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Vector spaces Vector spaces For all α and β ∈ F and all x and y ∈ V,
M.1 αx ∈ V M.2 (αβ)x = α(βx) M.3 α(x + y) = αx + αy M.4 (α + β)x = αx + βx M.5 There exists 1 ∈ F such that 1x = x
Some examples
x =
x 1 ... xn
(^) , xi ∈ R (or C), V = Rn^ (or Cn)
A =
a (^11). · · · a 1 n ..... ... am 1 · · · amn
(^) , xi ∈ R (or C), V = Rm×n^ (or Cm×n)
Vector spaces Vector spaces A vector is a set of elements taken from a field, F. Our most common choices for the field are R (real numbers) or C (complex numbers). Addition: x + y (where x and y are vectors) Scalar multiplication: αx, where α ∈ F and x is a vector. Defining properties A vector space, V, satisfies the following properties: For all α ∈ F and all x and y ∈ V, A.1 x + y ∈ V (closure) A.2 (x + y) + z = x + (y + z) (distributive) A.3 x + y = y + x (commutative) A.4 There exists a unique 0 ∈ V such that x + 0 = x A.5 There exists −x ∈ V such that x + (−x) = 0
Roy Smith: ECE 210a: 2.
Vector spaces Span
Given a set of vectors,
S = { v 1 , v 2 ,... , vn } , vi ∈ V, the span of S is the set of all linear combinations.
span(S) = { α 1 v 1 + α 2 v 2 + · · · + αnvn | αi ∈ F }
Properties
Clearly x ∈ span S =⇒ x ∈ V.
Furthermore, the span S is also closed and is therefore a subspace.
Proof:
Vector spaces More examples
- Functions, f : R[0, 1] −→ R, mapping [0, 1] on R to R.
- Real-valued polynomials of order n.
Subspaces
A subspace, S ⊆ V (with the same field, addition and multiplication definitions), is also a vector space.
Therefore,
- x, y ∈ S =⇒ x + y ∈ S
- α ∈ F and x ∈ S =⇒ αx ∈ S
The key property of subspaces is closure. Other properties are inherited from the vector space.
S = { 0 } is the trivial subspace.
Roy Smith: ECE 210a: 2.
Vector spaces Example 3. Vandermonde matrices.
Vm×n :=
1 x 1 x^21 · · · xn 1 −^1 (^1). x 2 x^22 · · · xn 2 −^1 .. ... ... ... 1 xm x^2 m · · · xn m−^1
,^ with^ xi 6 =^ xj for all^ i^6 =^ j.
Show that if n = m, det(Vm×n) 6 = 0. (In the case where n < m an analogous result can be stated and proven identically).
Proof. If det(Vm×n) = 0 then there exists a vector,
α. 0 .. αn− 1
(^6) = 0, such that Vm×n
α. 0 .. αn− 1
Therefore, for each i = 1,... , m, α 0 + α 1 xi + α 2 x^2 i + · · · + αn− 1 xn i −^1 = 0.
This implies that xi is a root of the n − 1 order polynomial, α 0 + α 1 x + α 2 x^2 + · · · αn− 1 xn−^1. But such a polynomial can have at most n − 1 roots which is a contradiction as n − 1 < m.
Vector spaces Example 2. Consider a square matrix as a collection of column vectors,
A ∈ Rn×n^ =
a∗ 1 a∗ 2... a∗n
(^) Column vector notation: a∗ 3 :=
a 13 ... am 3
Now consider the set of column vectors in A,
S =
a∗ 1
a∗ 2
a∗n
Show that S is a linearly independent set if and only if det(A) 6 = 0.
Basic idea of the proof:
If det(A) = 0 then there exists an eigenvector, ˆx 6 = 0, such that,
A ˆx =
∑^ n j=
x ˆj
a∗j
As there exists at least one ˆxj 6 = 0, this equation has a non-trivial solution, implying linear dependence of the vectors.
Roy Smith: ECE 210a: 2.
Vector spaces Example 4 The space of solutions, x(t) ∈ C[0, 1] (continuous functions on [0,1]), that satisfy, d^2 dt^2 x(t) +^ x(t) = 0.
What is the dimension of this space? Find a basis.
Unique representations
Given a basis, B = { b 1 ,... , bn } ∈ V.
