Ludium
Sign In
Linear Algebra Done Right
Vector Spaces
01Complex Numbers as Pairs of Real Numbers02Lists and Fⁿ: Coordinate Addition and a First Proof03Vectors in R², Scalar Multiplication, and FieldsProblem set0/10Problem set 20/10
01Definition of a Vector Space: Fⁿ, F^∞ and Function Spaces02Unique Zero and Inverses, 0v = 0, and the Subspace Test03Sums of Subspaces: The Smallest Subspace Containing Them04Direct Sums: Only One Way to Write ZeroProblem set0/10Problem set 20/10
Finite-Dimensional Spaces
01Linear Combinations, Span, and Finite-Dimensional Spaces02Linear Independence and the Linear Dependence Lemma03Independent Lists Are No Longer Than Spanning ListsProblem set0/10Problem set 20/10
01Bases: Reduce a Spanning List, Extend an Independent List02Dimension: Why Every Basis Has the Same Length03The Dimension of a Sum of Two SubspacesProblem set0/10Problem set 20/10
Linear Maps
01Linear Maps and the Linear Map Lemma: Values on a Basis02Sums, Scalar Multiples and Products of Linear Maps03Null Space, Range, Injectivity and Surjectivity04The Fundamental Theorem of Linear Maps and Linear SystemsProblem set0/10Problem set 20/10
01The Matrix of a Linear Map: Each Column Is Tvₖ02Matrix Operations and Why Multiplication Is Row Times Column03Rows, Columns and Linear Combinations in Matrix Products04Column Rank Equals Row Rank: Column-Row FactorizationProblem set0/10Problem set 20/10
01Invertible Linear Maps: Injective and Surjective02Equal Dimensions: Injective or Surjective Means Invertible03Isomorphic Spaces and the Dimension of L(V, W)04The Matrix of a Vector: M(Tv) = M(T)M(v)05Change of Basis for an Operator: A = C⁻¹BCProblem set0/10Problem set 20/10
01Products of Vector Spaces: When Dimensions Add02Quotient Spaces V/U: Parallel Lines as Vectors03The Quotient Map and V/null T ≅ range TProblem set0/10Problem set 20/10
01The Dual Basis Reads Coefficients and T′ Reverses Arrows02The Annihilator U⁰ and dim U⁰ = dim V − dim U03Null T′ and Range T′ as Annihilators; M(T′) Is the TransposeProblem set0/10Problem set 20/10
Polynomials and Eigenvalues
01Conjugates, |z| and the Complex Triangle Inequality02Zeros, Factors and the Division Algorithm via a Basis03Fundamental Theorem of Algebra and Complete Factorization04Real Polynomials: Linear and Irreducible Quadratic FactorsProblem set0/10Problem set 20/10
01Invariant Subspaces and Eigenvalues: Tv = λv02Eigenvectors, T − λI and At Most dim V Eigenvalues03p(T) and Why Every Complex Operator Has an Eigenvalue04The Minimal Polynomial: Existence, Uniqueness and Computation05Eigenvalues Are the Zeros of the Minimal Polynomial06Why Odd Dimension Forces a Real EigenvalueProblem set0/10Problem set 20/10
01Upper-Triangular Matrices and Invariant Spans02Eigenvalues Sit on the Upper-Triangular Diagonal03The Minimal Polynomial Decides Upper-TriangularityProblem set0/10Problem set 20/10
01Diagonalizable Operators and Direct Sums of Eigenspaces02Four Equivalent Tests for Diagonalizability03Distinct Linear Factors Decide Diagonalizability04The Gershgorin Disk Theorem: Where Eigenvalues Live05Commuting Operators Share a Diagonalizing Basis06Commuting Operators Over C: One Upper-Triangular BasisProblem set0/10Problem set 20/10
Inner Product Spaces
01Inner Products: Why Complex Spaces Need a Conjugate02Norms, Orthogonality and the Pythagorean Theorem03Cauchy-Schwarz and the Triangle InequalityProblem set0/10Problem set 20/10
01Orthonormal Bases: Bessel's Inequality and Parseval02Gram-Schmidt, Orthonormal Bases and Schur's Theorem03Linear Functionals and the Riesz Representation TheoremProblem set0/10Problem set 20/10
01Orthogonal Complements and the Direct Sum V = U ⊕ U⊥02Orthogonal Projection onto U and Why U⊥⊥ = U03Riesz Representation Without a Basis: v ⊥ null φProblem set0/10Problem set 20/10
01The Closest Point Is the Projection: Fitting Sine on [−π, π]02The Pseudoinverse T† and Its Three Inverse-Like Properties03T†w: The Best Approximate Solution of Smallest NormProblem set0/10Problem set 20/10
Operators on Inner Product Spaces
01The Adjoint T*: Moving T into the Second Slot of ⟨Tv, w⟩02Null T* = (range T)⊥ and M(T*) as the Conjugate Transpose03Self-Adjoint Operators: Real Eigenvalues and Real ⟨Tv, v⟩04Normal Operators: ‖Tv‖ = ‖T*v‖ and Orthogonal EigenvectorsProblem set0/10Problem set 20/10
01Real Spectral Theorem: Self-Adjoint ⇔ Orthonormal Eigenbasis02The Complex Spectral Theorem: Normal Is Enough03Positive Operators: Six Equivalent Characterizations04The Unique Positive Square Root √TProblem set0/10Problem set 20/10
01Isometries: Five Equivalent Ways to Preserve Length02Unitary Operators: S⁻¹ = S*, Eigenvalues on |λ| = 103QR Factorization: Gram–Schmidt Recorded as A = QR04Cholesky Factorization: Positive Definite B = R*RProblem set0/10Problem set 20/10

