Linear Algebra

Section VI. Projection 255
first one in a way that is bound to the coordinates (that is, it fixes a basis and
then computes) and the second in a way that is more conceptual.
(a) Produce a matrix that describes the function’s action.
(b) Show also that this map can be obtained by first rotating everything in the
plane π/4 radians clockwise, then projecting into the x-axis, and then rotating
π/4 radians counterclockwise.
1.21 For �a, �b ∈ R n let �v1 be the projection of �a into the line spanned by �b,let�v2 be
the projection of �v1 into the line spanned by �a, let�v3 be the projection of �v2 into
the line spanned by �b, etc., back and forth between the spans of �a and �b.Thatis, �vi+1 is the projection of �vi into the span of �a if i + 1 is even, and into the span of �b if i + 1 is odd. Must that sequence of vectors eventually settle down—must there
be a sufficiently large i such that �vi+2 equals �vi and �vi+3 equals �vi+1? If so, what
is the earliest such i?
3.VI.2 Gram-Schmidt Orthogonalization
This subsection is optional. It requires material from the prior, also optional,
subsection. The work done here will only be needed in the final two sections of
Chapter Five.
The prior subsection suggests that projecting into the line spanned by �s
decomposes a vector �v into two parts
�v
�v − proj [�s ] (�v )
proj [�s ] (�v )
�v =proj [�s ](�v) + � �v − proj [�s ](�v) �
that are orthogonal and so are “not interacting”. We now make that phrase
precise.
2.1 Definition Vectors �v1,...,�vk ∈ R n are mutually orthogonal when any two
are orthogonal: if i �= j then the dot product �vi �vj is zero.
2.2 Theorem If the vectors in a set {�v1,...,�vk} ⊂R n are mutually orthogonal
and nonzero then that set is linearly independent.
Proof. Consider a linear relationship c1�v1 + c2�v2 + ···+ ck�vk = �0. If i ∈ [1..k]
then taking the dot product of �vi with both sides of the equation
�vi (c1�v1 + c2�v2 + ···+ ck�vk) =�vi �0
ci · (�vi �vi) =0
shows, since �vi is nonzero, that ci is zero. QED