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Following on from a previous post of mine about linear operators, I'm trying to firm up my understanding of changing between bases for a given vector space.
For a given vector space V over some scalar field \mathbb{F}, and two basis sets \mathcal{B} = \lbrace\mathbf{e}_{i}\rbrace_{i=1,\ldots , n} and \mathcal{B}' = \lbrace\mathbf{e}'_{i}\rbrace_{i=1,\ldots , n} which form two given bases for V, we can express any vector \mathbf{v} \in V as a unique linear combination with respect to each of these bases, i.e. \qquad\qquad\qquad\qquad\qquad\qquad\qquad \mathbf{v}=\sum_{i=1}^{n}v_{i}\mathbf{e}_{i} \quad\text{and}\quad \mathbf{v} = \sum_{i=1}^{n}v'_{i}\mathbf{e}'_{i} Now, suppose we have some linear operator \mathcal{S}:V \longrightarrow V which maps the basis \mathcal{B} to the basis \mathcal{B}', defined in the following manner \qquad\qquad\qquad\qquad\qquad\qquad\qquad\mathcal{S} \left(\mathbf{e}_{j}\right)= \mathbf{e}'_{j} As \mathcal{S} \left(\mathbf{e}_{j}\right) \in V is itself a vector in V we can express it as a linear combination of the basis vectors \mathbf{e}_{i} \;\; \left(i=1, \ldots , n\right), \qquad\qquad\qquad\qquad\qquad\qquad\qquad\mathbf{e}'_{j}= \sum_{i=1}^{n}\left(\mathcal{S} \left(\mathbf{e}_{j}\right)\right)_{i}\mathbf{e}_{i}= \sum_{i=1}^{n}S_{ij}\mathbf{e}_{i} where S_{ij}\equiv \left(\mathcal{S} \left(\mathbf{e}_{j}\right)\right)_{i} is the i^{th} component of the j^{th} basis vector \mathbf{e}'_{j} with respect to the basis \mathcal{B}. The S_{ij}\equiv \left(\mathcal{S} \left(\mathbf{e}_{j}\right)\right)_{i} are the components of the linear operator \mathcal{S} with respect to the basis \mathcal{B}. The columns of \left[\mathcal{S}\right]_{\mathcal{B}} are the column vector representations of the vectors \mathbf{e}'_{j} \in \mathcal{B}' with respect to the basis \mathcal{B}.
Would this be a correct description? Thanks for your time.
For a given vector space V over some scalar field \mathbb{F}, and two basis sets \mathcal{B} = \lbrace\mathbf{e}_{i}\rbrace_{i=1,\ldots , n} and \mathcal{B}' = \lbrace\mathbf{e}'_{i}\rbrace_{i=1,\ldots , n} which form two given bases for V, we can express any vector \mathbf{v} \in V as a unique linear combination with respect to each of these bases, i.e. \qquad\qquad\qquad\qquad\qquad\qquad\qquad \mathbf{v}=\sum_{i=1}^{n}v_{i}\mathbf{e}_{i} \quad\text{and}\quad \mathbf{v} = \sum_{i=1}^{n}v'_{i}\mathbf{e}'_{i} Now, suppose we have some linear operator \mathcal{S}:V \longrightarrow V which maps the basis \mathcal{B} to the basis \mathcal{B}', defined in the following manner \qquad\qquad\qquad\qquad\qquad\qquad\qquad\mathcal{S} \left(\mathbf{e}_{j}\right)= \mathbf{e}'_{j} As \mathcal{S} \left(\mathbf{e}_{j}\right) \in V is itself a vector in V we can express it as a linear combination of the basis vectors \mathbf{e}_{i} \;\; \left(i=1, \ldots , n\right), \qquad\qquad\qquad\qquad\qquad\qquad\qquad\mathbf{e}'_{j}= \sum_{i=1}^{n}\left(\mathcal{S} \left(\mathbf{e}_{j}\right)\right)_{i}\mathbf{e}_{i}= \sum_{i=1}^{n}S_{ij}\mathbf{e}_{i} where S_{ij}\equiv \left(\mathcal{S} \left(\mathbf{e}_{j}\right)\right)_{i} is the i^{th} component of the j^{th} basis vector \mathbf{e}'_{j} with respect to the basis \mathcal{B}. The S_{ij}\equiv \left(\mathcal{S} \left(\mathbf{e}_{j}\right)\right)_{i} are the components of the linear operator \mathcal{S} with respect to the basis \mathcal{B}. The columns of \left[\mathcal{S}\right]_{\mathcal{B}} are the column vector representations of the vectors \mathbf{e}'_{j} \in \mathcal{B}' with respect to the basis \mathcal{B}.
Would this be a correct description? Thanks for your time.
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