Summary Advanced Lineaire Algebra (ALA) | EOR Year 2 Tilburg University

A Summary On Advanced Linear Algebra


Chapter 4.4 Coordinate Systems



Definition Basis
Let 𝐻 ⊂ 𝑉. An indexed set of vectors ℬ = {𝐛1 , … , 𝐛𝑝 } in 𝑉 is a basis for 𝐻 if

a) ℬ is a linearly independent set
b) The subspace spanned by ℬ coincides with 𝐻 (so {𝐛1 , … , 𝐛𝑝 } ∈ 𝐻) ,that is 𝐻 = Span⁡{𝐛1 , … , 𝐛𝑝 }


Theorem The Unique Representation Theorem

Let ℬ = {𝐛1 , … , 𝐛𝑛 } be a basis for a vector space 𝑉. Then ∀𝑥 ∈ 𝑉 there exist unique 𝑐1 , … , 𝑐𝑛 such
that: 𝑥 = 𝑐1 𝒃1 + ⋯ + 𝑐𝑛 𝒃𝑛


Proof

Let 𝑥 ∈ 𝑉, then ∃𝑑1 … 𝑑𝑛 ∈ ℝ such that:

𝑥 = 𝑑1 𝒃1 + ⋯ + 𝑑𝑛 𝒃𝑛
Also: 𝟎 = 𝒙 − 𝒙 = 𝑑1 𝒃1 + ⋯ + 𝑑𝑛 𝒃𝑛 − 𝑐1 𝒃1 + ⋯ + 𝑐𝑛 𝒃𝑛 = (𝑑1 − 𝑐1 )𝒃1 + ⋯ + (𝑑𝑛 − 𝑐𝑛 )𝒃𝑛
Since ℬ = {𝐛1 , … , 𝐛𝑛 } spans 𝑉, ℬ is linear independent, hence 𝒃𝑖 ≠ 𝟎 for 𝑖 = 1, … , 𝑛 so
𝑑𝑖 − 𝑐𝑖 = 0 for⁡𝑖 = 1, … , 𝑛 , so 𝑑𝑖 = 𝑐𝑖 for 𝑖 = 1, … , 𝑛.



Definition

Suppose ℬ = {𝐛1 , … , 𝐛𝑛 } is a basis for 𝑉 and 𝒙 is in 𝑉. The coordinates of 𝒙 relative to the basis ℬ
(or the ℬ-coordinates of 𝒙) are the weights 𝑐1 , … , 𝑐𝑛 such that 𝒙 = 𝑐1 𝐛1 + ⋯ + 𝑐𝑛 𝐛𝑛


So if 𝑐1 , … , 𝑐𝑛 are the ℬ-coordinates of 𝒙 then the vector in ℝ𝑛 :
𝑐1
[𝐱]ℬ = [ ⋮ ]
𝑐𝑛
Is the coordinate vector of 𝒙 (relative to ℬ), or the ℬ-coordinate vector of 𝒙. The mapping 𝐱 ↦ [𝐱]ℬ
is the coordinate mapping (determined by ℬ).

Notation note: 𝑃ℬ = [𝐛1 𝐛2 ⋯ 𝐛𝑛 ]

Here we call 𝑃ℬ the change-of-coordinates matrix form ℬ to the standard basis in ℝ𝑛 .
Left multiplication by 𝑃ℬ transforms the coordinate vector [𝐱]ℬ into 𝒙.

,Theorem

Let ℬ = {𝐛1 , … , 𝐛𝑛 } be a basis for a vector space 𝑉. Then the coordinate mapping 𝐱 ↦ [𝐱]ℬ is a one-
to-one linear transformation from 𝑉 onto ℝ𝑛 .


