Types Template

20.04.2025 00:00

# Section 1 Hello Typst 2+2=41∫50𝑥𝑑𝑥 # Display 𝜃 𝜑 A B C # Linear maps 𝑇:𝑈⟶𝑉. Also a homomorphism. 𝑈:𝑉⟶𝑉 is a linear opereator or endomorphism of 𝑉. A linear transformation represents a specific geometric transformation, while the matrix representation is a set of coefficients that describe it in a particular basis. 𝑇:ℝ3⟶ℝ2 𝑇[𝑥1𝑥2𝑥3]=[2𝑥1+𝑥3𝑥1−𝑥2+𝑥3] Is 𝑇 linear? Let 𝑥,𝑦∈ℝ3 𝑇(𝑥)+𝑇(𝑦)=[2𝑥1+𝑥3𝑥1−𝑥2+𝑥3]+[2𝑦1+𝑦3𝑦1−𝑦2+𝑦3]=[2(𝑥1+𝑦1)+(𝑥3+𝑦3)(𝑥1+𝑦1)−(𝑥2+𝑦2)+(𝑥3+𝑦3)]=𝑇(𝑥+𝑦) It must hold that 𝑇(0)=0 𝑅(𝑇)={[2𝑥1+𝑥3𝑥1−𝑥2+𝑥3]:𝑥1,𝑥2,𝑥3∈ℝ}={𝑥1[21]+𝑥2[0−1]+𝑥3[11]:𝑥1,𝑥2,𝑥3∈ℝ}= Span{[21],[0−1],[11]}=ℝ2 𝑅(𝑇)=Span(𝑇(𝑏1),𝑇(𝑏2),…,𝑇(𝑏𝑛)) If 𝑇:𝑈⟶𝑉 is a linear map from a finite-dimensional vector space 𝑈, dim𝑁(𝑇)+ dim𝑅(𝑇)=dim𝑈 Rank of T = dim𝑅(𝑇)=dimcol𝑇 # Subspace To prove the existance of a subspace, it must contain 0, and be closed under addition and scalar multiplication # Bases A subset 𝔅 of a vector space 𝑉 is a basis if all 𝑣∈𝑉 can be uniquely expressed as a linear combination of finite elements in 𝔅. # Matrix representation of a linear map We have finite dimensional vector space 𝑉 over 𝐹 with basis 𝛽={𝑣1,…,𝑣𝑛}. For example, in 𝑃2(ℝ) we have basis 𝛽1={1,𝑥,𝑥2}, 𝛽2={𝑥2,𝑥,1}, 𝛽3={1,1+ 𝑥,1+𝑥+𝑥2}. Let 𝑝(𝑥)=3𝑥2+5𝑥+4. For 𝛽1 we have [𝑝(𝑥)]𝛽1=[453], for 𝛽2 we have [𝑝(𝑥)]𝛽2= [354], for 𝛽3 we have [𝑝(𝑥)]𝛽3=[−123]. 𝑇:𝑉⟶𝐹𝑛 s.t. 𝑇(𝑣)=[𝑣]𝛽. T is a linear map converting a vector to its coordinates under basis 𝛽 # Matrix vector product A matrix vector product maps the coordinates of a vectur under a (finite) basis to the coordinates of the vector under another (finite) basis. Every linear map between finite vector spaces has a unique matrix representation for every choice of basis in the domain and codomain. Examplefor 𝑇:𝐼⟶𝑉 over 𝐹, we choose 𝛽 and 𝛾 as respective bases. 