Lin Alg 2 Flashcards
the “dot product” on Rⁿ (column vectors)
:= [ ]
vᵗw
There is also a related “dot product” on Cⁿ
:= [ ]
⁻vᵗw (v conjugate transpose)
Let V be a vector space over a field F. A bilinear form on V is a map
F : [ ]
such that for all u, v, w ∈ V, λ ∈ F: [ ]
A bilinear form on V is a map F : V × V → F such that for all u, v, w ∈ V, λ ∈ F: (i) F(u + v, w) = F(u, w) + F(v, w) (ii) F(u, v + w) = F(u, v) + F(u, w) (iii) F(λv, w) = λF(v, w) = F(v, λw).
We say a bilinear form F is symmetric if [ ]
We say a bilinear form F is non-degenerate if [ ]
We say a bilinear form F is positive definite if [ ]
We say,
F is symmetric if: F(v, w) = F(w, v) for all v, w ∈ V .
F is non-degenerate if: F(v, w) = 0 for all v ∈ V implies w = 0.
When F = R we’ll say F is positive definite if for all v =/= 0 ∈ V : F(v, v) > 0.
Note that a positive definite form is always non-degenerate (since F(v, v) cannot
be 0 for v =/= 0).
A real vector space V endowed with a bilinear, symmetric positive definite form
F(·, ·) is called an [ ]
inner product space
Let V be a vector space over C. A sesquilinear form on V is a map F: [ ]
such that for all u, v, w ∈ V, λ ∈ C : [ ]
A sesquilinear form on V is a map F : V × V → C such that for all u, v, w ∈ V, λ ∈ C : (i) F(u + v, w) = F(u, w) + F(v, w) (ii) F(u, v + w) = F(u, v) + F(u, w) (iii) F(¯λv, w) = λF(v, w) = F(v, λw). (lambda conjugate)
We say a bilinear form F is conjugate symmetric if [ ], and, if so, F(v, v)…
F(v, w) = ¯F(w, v) for all v, w ∈ V, (conjugate)
and, if so, F(v, v) ∈ R as F(v, v) = ¯F(v, v).
A complex vector space V with a sesquilinear, conjugate symmetric, positive definite form F = < ·, · >
is called…
A complex vector space V with a sesquilinear, conjugate symmetric, positive definite form F = < ·, · >
is called a (complex) inner product space
Given a real or complex inner product space, we say {w1, · · · , wn} are mutually orthogonal
if…
and are orthonormal if…
< wi, wj > = 0 for all i =/= j,
they are orthonormal if they are mutually orthogonal and < wi, wj > = 1 for each i.
Let V be an inner product space over K (equal R or C) and {w1, · · · , wn} ⊂ V
be orthogonal with wi =/= 0 for all i. Then w1, · · · , wn are…
linearly independent
Let V be an inner product space over K (equal R or C) and {w1, · · · , wn} ⊂ V
be orthogonal with wi =/= 0 for all i. Then w1, · · · , wn are linearly independent.
Prove it.
bottom of pg 37
Perform the Gram-Schmidt orthonormalisation process on B = {v1, · · · , vn}, a basis of the inner product space V over K = R, C.
Top of pg 38
Every finite dimensional inner product space V over K = R, C has an orthonormal [ ]
Prove it.
basis
done by Gram-Schmidt orthonormalisation process
Let V be an inner product space over K = R, C. Then for all v ∈ V ,
< v, · > : V → K
w → < v, w >
is a…
linear functional, as < , > is linear in the second co-ordinate
The map defined by v → < v, · > is a natural…
every complex vector space V is in particular a real vector space, and if it is finite dimensional then…
natural injective R-linear map φ : V → V′,
which is an isomorphism when V is finite dimensional.
2 dim𝒸 V = dimᵣ V.
The map defined by v → < v, · > is a natural injective R-linear map φ : V → V′,
which is an isomorphism when V is finite dimensional.
Prove it.
Note φ : v → < v, · >, so and we must first show φ(v + λw) = φ(v) + λφ(w) for all v, w ∈
V, λ ∈ R, i.e.
< v + λw, ·> = < v, · > + λ< w, · >.
And this is true. So φ is R-linear. (Note it is conjugate linear for λ ∈ C.) As < ·, · > is non-degenerate, < v, · > = ⁻< ·, v > is not the zero functional unless v = 0. Hence, φ is injective.
If V is finite dimensional, then
dimᵣ V = dimᵣ V′, and hence Im φ = V′.
Thus, φ is surjective and hence
an R-linear isomorphism.
Let U ⊆ V be a subspace of an inner product space V . The orthogonal complement is defined as…
U⊥ := {v ∈ V | < u, v > = 0 for all u ∈ U}
We have that U⊥ is a subspace of…
V
We have that U⊥ is a subspace of V
Prove it.
First 0 ∈ U⊥. Now let v, w ∈ U⊥ and λ ∈ K. Then, for all u ∈ U,
< u, v + λw > = < u, v > + λ< u, w > = 0 + 0 = 0.
- U ∩ U⊥ = …
- U ⊕ U⊥ = … (and so dim U⊥ = …)
- (U + W)⊥ = …
- (U ∩ W)⊥ ⊇ … (with equality if …)
- U ⊆ … (with equality if …)
- U ∩ U⊥ = {0}
- U ⊕ U⊥ = V if V is finite dimensional (and so dim U⊥ = dim V − dim U)
- (U + W)⊥ = U⊥ ∩ W⊥
- (U ∩ W)⊥ ⊇ U⊥ + W⊥ (with equality if dim V < ∞)
- U ⊆ (U⊥)⊥ (with equality if V is finite dimensional)
Prove
- U ∩ U⊥ = {0}
- U ⊕ U⊥ = V if V is finite dimensional (and so dim U⊥ = dim V − dim U)
- (U + W)⊥ = U⊥ ∩ W⊥
- (U ∩ W)⊥ ⊇ U⊥ + W⊥ (with equality if dim V < ∞)
- U ⊆ (U⊥)⊥ (with equality if V is finite dimensional)
pg 40
- and 4. are exercises
Let V be finite dimensional. Then, under the R-linear isomorphism φ : V → V′
given by v → < v, · >, the space U⊥ maps…
Let V be finite dimensional. Then, under the R-linear isomorphism φ : V → V′
given by v → < v, · >, the space U⊥ maps isomorphically to U⁰
(considered as R vector spaces).
Let V be finite dimensional. Then, under the R-linear isomorphism φ : V → V′
given by v → < v, · >, the space U⊥ maps isomorphically to U⁰
(considered as R vector spaces).
Prove it.
Let v ∈ U⊥. Then for all u ∈ U,
< u, v > = ¯< v, u > = 0, and hence < v, · > ∈ U⁰.
Hence Im(φ|ᵤ⊥ ) ⊆ U⁰.
We also have that dim U⊥ = dim V − dim U = dim U⁰, and so φ(U⊥) = U⁰.
(Note that φ has trivial kernel so we need only check equality of dimensions.)
Given a linear map T : V → V , a linear map T∗: V → V is its adjoint if…
Given a linear map T : V → V , a linear map T∗: V → V is its adjoint if for all v, w ∈ V ,
< v, T(w) > = < T∗(v), w >.