Proof

Because of the Unique Representation Theorem we know that⁡𝐱 ↦ [𝐱]ℬ ⁡one-to-one.
Let 𝐮 = 𝑐1 𝐛1 + ⋯ + 𝑐𝑛 𝐛𝑛 and 𝐰 = 𝑑1 𝐛1 + ⋯ + 𝑑𝑛 𝐛𝑛 be vectors in 𝑉.
Now 𝐱 ↦ [𝐱]ℬ is a linear transformation iff. :

- [𝐮 + 𝐰]ℬ = [𝐮]ℬ + [𝐰]ℬ
o 𝐮 + 𝐰 = (𝑐1 + 𝑑1 )𝐛1 + ⋯ + (𝑐𝑛 + 𝑑𝑛 )𝐛𝑛 so
𝑐1 + 𝑑1 𝑐1 𝑑1
[𝐮 + 𝐰]ℬ = [ ⋮ ] = [ ⋮ ] + [ ⋮ ] = [𝐮]ℬ + [𝐰]ℬ
𝑐𝑛 + 𝑑𝑛 𝑐𝑛 𝑑𝑛
- [𝑟𝐮]ℬ = 𝑟[𝐮]ℬ
o 𝑟𝐮 = 𝑟(𝑐1 𝐛1 + ⋯ + 𝑐𝑛 𝐛𝑛 ) = (𝑟𝑐1 )𝐛1 + ⋯ + (𝑟𝑐𝑛 )𝐛𝑛 so
𝑟𝑐1 𝑐1
[𝑟𝐮]ℬ = [ ] = 𝑟 [ ⋮ ] = 𝑟[𝐮]ℬ
⋮
𝑟𝑐𝑛 𝑐𝑛
∎



Definition Isomorphism

A one-to-one linear transformation from a vector space 𝑉 onto a vector space 𝑊 is called an
isomorphism. ‘iso’ is Greek for ‘the same’, which is an important fact as it explains that although the
terminology for 𝑉 and 𝑊 may differ, the two spaces are indistinguishable as vectorspaces.

, Chapter 5.4 Eigenvectors and Linear Transformations


Let 𝑉 ⊆ ⁡ ℝ𝑛 ⁡and 𝑊 ⊆ ⁡ ℝ𝑚 and 𝑇:⁡⁡ℝ𝑛 → ℝ𝑚 . Now choose ordered bases ℬ and 𝒞 for 𝑉 and 𝑊
respectively. Then ∀𝑥 ∈ 𝑉, [𝐱]ℬ ∈ ℝ𝑛 and [𝑇(𝐱)]𝒞 ∈ ℝ𝑚 because 𝑇:⁡⁡ℝ𝑛 → ℝ𝑚 and
∀𝒚 ∈ 𝑊:⁡[𝒚]𝒞 ∈ ℝ𝑚 . Let 𝒙 = 𝑟1 𝐛1 + ⋯ + 𝑟𝑛 𝐛𝑛 then [𝐱]ℬ = [𝑟1 ⁡…⁡⁡𝑟𝑛 ]′.
So also

[𝑇(𝐱)]𝒞 = [𝑇(𝑟1 𝐛1 ) + ⋯ + 𝑇(𝑟𝑛 𝐛𝑛 )]𝒞 = 𝑟1 [𝑇(𝐛1 )]𝒞 + ⋯ + 𝑟𝑛 [𝑇(𝐛𝑛 )]𝒞
= [[𝑇(𝐛1 )]𝒞 … [𝑇(𝐛𝑛 )]𝒞 ][𝑟1 ⁡…⁡⁡𝑟𝑛 ] = [[𝑇(𝐛1 )]𝒞 … [𝑇(𝐛𝑛 )]𝒞 ]⁡[𝐱]ℬ
Hence we can say that [𝑇(𝐱)]𝒞 = 𝑀[𝐱]ℬ where 𝑀 = [[𝑇(𝐛1 )]𝒞 … [𝑇(𝐛𝑛 )]𝒞 ].
The matrix 𝑀 is a matrix representation of 𝑇 and is called the matrix for 𝑻 relative to the bases
𝓑 and 𝓒.



