Page 1

Review

Gram-Schmidt orthogonalization and orthogonal complement

⌅⌅ Theorem 6.3: V is an inner product space; S = {v1, · · · , vk} ✓ V is orthogonal. Then

y = P_{k}

i=1 a_{i}v_{i} ) aj = ^{hy,v}^{j}^{i}

kvjk^{2}, j = 1, · · · , k.

⌅⌅ Corollary 6.3.1: If S is orthonormal,
y = P_{k}

i=1 a_{i}v_{i} ) ai = hy, vii ) y = P_{k}

i=1hy, viivi.

⌅⌅ Corollary 6.3.2: Orthogonality implies linear independence.

Page 2

⌅⌅ Gram-Schmidt orthogonalization:

v_{1} = w_{1}

v_{2} = w_{2} ^{hw}^{2}^{,v}^{1}^{i}

kv1k^{2} v_{1}
v_{3} = w_{3} ^{hw}^{3}^{,v}^{1}^{i}

kv1k^{2} v_{1} ^{hw}^{3}^{,v}^{2}^{i}

kv2k^{2} v_{2}

· · ·

⌅⌅ Theorem 6.4: V is an inner product space; S = {w1, · · · , w^{n}} ✓ V
is linearly independent. Then letting v_{1} = w_{1} and v_{i} = w_{i}
P_{i 1}

j=1

hwi,v_{j}i

kvjk^{2} v_{j}, i = 2, · · · , n,

makes S^{0} = {v1, · · · , v^{n}} orthogonal and span(S^{0}) = span(S).

Page 3

⌅⌅ Theorem 6.5: V is an inner product space; dim(V ) < 1. Then
1. V has an orthonormal basis = {v1, · · · , v^{n}}.

2. 8x 2 V, x = P_{n}

i=1hx, viivi

⌅ That is, [x] = (hx, v1i, · · · , hx, v^{n}i)^{t}.

⌅ 2 is called a Fourier series expansion.

⌅ hx, vii are called Fourier coefficients.

⌅ kxk^{2} = hP_{n}

i=1hx, viivi, P_{n}

j=1hx, vjivji

= P_{n}

i=1 P_{n}

j=1hx, viihx, vjihvi, v_{j}i = P_{n}

i=1 |hx, vii|^{2}
: Parseval’s relation

⌅ example: C([0, 2⇡), C) with hf, gi = _{2⇡}^{1} R _{2⇡}

0 f (t)g(t)dt

= {fk(t) = e^{ikt} : k 2 Z} is an orthonormal basis.

hf, f_{k}i = _{2⇡}^{1} R _{2⇡}

0 f (t)e ^{ikt}dt : k-th Fourier coefficient of f

Page 4

⌅⌅ Corollary 6.5: V is an inner product space; dim(V ) < 1;

= {v1, · · · , v^{n}} is an orthonormal basis for V ; T is a linear
operator on V ; A = [T ] . Then A_{ij} = hT (vj), v_{i}i.

⌅⌅ orthogonal complement S^{?} of S :
S^{?} = {x 2 V : hx, yi = 0, 8y 2 S}

⌅ Do not confuse this with the set complement

⌅ S^{?} is a subspace for any set S.

⌅ S ✓ (S^{?})^{?}, but if S is a subspace, then S = (S^{?})^{?}.

⌅ {0}^{?} = V, V ^{?} = {0}

⌅ [End of Review]

Page 5

⌅⌅ Theorem 6.6. and Corollary 6.6: V is an inner product space; W
is a subspace; dim(W ) < 1; {v1, · · · , v_{k}} is an orthonormal basis
for W ;y 2 V. Then

1. y = u + z, where u 2 W is unique and closest to y, and if dim(V ) < 1,

z 2 W^{?} is unique and closest to y.

2. u = P_{k}

i=1hy, viivi.

⌅ proof:

Let u = P_{k}

i=1hy, viivi. and z = y u.

) u 2 W

To show z 2 W^{?}, we only need to
show hz, vii = 0, j = 1, · · · , k.

