No idea why this is so long!!!

Key Sequence

Firehose of a chapter.

• We first began an unrelated exploration in Product of Vector Spaces (“tuples”):
  • we show that the Product of Vector Spaces is a vector space
  • because you can build a list out of zeroing every element except each one on each basis of each element of the tuple sequentially, we learned that the dimension of the Product of Vector Spaces is the sum of the spaces’ dimension.
  • we defined the product-to-sum map \(\Gamma\)
    • \(U_1 + \dots + U_{m}\) is a direct sum IFF \(\Gamma\) is injective
    • and, as a result, \(U_1 + \dots + U_{m}\) is a direct sum IFF \(\dim (U_1 + \dots + U_{m}) = \dim U_1 + \dots + \dim U_{m}\)
• We then tackled the fun part of this chapter, which is affine subsets, parallel structures, quotient spaces, quotient map (affine subsetification maps)
  • we learned an important and useful result that two affine subsets parallel to \(U\) are either equal or disjoint (\(v-w \in U\) means \(v+U = w+U\) means \(v+U \cap w+U \neq \emptyset\), means the first thing)
  • we defined the operations on quotient space, and showed that quotient space operations behave uniformly on equivalent affine subsets. This, and the usual closer proof, demonstrates that quotient spaces is a vector space
  • with the help of the affine subsetification map (the quotient map \(\pi\)), we show that the dimension of a quotient space is the difference between dimensions of its constituents essentially by invoking rank-nullity theorem after knowing the fact that \(null\ \pi = U\) (because \(u+U\) is an affine subset that has not been shifted (think about a line moving along itself… it doesn’t move))
• Then, and I’m not quite sure why, we defined \(\widetilde{T}: V / null\ T \to W\), for some \(T: V\to W\), defined as \(\widetilde{T}(v+null\ T) = Tv\).
  • We show that the map is Linear, injective, its range is \(range\ T\), and so it forms an isomorphism between \(V / null\ T\) and \(range\ T\).

Here’s something: products and quotients, the intuition

Because duality is fun and I’m bored and houjun-being-obtuse.

Key Sequence

New Definitions

• linear functional
• dual space

Results and Their Proofs

• dimension of dual space is equivalent to the original space

Questions for Jana

Interesting Factoids

Hello from onboard NH107! Or perhaps my next connecting flight, or from China.

EIGENSTUFF and OPERATORS! Invariant subspaces are nice.

Sometimes, if we can break the domain of a linear map down to its eigenvalues, we can understand what its doing on a component-wise level.

Key Sequence

• we defined an invariant subspace, and gave a name to 1-D invariant subspaces: the span of eigenvectors
• we showed some properties of eigenvalues and showed that a list of eigenvectors are linearly independent
  • a correlate of this is that operators on finite dimensional V has at most dim V eigenvalues
• finally, we defined map restriction operator and quotient operator, and showed that they were well-defined

New Definitions

• invariant subspace
  • conditions for nontrivial invariant subspace
• eigenvalues + eigenvectors + eigenspace
• two new operators: map restriction operator and quotient operator

Results and Their Proofs

• properties of eigenvalues
• list of eigenvectors are linearly independent
  • eigenspaces are disjoint
  • operators on finite dimensional V has at most dim V eigenvalues
• quotient operator is well-defined

Questions for Jana

Interesting Factoids

“eigenvalue” is sometimes called the “characterizing value” of a map

Key Sequence

• we began the chapter defining \(T^m\) (reminding ourselves the usual rules of \(T^{m+n} = T^{m}T^{n}\), \((T^{m})^{n} = T^{mn}\), and, for invertible maps, \(T^{-m} = (T^{-1})^{m}\)) and \(p(T)\), wrapping copies of \(T\) into coefficients of a polynomial, and from those definitions showed that polynomial of operator is commutative
• we then used those results + fundamental theorem of algebra to show that operators on complex vector spaces have an eigenvalue
• that previous, important result in hand, we then dove into upper-triangular matricies
  • specifically, we learned the properties of upper-triangular matrix, that if \(v_1 … v_{n}\) is a basis of \(V\) then \(\mathcal{M}(T)\) is upper-triangular if \(Tv_{j} \in span(v_1, … v_{j})\) for all \(j \leq n\); and, equivalently, \(T\) in invariant under the span of \(v_{j}\)
  • using that result, we show that every complex operator has an upper-triangular matrix
  • using some neat tricks of algebra, we then establish that operator is only invertible if diagonal of its upper-triangular matrix is nonzero, which seems awfully unmotivated until you learn that…
  • eigenvalues of a map are the entries of the diagonal of its upper-triangular matrix, and that basically is a direct correlary from the upper-triangular matrix of \(T-\lambda I\)

New Definitions

• \(T^m\)
• \(p(T)\)
  • technically also product of polynomials
• matrix of an operator
  • diagonal of a matrix
  • upper-triangular matrix

Results and Their Proofs

• \(p(z) \to p(T)\) is a linear function
• polynomial of operator is commutative
• operators on complex vector spaces have an eigenvalue
• properties of upper-triangular matrix
• every complex operator has an upper-triangular matrix
• operator is only invertible if diagonal of its upper-triangular matrix is nonzero
• eigenvalues of a map are the entries of the diagonal of its upper-triangular matrix

Questions for Jana

• why define the matrix of an operator again?? just to stress that its square
• for the second flavor of the proof that every complex operator has an upper-triangular matrix, why is \(v_1 … v_{j}\) a basis of \(V\)?

Interesting Factoids

Its 12:18AM and I read this chapter for 5 hours. I also just got jumpscared by my phone notification. What’s happening?

Key Sequence

• we defined an eigenspace, which is the space of all eigenvalues of a distinct eigenvector, and show that
  • they form a direct sum to the whole space
  • and, as a correlate to how direct sums are kind of like disjoint sets, we have the perhaps expected result of dimension, that the sum of the eigenspace’ dimensions must be smaller than or equal than that of \(V\)
• we defined a Diagonal Matrix, which by its structure + calculation can be shown to require that it is formed by a basis of eigenvalues
  • and from there, and the properties of eigenspaces above, we deduce some conditions equal to diagonalizability
  • a direct correlary of the last point (perhaps more straightforwardly intuited by just lining eigenvalue up diagonally in a matrix) is that enough eigenvalues implies diagonalizability

New Definitions

• diagonal matrix
  • properties of diagonal matrices
• eigenspace

Results and Their Proofs

• eigenspaces are a direct sum
• dimension of sum of eigenspaces is smaller than or equal to the dimension of the whole space
• conditions equal to diagonalizability
• enough eigenvalues implies diagonalizability

Questions for Jana

• for diagonalizability, shouldn’t \(n\) be \(m\) on item 3?

Interesting Factoids

Short!