For every x ∈ V, there exists αi, i = 1,... , n, such that,
x =
∑^ n i=
αi bi and the αi are unique.
This choice of basis for any given space is not unique.
Vector spaces Bases
A basis of a vector space is a set of linearly independent vectors which span the vector space.
Every nontrivial subspace has a basis.
The number of vectors in a basis set is the dimension of the vector space.
Example 1. The “standard” basis for Rn.
^ =:^ {^ e^1 , e^2 ,... , en^ }^.
Example 2. M^2 ×^2 : space of 2 × 2 real-valued matrices. { [ 1 0 0 0
]
[ 0 1
]
[ 0 0
]
[ 0 0
] }
or
{ [ 1 0
]
[ 0 0
]
[ 0
]
[ 0 0
] }
Example 3. S^2 ×^2 : space of symmetric 2 × 2 real-valued matrices.
What is the dimension of this space?
Roy Smith: ECE 210a: 2.
Linear transformations Example 2.
Consider U as the space of continuous functions, f (t) : R[0, ∞) −→ R[0, ∞).
Define the operator: T (f ) :=
∫ (^) x 0 f^ (t)dt.^ (more precisely written as:^ T^ (f^ )(x)) This linear because, ∫ (^) x 0 [αf^ (t) +^ g(t)]dt^ =^ α
∫ (^) x 0 f^ (t)dt^ +
∫ (^) x 0 g(t)dt,^ for all^ f^ (t) and^ g(t)^ ∈ U.
Example 3.
Define Qθ : R^2 −→ R^2 as a rotation of a vector in R^2 by a fixed angle of θ.
If u =
[x y
]
, then Qθ(u) :=
[x cos θ − y sin θ x sin θ + y cos θ
]
[cos θ − sin θ sin θ cos θ
] [x y
]
This is linear: Qθ(αu + v) = αQθ(u) + Qθ(v).
Note the correspondence with matrix multiplication.
Linear transformations Linear transformations
Given two vector spaces, U and V, over a field, F, a linear transformation is a mapping, T , from U to V (written T : U −→ V) satisfying,
- T (x + y) = T (x) + T (y) for all x and y ∈ U.
- T (αx) = αT (x) for all x ∈ U and α ∈ F.
In the case where T : U −→ U (i.e. mapping back into the same space), T is called a linear operator.
The zero transformation is: 0(x) = 0, or more precisely 0([x]U ) = [0]V.
The identity operator is: I(x) = x.
Example 1. Matrix multiplication: T : Rn^ −→ Rm. T (x) = Ax, where A ∈ Rm×n.
This is obviously linear; A(αx + y) = αAx + Ay.
Roy Smith: ECE 210a: 2.
Linear transformations Coordinate matrix representations
So now,
[T ([x]BU )]BV =
[
T
( (^) ∑n
j=
xj uj
)]
BV
∑^ n j=
[T (xj uj)]BV =
∑^ n j=
xj [T (uj)]BV =
∑^ n j=
xj
∑m i=
αij vi
We now express a vector y ∈ V, the result of a linear transformation, y = T (x), x ∈ U, in terms of the basis BV ,
[y]BV =
∑^ m i=
yivi =
y 1 ... ym
From above, we can identify each of the yi values,
yi =
∑^ n j=
xj αij =
∑^ n j=
αij xj.
This has an obvious matrix interpretation:
y. 1 .. ym
α. 11 · · · α 1 n .. ... αm 1 · · · αmn
x. 1 .. xn
=: BV [T ] BU
x. 1 .. xn
(^). (Meyer’s notation: [T ]BU BV )
Linear transformations Space of linear transformations
Given vector spaces U and V over F, the set of all linear transformations, L(U, V) := { T : U −→ V | T is linear } is itself a vector space over F.
Basis for U : BU = { u 1 ,... , un } (dim(U) = n)
Basis for V : BV = { v 1 ,... , vm } (dim(V) = m)
We can define a linear transformation T : U −→ V by its action on each of the basis vectors in U.
T will map uj (the jth basis vector in U ) to some vector in V,
[T (uj )]BV =
∑^ m i=
αij vi =
α. 1 j .. αmj
Now any x ∈ U is expressed in terms of the basis BU ,
[x]BU = ∑^ n j=
xjuj =
x. 1 .. xn
Roy Smith: ECE 210a: 2.