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Page 6

hz, vji = h(y P_{k}

i=1hy, viivi), v_{j}i

= hy, vji hP_{k}

i=1hy, viivi, v_{j}i = hy, vji P_{k}

i=1hy, viihvi, v_{j}i

= 0 [hvi, v_{j}i = ij]
) z 2 W^{?}

“unique”: Let y = u^{0} + z^{0}, u^{0} 2 W, z^{0} 2 W^{?}.

) u u^{0} = z^{0} z, u u^{0} 2 W, z^{0} z 2 W^{?} [y = u + z]

) u u^{0} = z^{0} z 2 W \ W^{?} = {0}

) u^{0} = u and z^{0} = z

“closest”: For any x in W,

ky xk^{2} = k(y u) + (u x)k^{2}

= h(y u) + (u x), (y u) + (u x)i

= ky uk^{2}+ku xk^{2}+hy u, u xi+hu x, y ui
iii. _{utz}

Page 7

) ky xk^{2} = ky uk^{2} + ku xk^{2}
[y u 2 W ^{?}, u x 2 W ]

) ky xk^{2} ky uk^{2}

A similar proof can show that z is unique and closest to y.

⌅⌅ orthogonal projection of y on W : P_{W}^{?}(y) = P_{k}

i=1hy, viivi, where {v1, · · · , vk} is an orthonormal basis for W .

⌅ A non-orthogonal projection is shown above. V = W_{1} W_{2}

⌅⌅ example: Consider the inner product space PR _{1} _{2}(R) withhf, gi =

1 f (x)g(x)dx.

We found earlier the orthonormal basis by G-S process.

= {v1, v_{2}, v_{3}} = ⇢q

12,

q3 2x,

q5

8(3x^{2} 1) .

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Page 8

Let f(x) = a + bx + cx^{2}.
hf, v1i = R _{1}

1(a + bx + cx^{2})
q1

2dx =

p2

3 (3a + c)
hf, v2i = R _{1}

1(a + bx + cx^{2})
q3

2xdx =

q2
3b
hf, v3i = R _{1}

1(a + bx + cx^{2})
q5

8(3x^{2} 1)dx = ^{2}_{3}
q2

5c f (x) = hf, v1iv1 + hf, v2iv2 + hf, v3iv3

=

p2

3 (3a+c)· q1

2+ q2

3b· q3

2x+^{2}_{3}
q2

5c· q5

8(3x^{2} 1)) = a+bx+cx^{2}
Let W = span({v1, v_{3}}) =span

✓⇢q1 2,

q5

8(3x^{2} 1)

◆

=span({1, x^{2}})
P_{W}^{?}(f ) = hf, v1iv1 + hf, v3iv3

=

p2

3 (3a + c) ·

q1

2 + ^{2}_{3}
q2

5c ·

q5

8(3x^{2} 1) = a + cx^{2}
P ^{?}

W^{?}(f ) = hf, v2iv2 = bx

Page 9

If instead we let W =^{⇠} span({v1, v_{2}}) =span✓⇢q

12, q3

2x

◆

=span({1, x})
P ^{?}_{⇠}

W(f ) = hf, v1iv1 + hf, v2iv2 = (a + ^{c}_{3}) + bx
P ^{?}_{⇠}

W^{?}

(f ) = hf, v3iv3 = c(x^{2} ^{1}_{3})

So (a + ^{c}_{3}) + bx is the polynomial in W^{⇠} that is “closest” to f, and
c(x^{2} ^{1}_{3}) is the polynomial in W^{⇠} ^{?} that is “closest” to f.

Note that span({1, x}) and span({x^{2}}) are not orthogonal comple-
ments of each other, while span({(1, 0, 0), (0, 1, 0)}) and span({0, 0, 1})
are . Why?

