# Linear operators and change of basis

1. Sep 4, 2014

### "Don't panic!"

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.

Last edited: Sep 4, 2014
2. Sep 4, 2014

### Fredrik

Staff Emeritus
Looks great. I would just say "field" instead of "scalar field", because the latter term makes me think of functions defined on a manifold.

3. Sep 4, 2014

### "Don't panic!"

Ah ok. thanks for your help Fredrik.