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Thus the unique solution has ci = ui · x for i = 1, . . . , m, which means that
xk = (u1 · x)u1 + · · · + (um · x)um
and
x⊥ = x − (u1 · x)u1 − · · · − (um · x)um .
For linearity, take x, y ∈ Rn and k ∈ R. Then
x + y = (xk + x⊥ ) + (yk + y⊥ ) = (xk + yk ) + (x⊥ + y⊥ ),
with xk + yk in V and x⊥ + y⊥ orthogonal to V , so
projV (x + y) = xk + yk = projV (x) + projV (y).
Similarly,
kx = k(xk + x⊥ ) = kxk + kx⊥ ,
with kxk in V and kx⊥ orthogonal to V , so
projV (kx) = kxk = k projV (x).
Note 128. The orthogonal projection of x onto a subspace V ⊂ Rn is obtained by summing the orthogonal
projections (ui · x)ui of x onto the lines spanned by the orthonormal basis vectors u1 , . . . , um of V .
Theorem 129 (5.1.6, coordinates via orthogonal projection). For any orthonormal basis B =
(u1 , . . . , un ) of Rn ,
x = (u1 · x)u1 + · · · + (un · x)un
for all x ∈ Rn , so the B-coordinate vector of x is given by


u1 · x


[x]B =  ...  .
un · x
Proof. If V = Rn in Theorem 127, then clearly x = x + 0 is a decomposition of x with x in V and 0
orthogonal to V . Thus
x = xk = projV (x) = (u1 · x)u1 + · · · + (un · x)un .
Definition 130 (5.1.7). The orthogonal complement V ⊥ of a subspace V ⊂ Rn is the set of all vectors
x ∈ Rn which are orthogonal to all vectors in v ∈ V :
V ⊥ = {x ∈ Rn : v · x = 0 for all v ∈ V }.
Theorem 131. For a subspace V ⊂ Rn , the orthogonal complement V ⊥ is the kernel of projV . The image
of projV is V itself.
Proof. Note that x ∈ V ⊥ if and only if x⊥ = x, i.e. projV (x) = xk = 0.
Any vector in the image of projV is contained in V by definition. Conversely, for any v ∈ V , projV (v) = v,
so v is in the image of projV .
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