𝑥⟶𝑇(𝑥) [𝑥]𝛽⟶mat 𝐴[𝑇(𝑥)]𝛾 The transformation matrix from 𝛽 to 𝛾 is defined as 𝐴=[𝑇]𝛾𝛽 (or [𝑇]𝛾←𝛽). This expands to: [[𝑇(𝑢1)]𝛾[𝑇(𝑢2)]𝛾…[𝑇(𝑢𝑛)]𝛾] In summary, transform each basis vector of 𝛽 and express it in 𝛾 Exercise from bookConstruct a matrix that mirrors vectors to the line 𝑦=2𝑥. We have 𝑇:ℝ22⟶ℝ2. Since 𝑇 is linear ∃𝐴:𝑇(𝑥)=𝐴𝑥. We know 𝐴=[𝑇]𝛽1𝛽1= [.𝑇.]𝛽1 Note that 𝑏1=[12] s.t. 𝑇(𝑏)1=𝑏1. Also, 𝑏2=[2−1] s.t. 𝑇(𝑏2)=−𝑏2. Since they are independent in ℝ2, they can be taken as a basis 𝛽. Now, we have [𝑇]𝛽𝛽= [[𝑇(𝛽1)]𝛽[𝑇(𝛽2)]𝛽]=[[10][0−1]]=[100−1]. If you want to transform 𝑥 in standard basis, you need to transform it to the basis 𝛽, apply the matrix, and transform back to standard basis. You can also construct matrix [𝑇]𝛽1𝛽=[1−221] # Matrix algebra 𝑇𝑛:𝑈⟶𝑉 (𝑇1+𝑇2)(𝑥)=𝑇1(𝑥)+𝑇2(𝑥) 𝑎∈𝐹:(𝑎𝑇1)(𝑥)=𝑎𝑇1(𝑥) 𝑇𝑛 is therefore a vector space. 𝑇1:𝑈⟶𝑉 and 𝑇2:𝑉⟶𝑊 𝑥⟶(𝑇2∘𝑇1)(𝑥) 𝐵=[𝑇1]𝛽𝛼 𝐴=[𝑇2]𝛾𝛽 𝐶=[𝑇2∘𝑇1]𝛾𝛼=𝐴𝐵 # Invertibility For a linear map 𝑇:𝑈→𝑉 we know: 𝑇is injective⟺𝑁(𝑇)={0} 𝑇is surjective⟺𝑅(𝑇)=𝑉 (𝑇−1)−1=𝑇 (𝑇2∘𝑇1)−1=(𝑇1)−1∘(𝑇2)−1 IsomorphismInvertible linear map between isomorphic vector spaces. This isomorphism can be represented by an invertible square matrix 𝐴 s.t. there exists a unique matrix 𝐵 where 𝐵𝐴=𝐼 and 𝐴𝐵=𝐼. Note that (𝐴𝐵)−1=𝐵−1𝐴−1 The following statements are equivalent: • 𝐴is invertible • The columns of 𝐴 are linearly independent • The columns of 𝐴 span 𝐹𝑛 # Determinant Read §4.5 A determaninant is a scalar computed from a square matrix. A square matrix is invertible if the columns are independent, but also if the determinant is not 0. For matrix [𝑎11𝑎21𝑎12𝑎22] if 𝑎11=0 we have [𝑎110𝑎12𝑎11𝑎22−𝑎21𝑎12] so if det(𝐴)≠0 then 𝐴 is invertible. det(𝐴𝐵)=det𝐴det𝐵 det𝐴=det𝐴𝑇 det(𝐴𝐴−1)=1 𝐴−1=1det𝐴[𝑎22−𝑎21−𝑎12𝑎11] Column linearitydet[𝑐𝒂+𝑑𝒃𝒄]=𝑐[𝒂𝒄]+𝑑det[𝒃𝒄] # Rank The following properties hold for 𝑛×𝑛matrices𝐴,𝐵: • 𝑅(𝐴)=𝑅(𝐴𝑇) • 𝑅(𝐴𝐵)≤𝑅(𝐴) • 𝑅(𝐴𝐵)≤𝑅(𝐵) • 𝐴is invertible⟹matrix multiplication preserves rank • 𝑛×𝑛matrix𝐴is invertible⟺𝑅(𝐴)=𝑛 # Diagonalization Given a linear operator 𝑇 on vector space 𝑉 is diagonalizable if there is an ordered basis𝛽 for 𝑉 such that [𝑇]𝛽 is a diagonal matrix. If we let 𝛽={𝑣𝑗}, s.t. [𝑇]𝛽 is a diagonal matrix. 