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Mathematics
Linear and Abstract Algebra
Dimension of a Linear Transformation Matrix
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[QUOTE="fresh_42, post: 6507820, member: 572553"] If you use them as a linear transformation of ##\mathbb{R}^2## then you simply consider the ##2\times 2## matrix. If you want to deal with all those linear transformations and deal with ##\mathbb{M}_2(\mathbb{R})## as vector space instead, then you have to choose some ordered basis to make those matrices a linear vector. E.g. if ##e_{ij}## is the matrix with ##1## in the ## i ##-th row and ##j##-th column, and zeros otherwise, then ##(e_{11},e_{12},e_{21},e_{22})## would be such a bases. A ##2 \times 2## matrix would be ##\begin{bmatrix} 2&3\\4&5 \end{bmatrix}=2e_{11}+3e_{12}+4e_{21}+5e_{22}=(2,3,4,5)## A linear transformation between those, e.g. ##M_{\Phi}=\begin{bmatrix} 2&3&0&1\\0&0&0&5\\0&0&1&4\\0&0&0&0 \end{bmatrix}## would map \begin{align*} e_{11}&\longmapsto 2e_{11}+3e_{12}+1e_{22}\\ e_{12}&\longmapsto 5e_{22}\\ e_{21}&\longmapsto 1e_{21}+4e_{22}\\ e_{22}&\longmapsto 0 \end{align*}Edit: This interpretation corresponds to ##\vec{v} \longmapsto \vec{v}\cdot M_{\Phi}##. If we choose matrix multiplication from the left, i.e. ##\vec{v} \longmapsto M_{\Phi}\cdot \vec{v}## then \begin{align*} e_{11}&\longmapsto 2e_{11}\\ e_{12}&\longmapsto 3e_{11}\\ e_{21}&\longmapsto 1e_{21}\\ e_{22}&\longmapsto 1e_{11}+5e_{12}+4e_{21} \end{align*} [/QUOTE]
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Mathematics
Linear and Abstract Algebra
Dimension of a Linear Transformation Matrix
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