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Page 10

⌅⌅ Theorem 6.7: V is an inner product space; dim(V ) = n; S =
{v1, · · · , v_{k}} is an orthonormal set. Then

1. S can be extended to an orthonormal basis
{v1, · · · , v_{k}, v_{k+1}, · · · , v^{n}} for V .

2. W = span(S) ) {v_{k+1}, · · · , vn} is a basis for W^{?}.

3. For any subspace W of V , dim(V ) = dim(W )+ dim(W^{?}).

⌅ proof of 2: 8x 2 W^{?}, x = P_{n}

i=1 a_{i}v_{i} [basis for V ]

) hx, vii = ai = 0, i = 1, · · · , k [orthogonal complement]

) x = P_{n}

i=k+1 a_{i}v_{i} ^{↳} ^{View}

Page 11

Adjoint

⌅⌅ The adjoint T^{⇤} of a linear operator T on a finite-dimensional inner
product space V is another operator on V such that for any x and y
in V, hT (x), yi = hx, T^{⇤}(y)i.

⌅ On F^{n}, this correspond to the conjugate transpose A^{⇤} = A^{t} of
the matrix A. A^{⇤} is called the adjoint of A.

hx, yi = P_{n}

i=1 x_{i}y_{i} = y^{t}x = y^{⇤}x

hAx, yi = y^{⇤}(Ax) = (A^{⇤}y)^{⇤}x = hx, A^{⇤}yi
That is, (L_{A})^{⇤} = L_{A}⇤.

⌅ In R^{2}, if T is rotation by ✓, then T ^{⇤} = T ^{1} is rotation by ✓;

and if T is projection onto x axis, then T^{⇤} = T.

✓cos sin sin cos

◆

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Page 12

⌅⌅ For the derivation of adjoint, consider a scalar-valued linear trans- formation g : V ! F, which is also called a functional.

⌅⌅ Theorem 6.8: V is an inner product space; dim(V ) < 1; g : V ! F is a linear transformation. Then 9y, unique in V, such that 8x 2 V, g(x) = hx, yi.

proof: Let = {v1, · · · , v^{n}} be an orthonormal basis for V, and,
for the given g, let y = P_{n}

i=1 g(v_{i})v_{i}. Then 8x 2 V,
g(x) = g(P_{n}

i=1hx, viivi) [orthonormal basis]

= P_{n}

i=1hx, viig(vi) [linear]

= P_{n}

i=1hx, g(vi)v_{i}i = hx, P_{n}

i=1 g(v_{i})v_{i}i = hx, yi

“uniqueness” : 8x 2 V, g(x) = hx, yi = hx, y^{0}i ) y = y^{0} [Thm
6.1-5]

⌅ Note that this y is not affected by the choice of the basis.

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Page 13

⌅⌅ example: g : R^{3} ! R such that g((a1, a_{2}, a_{3})) = 2a_{1} a_{2} + 4a_{3}.
y = P_{n}

i=1 g(v_{i})v_{i} = g(e_{1})e_{1} + g(e_{2})e_{2} + g(e_{3})e_{3} = (2, 1, 4)
g((a_{1}, a_{2}, a_{3})) = h(a1, a_{2}, a_{3}), (2, 1, 4)i = 2a1 a_{2} + 4a_{3}

⌅⌅ Theorem 6.9: V is an inner product space; dim(V ) < 1; T is a
linear operator on V. Then 9T^{⇤}, a unique operator on V, such that

1. 8x, y 2 V, hT (x), yi = hx, T^{⇤}(y)i.

2. T^{⇤} is linear.

⌅ proof: Fix y and let g(x) = hT (x), yi.

) g is linear. [T is linear; h·, yi is linear.]

) 8x 2 V, g(x) = hx, y^{0}i for some y^{0}, which is unique in V. [Thm
6.8]

Define T^{⇤} such that T^{⇤}(y) = y^{0}.

