Suggest edit — Axler 5.B

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title: "Axler 5.B"
source: https://www.jemoka.com/posts/kbhaxler_5_b/
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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 \(v_1 &hellip; v_{n}\) is a basis of \(V\) then \(\mathcal{M}(T)\) is upper-triangular if \(Tv_{j} \in span(v_1, &hellip; v_{j})\) 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&hellip; 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 \(v_1 &hellip; v_{j}\) 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&rsquo;s happening?