[lin.alg] [tricks]

A = \begin{bmatrix}0 & 1 & 0 & \cdots & 0 \\ \vdots & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 & 1 \\ -a_{n-1} & -a_{n-2} & \cdots & \cdots & -a_{0} \end{bmatrix}

has characteristic polynomial z^n + a_{n-1}z^{n-1} + \dotsb + a_0.

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[lin.alg]

Let V be a K-vector space, then A := \mathrm{span}\left\{ \text{linear maps}\ T : K \rightarrow V \right\} \cong V as vector spaces. (Abusing notation and treating K as a 1D vector space over K in the definition of A).

A is a vector space. Easy to check that \varphi : A \rightarrow V : T \mapsto Te where e is the single basis vector in K is a linear transformation. It is surjective: \forall v \in V,\ T_v : K \rightarrow V,\ T_v(e) = v is a linear map. It is injective: if Te = Se then Ta = Tke = kTe = kSe = Sa\ \forall a \in K, i.e. S = T.