) hT (x), yi = hx, y^{0}i = hx, T^{⇤}(y)i :“1”

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Page 14

“linearity”: 8x 2 V,

hx, T^{⇤}(c_{1}y_{1} + c_{2}y_{2})i = hT (x), c1y_{1} + c_{2}y_{2}i [1]

= c_{1}hT (x), y1i + c2hT (x), y2i

= c_{1}hx, T^{⇤}(y_{1})i + c2hx, T^{⇤}(y_{2})i [1]

= hx, c1T^{⇤}(y_{1}) + c_{2}T ^{⇤}(y_{2})i

) T^{⇤}(c_{1}y_{1} + c_{2}y_{2}) = c_{1}T^{⇤}(y_{1}) + c_{2}T ^{⇤}(y_{2}) [Thm 6.1-5]

⌅⌅ Theorem 6.10: V is an inner product space; dim(V ) < 1;

= {v1, · · · , v^{n}} is an orthonormal basis for V ; T is a linear
operator on V .

Then [T ^{⇤}] = [T ]^{⇤}.

proof: Let A = [T ] and B = [T ^{⇤}] .
) Bij = hT^{⇤}(v_{j}), v_{i}i [Corol 6.5]

= hvi, T^{⇤}(v_{j})i = hT (vi), v_{j}i = Aji = (A^{⇤})_{ij}

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Page 1

Review

⌅⌅ Theorem 6.6. and Corollary 6.6: V is an inner product space; W is a subspace; dim(W ) < 1; {v1, · · · , vk} is an orthonormal basis for W ;y 2 V. Then

1. y = u + z, where u 2 W is unique
and closest to y, and if dim(V ) < 1,
z 2 W^{?} is unique and closest to y.

2. u = P_{k}

i=1hy, viivi.

⌅⌅ orthogonal projection of y on W : P_{W}^{?}(y) = P_{k}

i=1hy, viivi, where {v1, · · · , vk} is an orthonormal basis for W .

Page 2

⌅⌅ Theorem 6.7: V is an inner product space; dim(V ) = n; S =
{v1, · · · , v_{k}} is an orthonormal set. Then

1. S can be extended to an orthonormal basis
{v1, · · · , vk, v_{k+1}, · · · , v^{n}} for V .

2. W = span(S) ) {v_{k+1}, · · · , v^{n}} is a basis for W^{?}.

3. For any subspace W of V , dim(V ) = dim(W )+ dim(W^{?}).

⌅⌅ The adjoint T^{⇤} of a linear operator T on a finite-dimensional inner
product space V is another operator on V such that for any x and y
in V, hT (x), yi = hx, T^{⇤}(y)i.

⌅ On F ^{n}, this correspond to the conjugate transpose A^{⇤} = A^{t} of the
matrix A. A^{⇤} is called the adjoint of A.

hAx, yi = y^{⇤}(Ax) = (A^{⇤}y)^{⇤}x = hx, A^{⇤}yi. That is, (LA)^{⇤} = L_{A}⇤.

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Page 3

⌅⌅ Theorem 6.8: V is an inner product space; dim(V ) < 1; g : V !
F is a linear transformation (functional). Then 9y, unique in V,
such that 8x 2 V, g(x) = hx, yi, where y = P_{n}

i=1 g(v_{i})v_{i}.

⌅⌅ example: g : R^{3} ! R such that g((a1, a_{2}, a_{3})) = 2a_{1} a_{2} + 4a_{3}.
y = P_{n}

i=1 g(v_{i})v_{i} = g(e_{1})e_{1} + g(e_{2})e_{2} + g(e_{3})e_{3} = (2, 1, 4)

⌅⌅ Theorem 6.9: V is an inner product space; dim(V ) < 1; T is a
linear operator on V. Then 9T^{⇤}, a unique operator on V, such that

1. 8x, y 2 V, hT (x), yi = hx, T^{⇤}(y)i.

2. T^{⇤} is linear.

⌅⌅ Theorem 6.10: V is an inner product space; dim(V ) < 1;

= {v1, · · · , v^{n}} is an orthonormal basis for V ; T is a linear
operator on V . Then [T ^{⇤}] = [T ]^{⇤}.

⌅⌅ [End of Review]

Page 4

⌅⌅ Corollary 6.10: A is an n ⇥ n matrix. Then LA^{⇤} = (L_{A})^{⇤}.
proof: Let be the standard basis for F ^{n}.

[(L_{A})^{⇤}] = [L_{A}]^{⇤} [Thm 6.10] = A^{⇤} = [L_{A}⇤] .

