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--- title: "Linear Algebra Errors" type: concept related: [Nus Math530 Similar To Diagonal, Isomorphism, Vector Space, Polynomial] source: https://www.jemoka.com/posts/kbhlinear_algebra_errors/ confidence: high status: active --- Gaussian Elimination Quiz Demonstrate that matrices’ multiplication are not commutative (error: didn’t consider \(m\times m\)) Which \(2\times 2\) matrices under multiplication form a group? (error: closure need to proved on invertable matrices under multiplication, not just \(2\times 2\)) Deriving Rotation matrices (error: clockwise vs counter-clockwise) Linear Independence Quiz Connection between linear independence and systems equations (error: beated around the bush) — the matrix of an nxn system of equations has a solution if the matrix’s column vectors is linearly independent Basis and Dimension Quiz put 0 into a basis AAAA not lin. indep; figure out what the basis for a polynomial with a certain root is: it is probably of dimension m (instead of m+1), because scalars doesn’t work in the case of p(3)=0; so basis is just the scalars missing some inequality about basis? — its just that lin.idp sets is shorter or equal to basis and spanning sets is longer or equal to basis Final, part 1 definition of vector space: scalar multiplication is not an operation straight forgot \(dim(U+V) = dim U + dim V - dim (U\cap V)\) plane containing \((1,0,2)\) and \((3,-1,1)\): math mistake proof: det A det B = det AB Final, part 2 Counterproof: If \(v_1 \dots v_4\) is a basis of \(V\), and \(U\) is a subspace of \(V\) with \(v_1, v_2 \in U\) and \(v_3, v_4\) not in \(U\), \(v_1, v_2\) is a basis of \(U\) Counterproof: if \(T \in \mathcal{L}(V,V)\) and \(T^{2}=0\), then \(T=0\) Counterproof: if \(s,t \in \mathcal{L}(V,V)\), and \(ST=0\), then \(null\ s\) is contained in \(range\ T\) Product Spaces Quiz Need more specific description: explain why we use product and quotient to describe product and quotient spaces? Prove that \(\mathcal{L}(V_1 \times V_2 \times \dots \times V_{m}, W)\) and \(\mathcal{L}(V_1, W) \times \dots \times \mathcal{L}(V_{m}, W)\) are isomorphic. Error: didn’t do it correctly for infinite dimensional Quotient Spaces Quiz Couldn’t prove that the list in linearly independent: the linear combinations is some \(c_1v_1 + \dots c_{m}v_{m} + U\); as \(v_1 \dots v_{m}\) is a basis of \(V / U\), \(c_1 \dots c_{m} = 0\), now the second part is also a basis so they are \(0\) too. The spanning proof: \(v + U =\) , rewrite as basis, etc. she graded wrong: what’s the importance of \(\widetilde{T}\)? Give two statements equivalent to \(v+U = w+U\), prove equivalence betewen this statement and the others didn’t prove both directions! Polynomials Quiz state the fundamental theorem of algebra; error: \(\mathcal{P}_{m}(\mathbb{F})\) is a vector space of polynomials with degree at most \(m\), and yet the FtOA requires exactly \(m\) Upper Triangular Quiz upper-triangular representation is findable when the space is 1) under complexes and 2) for finite-dimensional vector spaces; need BOTH conditions Upper Triangular Quiz UNCLEAR: Geometric Multipliicty is bounded by Algebric Multiplicity; Algebraic multiplicity (“real estate” taken on the upper-triangular diagonal) v. geometric multiplicity (amount of linearly independent eigenvectors included with that eigenvalue); so if geometric multiplicity < algebraic multiplicity, the map is not diagonalizable because its not bringing enough linearly independent eigenvectors Diagonalization Quiz enough eigenvalues go in only one direction: it existing means its diagonalizable, but the opposite isn’t true the proof for \(T\) is diagonalizable IFF the matrix \(T\) is similar to a diagonal matrix: NUS-MATH530 Similar to Diagonal Final, part 1 State the complex spectral theorem (error: the condition of normality is a PARALLEL result) Final, Part 2 Said this was true, but its not; \(null\ T \bigoplus range\ T = V\), \(T\) is diagonalizable; Said this was true, but its false \(T^{2}= 0\) IFF \(null\ T = range\ T\) suppose \(T=0\), \(T^{2} = 0\). \(null\ T = V\), \(range\ T = 0\). Spectral theorem doesn’t define diagonalizability, it defines diagonalibility for ORTHONORMAL missing derivation of the pseudoinverse