𝑇(𝑣𝑗)=Φ−1𝐵(𝐿𝐷(Φ𝐵(𝑣𝑗)))= 𝐷𝑗𝑗𝑣𝑗=𝜆𝑗𝑣𝑗. The basis vectors consist of the eigenvectors of 𝑇 and 𝜆𝑖 is the eigenvalue corresponding to 𝑣𝑗. All non-zero elements of ker(𝑇) are eigenvectors with eigenvalue 0. Eigenvalues are invariant under coordinate transformation. Eigenvalues are distinct ⇒ eigenvectors are linearily independent ⇒ 𝑇 is diagonalizable # Polynomial splitting 𝑓(𝑡)=(𝑡2+1)(𝑡−2) only splits over ℂ, so in ℝ the only eigenvalue is 2. If you have a diagonalizable matrix, the characteristic polynomial splits. # Eigen If we have linear operator 𝐴 and vector 𝑣≠0 s.t. 𝐴𝑣=𝜆𝑣, 𝑣 is an eigenvector of 𝐴 with eigenvalue 𝜆. A vector 𝑣∈𝑉 with 𝑣≠0 is an eigenvector of 𝑇 corresponding to 𝜆 iff 𝑣∈𝑁(𝑇−𝜆𝐼) # Orthonormality We have space 𝐻, the space of continuous complex valued functions over [0,2𝜋]. Let𝑓𝑛(𝑡)=𝑒𝑖𝑛𝑡 We want to check if 𝑆={𝑓𝑛:𝑛is an integer} ⟨𝑓𝑚,𝑓𝑛⟩=12𝜋∫2𝜋0𝑒𝑖𝑚𝑡𝑒𝑖𝑛𝑡𝑑𝑡=12𝜋∫2𝜋0𝑒𝑖(𝑚−𝑛)𝑡𝑑𝑡=[12𝜋(𝑚−𝑛)𝑒𝑖(𝑚−𝑛)𝑡]2𝜋0=𝛿𝑚𝑛 For orthogonal bases we get the components of a vector by: 𝐯=⟨𝐯,𝑒𝑖⟩‖𝑒𝑖‖2 Proof: ⟨𝑣,𝑒𝑗⟩=𝑎𝑖⟨𝑒𝑖,𝑒𝑗⟩=𝑎𝑖𝛿𝑗𝑖‖𝑒𝑗‖2=𝑎𝑗‖𝑒𝑗‖2 # Example Let 𝑥=(1,2,4,7)∈𝔽4 and basis𝛽={(12,12,12,12),(12,12,−12,−12),(12,−12,−12,12),(−12,12,−12,12)} [𝑥]𝛽=𝑎𝑖𝛽𝑖=⟨𝑥,𝛽𝑖⟩𝛽𝑖=7𝛽1−4𝛽2+𝛽3+2𝛽4 # Gram-Schmidt Suppose 𝑣1,…,𝑣𝑚 is a linearly independent list of vectors in 𝑉. Let 𝑓1=𝑣1. For 𝑘= 2,…,𝑚 define 𝑓𝑘: 𝑓𝑘=𝑣𝑘−⟨𝑣𝑘,𝑓1⟩‖𝑓1‖2𝑓1−…−⟨𝑣𝑘,𝑓𝑘−1⟩‖𝑓𝑘−1‖2𝑓𝑘−1 The orthonormal basis is 𝑒𝑖=𝑓𝑖‖𝑓𝑖‖ # Legendre polynomials Let 𝑉=𝑃2(ℝ) with inner product ⟨𝑓(𝑥),𝑔(𝑥)⟩=∫1−1𝑓(𝑡)𝑔(𝑡)𝑑𝑡. The Gram-Schmidt procedure can be used to compute an orthogonal basis. Let standard basis 𝑒={1,𝑥,𝑥2}. It is trivial to see that this basis is not orthonormal. We find the orthonomal basis by applying Gram-Schmidt on 𝑣𝑖=𝑒𝑖: 𝑓1=𝑣1=1 ‖𝑓1‖2=∫1−11=2 # Random definitions NormalA matrix or operator 𝖳 is normal if 𝖳∗=𝖳 Self-adjoint (Hermetian)𝑇∗=𝑇 If ‖𝑇(𝑥)‖=‖𝑥‖, if 𝔽=ℂ, 𝑇 is a unitary operator, if 𝔽=ℝ, 𝑇 is an orthogonal operator.