Then the uniqueness of the representation completes the proof.

⌅ This can directly be shown by taking conjugate transpose of the matrix.

hLA(x), yi = hAx, yi = y^{⇤}(Ax)

= (A^{⇤}y)^{⇤}x = hx, A^{⇤}yi = hx, LA^{⇤}(y)i ! (LA)^{⇤} = L_{A}⇤.

⌅⌅ example: T : C^{3} ! C^{3}; is the standard basis.

T (a, b, c) = (2a + ib, b 5ic, a + (1 i)b + 3c) [T ] =

0

@2 i 0

0 1 5i

1 1 i 3

1

A , [T^{⇤}] = [T ]^{⇤} =
0

@ 2 0 1

i 1 1 + i 0 5i 3

1
A
T ^{⇤}(a, b, c) = (2a + c, ia + b + (1 + i)c, 5ib + 3c)

=A

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Page 5

⌅⌅ example: Consider the inner product space PR _{1} _{2}(R) with hf, gi =

1 f (x)g(x)dx.

Let f(x) = a + bx + cx^{2} and g(x) = p + qx + rx^{2}.

When T (f) = f^{0} = b + 2cx, what would T^{⇤}(g) look like?

=

⇢q1 2,

q3 2x,

q5

8(3x^{2} 1) [orthonormal basis]

[g] = ⇣p

2p +

p2 3 r,

p6

3 q, ^{2}

p10

15 r⌘_{t}
[T ] =

0

@0p

3 0 0 0 p

15 0 0 0

1

A , [T^{⇤}] = [T ]^{⇤} =
0

@

0 0 0

p3 0 0 0 p

15 0 1 A

[T^{⇤}(g)] = [T^{⇤}] [g]

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Page 6

[T^{⇤}(g)] =
0

@

0 0 0

p3 0 0 0 p

15 0 1 A

0 BB

@

p2p +

p2 3 r

p6 3 q

2p 1510r

1 CC A =

0 BB

@ p 0

6(p + ^{r}_{3})
p10q

1 CC A

T ^{⇤}(p + qx + rx^{2}) =p

6(p + ^{r}_{3})
q3

2x +p 10q

q5

8(3x^{2} 1)

= ^{5}_{2}q + (3p + r)x + ^{15}_{2} qx^{2}
We can confirm this result by computing
hT (f), gi = R _{1}

1(b + 2cx)(p + qx + rx^{2})dx = 2bp + ^{2}_{3}br + ^{4}_{3}cq

= R _{1}

1(a + bx + cx^{2})( ^{5}_{2}q + (3p + r)x + ^{15}_{2} qx^{2})dx =
hf, T^{⇤}(g)i.^{=}

Page 8

⌅⌅ Theorem 6.11 and Corollary 6.11: V is an inner product space; T and U are linear operators on V ; A and B are n ⇥ n matrices; c is a scalar. Then the following hold.

(a) (T + U)^{⇤} = T^{⇤} + U^{⇤} (A + B)^{⇤} = A^{⇤} + B^{⇤}
(b) (cT )^{⇤} = cT^{⇤} (cA)^{⇤} = cA^{⇤}

(c) (T U )^{⇤} = U^{⇤}T^{⇤} (AB)^{⇤} = B^{⇤}A^{⇤}
(d) (T^{⇤})^{⇤} = T (A^{⇤})^{⇤} = A
(e) I^{⇤} = I I_{n}^{⇤} = I_{n}

proof of (c): 8x, y 2 V,

hx, (T U)^{⇤}(y)i = h(T U)(x), yi = hT (U(x)), yi

= hU(x), T^{⇤}(y)i = hx, U^{⇤}(T^{⇤}(y))i

= hx, (U^{⇤}T^{⇤})(y)i.