Definition [𝑇]ℬ

[𝑇]ℬ = [[𝑇(𝐛1 )]ℬ ⋯ [𝑇(𝐛𝑛 )]ℬ ]



Theorem Diagonal Matrix Representation

Suppose 𝐴 = 𝑃ℬ 𝐷𝑃ℬ −1 where 𝐷 is an diagonal 𝑛 × 𝑛 matrix (diagonal=square). If ℬ is the basis for
ℝ𝑛 formed from the columns of 𝑃ℬ , then 𝐷 is the ℬ-matrix for the transformation 𝐱 ↦ 𝐴𝐱.


Proof

Let 𝑃ℬ = [𝐛1 ⋯ 𝐛𝑛 ] , 𝐴 = 𝑃ℬ 𝐷𝑃ℬ −1 and ℬ = {𝐛1 , … , 𝐛𝑛 }. Recall 𝑃ℬ [𝐱]ℬ = 𝐱 ⇔ ⁡ [𝐱]ℬ = 𝑃ℬ −1 𝐱

If 𝑇(𝒙) = 𝐴𝒙 for 𝒙 ∈ ℝ𝑛 then we have:

[𝑇]ℬ = [[𝑇(𝐛1 )]ℬ ⋯ [𝑇(𝐛𝑛 )]ℬ ]
⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡= [[𝐴𝐛1 ]ℬ ⋯ [𝐴𝐛𝑛 ]ℬ ]
⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡= [𝑃ℬ −1 𝐴𝐛1 ⋯ 𝑃ℬ −1 𝐴𝐛𝑛 ]⁡⁡⁡⁡([𝐱]ℬ = 𝑃ℬ −1 𝐱)
⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡= 𝑃ℬ −1 𝐴[𝐛1 ⋯ 𝐛𝑛 ]
⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡= 𝑃ℬ −1 𝐴𝑃ℬ

Now because 𝐴 = 𝑃ℬ 𝐷𝑃ℬ −1 ⇒ [𝑇]ℬ = 𝑃ℬ −1 𝑃ℬ 𝐷𝑃ℬ −1 𝑃ℬ = 𝐷

∎



Efficiency note Computing [𝑇]ℬ

To compute [𝑇]ℬ efficiently compute 𝐴𝑃ℬ then compute: [𝑃ℬ ⁡⁡𝐴𝑃ℬ ] → [𝐼⁡⁡⁡𝑃ℬ −1 𝐴𝑃ℬ ]
This works because [𝑃ℬ ] → [𝐼] is the same as 𝑃ℬ −1 𝑃ℬ .

, Appendix B Complex Numbers


Definition Complex number

A complex number is a number written in the form:
𝑧 = 𝑎 + 𝑏𝑖

Where 𝑎, 𝑏 ∈ ℝ and 𝑖 is a formal symbol satisfying 𝑖 2 = −1.
Here 𝑎 is the real part of 𝑧 (Re⁡𝑧) and 𝑏 is the imaginary part of 𝑧 (Im⁡⁡𝑧)
The complex number system is denoted by ℂ.


It is important to note two complex numbers are considered equal iff. their real and imaginary parts
are equal.

The complex number system has the following operations of addition and multiplication:
(𝑎 + 𝑏𝑖) + (𝑐 + 𝑑𝑖) = (𝑎 + 𝑐) + (𝑏 + 𝑑)𝑖
(𝑎 + 𝑏𝑖)(𝑐 + 𝑑𝑖) = (𝑎𝑐 − 𝑏𝑑) + (𝑎𝑑 + 𝑏𝑐)𝑖
This shows that the usual laws of arithmetic for ℝ usually also hold for ℂ.


Definition 𝑧̅

The conjugate of 𝑧 defined above is 𝑧̅ (“z bar”) which is defined by: 𝑧̅ = 𝑎 − 𝑏𝑖


Note that 𝑧𝑧̅ = (𝑎 + 𝑏𝑖)(𝑎 − 𝑏𝑖) = 𝑎2 − 𝑏 2 𝑖 2 = 𝑎2 + 𝑏 2 > 0 hence 𝑧𝑧̅ has a square root.
Hence the absolute value or modulus of 𝑧 is:
|𝑧| = √𝑧𝑧̅ = √𝑎2 + 𝑏 2
Before the following define 𝑧, 𝑤 ∈ ℂ and let 𝑧 = 𝑎 + 𝑏𝑖 and 𝑤 = 𝑥 + 𝑦𝑖
The use of the conjugate and absolute values are listed below:

1. 𝑧̅ = z iff. 𝑧 is of the form: 𝑧 = 𝑎 + 0𝑖 (a real number)
2. ̅̅̅̅̅̅̅̅
𝑤+𝑧 =𝑤 ̅ + 𝑧̅
3. ̅̅̅̅ = 𝑤
𝑤𝑧 ̅𝑧̅; in particular, 𝑟𝑧
̅̅̅ = 𝑟𝑧̅ if 𝑟 is a real number.
4. 𝑧𝑧̅ = |𝑧|2 ≥ 0
5. |𝑤𝑧| = |𝑤||𝑧|
6. |𝑤 + 𝑧| ≤ |𝑤| + |𝑧|
̅̅̅̅̅̅̅̅
𝑤 + 𝑧 = (𝑎 + 𝑥) − (𝑏 + 𝑦)𝑖 = (𝑎 − 𝑏𝑖) + (𝑥 − 𝑦𝑖) = 𝑤 ̅ + 𝑧̅
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
2 ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
̅̅̅̅ = (𝑎 + 𝑏𝑖)(𝑥 + 𝑦𝑖) = 𝑎𝑥 + 𝑎𝑦𝑖 + 𝑏𝑥𝑖 + 𝑏𝑦𝑖 = 𝑎𝑥 + 𝑎𝑦𝑖 + 𝑏𝑥𝑖 − 𝑏𝑦
𝑤𝑧
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
= (𝑎𝑥 − 𝑏𝑦) + (𝑎𝑦 + 𝑏𝑥)𝑖 = (𝑎𝑥 − 𝑏𝑦) − (𝑎𝑦𝑖 + 𝑏𝑥𝑖) =
𝑎𝑥 − 𝑏𝑦 − 𝑎𝑦𝑖 − 𝑏𝑥𝑖 = (𝑎 − 𝑏𝑖)(𝑥 − 𝑦𝑖) = 𝑤 ̅𝑧̅
|𝑤𝑧| = √𝑤𝑧𝑤𝑧 ̅̅̅̅ = √𝑤𝑤 ̅𝑧𝑧̅ = √𝑤𝑤 ̅ √𝑧𝑧̅ = |𝑤||𝑧|
|𝑤 + 𝑧| = √(𝑤 + 𝑧)𝑤 ̅̅̅̅̅̅̅̅
+ 𝑧 = √𝑤 2 + z 2 =

1
If 𝑧 ≠ 0 then |𝑧| > 0 and 𝑧 has a multiplicative inverse, denoted by 𝑧 or 𝑧 −1 :

, 1 𝑧̅
= 𝑧 −1 = 2
𝑧 |𝑧|
Now each complex number 𝑧 = 𝑎 + 𝑏𝑖 corresponds to a point (𝑎, 𝑏) in the plane ℝ2 . We call the 𝑥-
axis the real axis and we call the 𝑦-axis the imaginary axis. The conjugate of 𝑧 is the mirror image of
𝑧 in the real axis and the absolute value of 𝑧 is the distance form (𝑎, 𝑏).
In ℝ2 we use polar coordinates where 𝜑 is the angle between 𝑧 and the real axis. 𝜑 is called the
argument of 𝑧, hence 𝜑 = arg 𝑧, where −𝜋 ≤ 𝜑 ≤ 𝜋.
𝑎 𝑏
Recall that cos(𝜑) = |𝑧| and sin(𝜑) = |𝑧| hence 𝑎 = |𝑧| cos(𝜑) and 𝑏 = |𝑧| sin(𝜑)

Now because 𝑧 = 𝑎 + 𝑏𝑖 ⇔ 𝑧 = |𝑧|(cos(𝜑) + 𝑖 sin(𝜑))

Let 𝑤 = |𝑤|(cos⁡ 𝜗 + 𝑖sin⁡ 𝜗), it then follows that:

𝑧𝑤 = |𝑧|(cos(𝜑) + 𝑖 sin(𝜑))|𝑤|(cos(𝜗) + 𝑖 sin(𝜗))
⁡⁡⁡⁡⁡⁡⁡= |𝑧||𝑤|(cos(𝜑) cos(𝜗) + ⁡𝑖 cos(𝜑) sin(𝜗) + ⁡𝑖 sin(𝜑) cos(𝜗) − sin(𝜑) sin(𝜗))
⁡⁡⁡⁡⁡⁡⁡ =
⏟ |𝑧||𝑤|(cos(𝜑 + 𝜗) + 𝑖 sin(𝜗 + ⁡𝜑))⁡
𝑡𝑟𝑖𝑔𝑜𝑛𝑜𝑚𝑒𝑡𝑟𝑖𝑐⁡𝑟𝑢𝑙𝑒𝑠




Theorem De Moivre’s Theorem

For the 𝑘th power of a complex number it holds that:

𝑧 𝑘 = 𝑟 𝑘 (cos(𝑘𝜑) + 𝑖 sin(𝑘𝜑))



Theorem Main Theorem of Algebra

Every polynomial with real coefficients can be factorized into linear factors over the complex numbers
That is, if 𝑝(𝑥) = 𝑥 𝑛 + 𝑎1 𝑥 𝑛−1 + ⋯ + 𝑎𝑛 , with 𝑎𝑖 ∈ ℝ, ∃𝜆𝑖 ∈ ℂ and 𝜇𝑖 ∈ ℝ such that:
𝑘 𝑚
𝑟𝑖
𝑝(𝑥) = ∏ (𝑥 − 𝜆𝑖 )𝑟𝑖 (𝑥 − 𝜆̅𝑖 ) ∏ (𝑥 − 𝜇𝑖 )𝑠𝑖
𝑖=1 𝑖=1

With 𝑟𝑖 , 𝑠𝑖 ∈ ℕ and 2𝑟1 + ⋯ + 2𝑟𝑘 + 𝑠1 + ⋯ + 𝑠𝑚 = 𝑛


Elaboration

Consider a polynomial 𝑝(𝑥) ∈ ℝ𝑛 then 𝑝(𝑥) = 0 has exactly 𝑛 roots, of which the roots are a
combination of 𝛼⁡real⁡roots + ⁡𝛽⁡complex⁡roots, where 𝛼 + 𝛽 = 𝑛 and 𝛽 is of the form 𝛽 = 2𝑘, 𝑘 ∈
ℕ. 𝛽 must be an even number as complex numbers always come in pairs. So there exists exactly 𝑛
times (𝑥 + 𝜆𝑖 ) 𝑖 = 1, . . , 𝑛⁡in which 𝑝(𝑥) can be factorized such that 𝑝(𝑥) = 0.
Consider 𝑝(𝑥) = 𝑥 − 5 then 𝑝(𝑥) ∈ ℝ1 and has exactly 1 root for which 𝑝(𝑥) = 0, namely 𝑥 = 5
Now consider 𝑝(𝑥) = 𝑥 2 − 4 , then we can factorize 𝑝(𝑥) into (𝑥 − 2)(𝑥 + 2) hence 𝑝(𝑥) has
exactly 2 roots: 𝑥 = 2 and 𝑥 = −2. Now consider 𝑝(𝑥) = 𝑥 2 + 4 then its roots are 2𝑖 and −2𝑖 or to
be exact 0 + 2𝑖 and 0 − 2𝑖. Now consider the polynomial 𝑝(𝑥) = 𝑥 3 − 2𝑥 2 + 𝑥 − 2 then the main
theorem of algebra says that we can have either 3 real roots and no complex roots OR 1 real root and
2 complex roots. Hence 𝑥 3 − 2𝑥 2 + 𝑥 − 2 = 0 ⇔ (𝑥 2 + 1)(𝑥 − 2) = 0 thus we have roots:
𝑥 = 2⁡, 𝑥 = 𝑖⁡, 𝑥 = −𝑖