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Linear operators and change of basis

  1. Sep 4, 2014 #1
    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 [itex]V[/itex] over some scalar field [itex]\mathbb{F}[/itex], and two basis sets [itex] \mathcal{B} = \lbrace\mathbf{e}_{i}\rbrace_{i=1,\ldots , n}[/itex] and [itex] \mathcal{B}' = \lbrace\mathbf{e}'_{i}\rbrace_{i=1,\ldots , n}[/itex] which form two given bases for [itex]V[/itex], we can express any vector [itex]\mathbf{v} \in V[/itex] as a unique linear combination with respect to each of these bases, i.e. [tex]\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} [/tex] Now, suppose we have some linear operator [itex]\mathcal{S}:V \longrightarrow V [/itex] which maps the basis [itex]\mathcal{B}[/itex] to the basis [itex]\mathcal{B}'[/itex], defined in the following manner [tex]\qquad\qquad\qquad\qquad\qquad\qquad\qquad\mathcal{S} \left(\mathbf{e}_{j}\right)= \mathbf{e}'_{j}[/tex] As [itex]\mathcal{S} \left(\mathbf{e}_{j}\right) \in V[/itex] is itself a vector in [itex]V[/itex] we can express it as a linear combination of the basis vectors [itex]\mathbf{e}_{i} \;\; \left(i=1, \ldots , n\right)[/itex], [tex] \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} [/tex] where [itex]S_{ij}\equiv \left(\mathcal{S} \left(\mathbf{e}_{j}\right)\right)_{i}[/itex] is the [itex]i^{th}[/itex] component of the [itex]j^{th}[/itex] basis vector [itex]\mathbf{e}'_{j}[/itex] with respect to the basis [itex]\mathcal{B}[/itex]. The [itex]S_{ij}\equiv \left(\mathcal{S} \left(\mathbf{e}_{j}\right)\right)_{i}[/itex] are the components of the linear operator [itex]\mathcal{S}[/itex] with respect to the basis [itex]\mathcal{B}[/itex]. The columns of [itex]\left[\mathcal{S}\right]_{\mathcal{B}}[/itex] are the column vector representations of the vectors [itex]\mathbf{e}'_{j} \in \mathcal{B}'[/itex] with respect to the basis [itex]\mathcal{B}[/itex].

    Would this be a correct description? Thanks for your time.
    Last edited: Sep 4, 2014
  2. jcsd
  3. Sep 4, 2014 #2


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    Looks great. I would just say "field" instead of "scalar field", because the latter term makes me think of functions defined on a manifold.
  4. Sep 4, 2014 #3
    Ah ok. thanks for your help Fredrik.
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