L_{(AB)}_{⇤} = (L_{AB})^{⇤} = (L_{A}L_{B})^{⇤} [Corol 6.10, Thm 2.15-6]

= (L_{B})^{⇤}(L_{A})^{⇤} = L_{B}⇤L_{A}⇤ = L_{B}⇤A^{⇤}

Page 9

proof of (d):

hT (x), yi = hx, T^{⇤}(y)i = hT^{⇤}(y), xi = hy, (T ^{⇤})^{⇤}(x)i

= h(T^{⇤})^{⇤}(x), yi
) (T^{⇤})^{⇤} = T

⌅⌅ least square approximation:

measurements: (t_{1}, y_{1}), (t_{2}, y_{2}), · · · , (t^{m}, y_{m})

approximation: Find a and b such that y = at + b, or

find a, b and c such that y = at^{2} + bt + c.

criterion: Minimize E = P_{m}

i=1[y_{i} (at_{i} + b)]^{2}, or
E = P_{m}

i=1[y_{i} (at^{2}_{i} + bt_{i} + c)]^{2}.

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Page 10

) E = ky Axk^{2}, where y = (y_{1}, · · · , y^{m})^{t}, and
A =

0

@ t_{1} 1
... ...

t_{m} 1
1

A , x =

✓a b

◆

, or A = 0

@ t^{2}_{1} t_{1} 1
... ... ...

t^{2}_{m} t_{m} 1
1

A , x = 0

@a b c

1 A

) Find x that minimizes ky Axk^{2}.

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Page 11

⌅⌅ orthogonality principle:

The minimizing x_{0} satisfies
8x, hAx, y Ax_{0}i = 0

W = R(L_{A})

= {Ax : x 2 F^{n}}

= {P

a_{i}x_{i} : x 2 F^{n}}

where a_{i} are the column vectors
of A.

So W is the column space of A.

The principle is quite general.

⌅⌅ Lemma 6.12.1: A 2 M^{m}_{⇥n}(F ); x 2 F^{n}; and y 2 F^{m}. Then
hAx, yim = hx, A^{⇤}yin.

proof: hAx, yim = y^{⇤}(Ax) = (y^{⇤}A)x = (A^{⇤}y)^{⇤}x = hx, A^{⇤}yi^{n}

⌅⌅ Lemma 6.12.2: A 2 M^{m}_{⇥n}(F ). Then rank(A^{⇤}A) = rank(A).

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nxm.mx ^{n} ^{=} ^{nxn}

Page 12

proof: (Ax = 0 ) A^{⇤}Ax = 0)
) N(LA) ✓ N(LA^{⇤}A)

) nullity(LA) nullity(LA^{⇤}A)

) rank(LA^{⇤}A) rank(LA) [dim thm: nullity + rank = n]

) rank(A^{⇤}A) rank(A) [also from Thm 3.7]

So we need only to show that N(L_{A}⇤A) ✓ N(LA).

x 2 N(LA^{⇤}A) ) A^{⇤}Ax = 0

) hA^{⇤}Ax, xi^{n} = hAx, Axi^{m} = 0
) Ax = 0 ) x 2 N(LA)

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⌅⌅ Theorem 6.12: A 2 Mm⇥n(F ); y 2 F^{m}. Then 9x0 2 F^{n} such that
1. A^{⇤}Ax_{0} = A^{⇤}y

2. 8x 2 F^{n}, kAx0 yk kAx yk
3. rank(A) = n ) x0 = (A^{⇤}A) ^{1}A^{⇤}y.

⌅ proof: Let W = {Ax : x 2 F^{n}} = R(LA) [col sp of A]

) y = u + z, u 2 W, z 2 W^{?} such that
u 2 W is unique and closest to y, and

z 2 W ^{?} is unique and closest to y. [Thm 6.6]

) 9x0 such that u = Ax_{0}. : “2”

) z = y Ax_{0} 2 W ^{?} ) 8x 2 F^{n}, hAx, y Ax_{0}i = 0
) 8x 2 F^{n}, hx, A^{⇤}(y Ax_{0})i = 0 : “1”, orthogonality

“3” follows from Lemma 6.12.2 and “1”.

⌅⌅ Corollary 6.12: A 2 Mm⇥n(F ); rank(A) = n. Then A^{⇤}A is invert-
ible.

( East Square ^{Solution})

= 0