𝑇𝑇∗=𝐼 Unitarily equivalent𝐷=𝑄−1𝐴𝑄, where the columns of 𝑄 are an orthonormal basis. Orthogonal matrix𝐴∗=𝐴 Let 𝑇 be a linear operator on finite-dimensional inner product space 𝑉. If ‖𝑇(𝛽)‖=‖𝛽‖ for all 𝛽 in some orthonormal basis for 𝑉, 𝑇 is not neccesarily unary. Proof: Assume a Euclidean norm. ‖𝑇(𝑥)‖= # Adjoint problem (note: adjoint is denoted by ∗ for operator and † for matrix. Abstract index notation is used for matrices) Disprove: For every linear operator T on V, and every ordered basis 𝛽 for V, we have[𝑇]†𝛽=[𝑇∗]𝛽 Let 𝑇(𝛽𝑗)=𝐴𝑘𝑗𝛽𝑘, 𝑇∗(𝛽𝑖)=𝐵𝑘𝑖𝛽𝑘, and 𝐺𝑖𝑗=⟨𝛽𝑖,𝛽𝑗⟩. We aim to disprove 𝐴†=𝐵. ⟨𝑇(𝛽𝑗),𝛽𝑖⟩=⟨𝐴𝑘𝑗𝛽𝑘,𝛽𝑖⟩ =𝐴𝑘𝑗⟨𝛽𝑘,𝛽𝑖⟩=(𝐴𝑗𝑘)†𝐺𝑘𝑖 =(𝐺𝐴†)𝑗𝑖 ⟨𝑇(𝛽𝑗),𝛽𝑖⟩=⟨𝛽𝑗,𝑇∗(𝛽𝑖)⟩ =⟨𝛽𝑗,𝐵𝑘𝑖𝛽𝑘⟩=𝐵𝑘𝑖𝐺𝑗𝑘 =(𝐵𝐺)𝑗𝑖 𝐺𝐴†=𝐵𝐺 𝐴†=𝐺−1𝐵𝐺 𝐺𝐴†𝐺−1=𝐵 For an orthonormal basis 𝑒, 𝐺=⟨𝑒𝑖,𝑒𝑗⟩=𝛿𝑖𝑗=𝐼 so 𝐴†=𝐺−1𝐵𝐺=𝐵, or [𝑇]†𝛽= [𝑇∗]𝛽, so the statement holds. For orthogonal basis 𝛾, 𝐺=𝛿𝑖𝑗⟨𝛾𝑖,𝛾𝑗⟩. 𝐺−1𝐵𝐺=def𝑃𝐵𝐺=𝑃𝑗𝑗𝐵𝑖𝑗𝐺𝑖𝑖=𝑃𝑗𝑗𝐺𝑖𝑖𝐵𝑖𝑗. This equals 𝐵 iff 𝐵𝑖𝑗=0 or 𝑃𝑗𝑗𝐺𝑖𝑖=1⇒ ⟨𝛾𝑖,𝛾𝑖⟩⟨𝛾𝑗,𝛾𝑗⟩=1⇒⟨𝛾𝑖,𝛾𝑖⟩=⟨𝛾𝑗,𝛾𝑗⟩. This means the statement also holds for orthogonal bases if 𝐵𝑖𝑗≠0⇒⟨𝛾𝑖,𝛾𝑖⟩=⟨𝛾𝑗,𝛾𝑗⟩. # Counterexample Let basis 𝛽={[10],[11]} on 𝑉=ℝ2. We have 𝐺=[⟨𝛽1,𝛽1⟩⟨𝛽2,𝛽1⟩⟨𝛽1,𝛽2⟩⟨𝛽2,𝛽2⟩]=[1112] and 𝐺−1= [2−1−11] We find operator T such that 𝐴=[𝑇]𝛽 and 𝐺𝐴†𝐺−1≠𝐴†. Let 𝑇(𝑣1𝛽1+𝑣2𝛽2)= 𝑣2𝛽1+𝑣1𝛽2. 𝐴=𝐴†=[0110], 𝐵=𝐺𝐴†𝐺−1=[1112][0110][2−1−11]=[130−1]≠𝐴† ∎

quaerat voluptatem. Ut enim aeque doleamus animo, cum corpore dolemus, fieri tamen permagna accessio potest, si aliquod aeternum et infinitum impendere malum nobis opinemur. Quod idem licet transferre in voluptatem, ut postea variari voluptas distinguique possit, augeri amplificarique non possit. At etiam Athenis, ut e patre audiebam facete et urban

Typst
Rust