇒ c′ ( β1 − β 2 ) = 0 ∀ β1 , β 2
That is, it is equivalent to 3) to write
Xd = 0 ⇒ c′d = 0 ∀ d
Then, the claim that 3) ⇒ 2) is the claim that
{c | [ Xd = 0 ⇒ c′d = 0 ∀ d] holds} ⊂ C ( X′)
Suppose that c is such that [ Xd = 0 ⇒ c′d = 0 ∀ d ] holds. Write
c = PX′c + ( I − PX′ ) c
and let d* = ( I − PX′ ) c . We can then argue that d* = 0 as follows. Clearly, d* ∈ C ( X′ ) so it
⊥
must be that c′d* = 0 from the condition
[ ].
But
c′d* = c′ ( I − Px′ ) c = c′c − c′Px′c
so that c′c = c′Px′c . But
c′c = c′ ( PX′ + ( I − PX′ ) ) c
= c′PX′c + c′ ( I − PX′ ) c
= c′PX′c + c′ ( I − PX′ )′ ( I − PX′ ) c
Then since c′ ( I − PX′ )′ ( I − PX′ ) c = d*′d* , we have d*′d* = 0 so that d* = 0 . Thus c = PX′c so that
c ∈ C ( X′ ) . ,