QR Factorization: Gram–Schmidt Recorded as A = QR

How does running Gram–Schmidt on a matrix's columns turn into a unique factorization A = QR?


Loading…

←Previous Unitary Operators: S⁻¹ = S*, Eigenvalues on |λ| = 1Next Cholesky Factorization: Positive Definite B = R*R →

Your summary note

    1. 1

      Unitary matrices and their characterizations

      Define a unitary matrix by orthonormal columns, and list the equivalent conditions: orthonormal rows, ∥Qv∥=∥v∥\|Qv\| = \|v\|∥Qv∥=∥v∥ for every vvv, and Q∗Q=QQ∗=IQ^*Q = QQ^* = IQ∗Q=QQ∗=I.

    2. 2

      Existence of A=QRA = QRA=QR with positive diagonal

      Apply Gram–Schmidt to the columns vkv_kvk​ of AAA, set Rj,k=⟨vk,ej⟩R_{j,k} = \langle v_k, e_j \rangleRj,k​=⟨vk​,ej​⟩, and show RRR is upper triangular with positive diagonal and QR=AQR = AQR=A.

    3. 3

      Uniqueness of the QR factorization

      Show that a second factorization A=Q^R^A = \widehat{Q}\widehat{R}A=Q​R gives span⁡(v1,…,vk)=span⁡(q1,…,qk)\operatorname{span}(v_1, \dots, v_k) = \operatorname{span}(q_1, \dots, q_k)span(v1​,…,vk​)=span(q1​,…,qk​) and ⟨vk,qk⟩>0\langle v_k, q_k \rangle > 0⟨vk​,qk​⟩>0, conclude Q^=Q\widehat{Q} = QQ​=Q, then R^=R\widehat{R} = RR=R.

    4. 4

      Worked example and solving Ax=bAx = bAx=b

      Carry the matrix with columns (1,0,0)(1, 0, 0)(1,0,0), (2,1,3)(2, 1, 3)(2,1,3), (1,−4,2)(1, -4, 2)(1,−4,2) through to QQQ and RRR, and reduce Ax=bAx = bAx=b to back substitution in Rx=Q∗bRx = Q^*bRx=Q∗b.

    Attempt 1 of 2