# Vectors
If the vectors $\mathcal{B} = \{\pmb{e}_i, i = 1, \ldots, n\}$ form a [[basis]] of an $n$-[[Dimension|dimensional]] [[vector space]] $V$, then any $\pmb{v} \in V$ can be written as:
$ \pmb{v} = \sum_{i = 1}^{n} v^i \pmb{e}_i$
The set of numbers $\{v^i, i = 1, \ldots n\}$ are the *components* of $\pmb{v}$ with respect to this [[basis]].
We can write the vector $\pmb{v}$ as a column vector in this basis, denoted by $[\pmb{v}]_{\mathcal{B}}$, as
$
[\pmb{v}]_{\mathcal{B}} =
\begin{pmatrix}
v^1 \\
v^2 \\
\vdots\\
v^n
\end{pmatrix}
$
We will often drop the subscript in $[.]_\mathcal{B}$ if the basis is obvious, its purpose is to distinguish components in different bases. Remember that vectors exists *independent of any chosen basis and their column vector expressions depend on the choice of basis*.
## Example
Note that, any set of basis vectors in the basis they define will look like [[Standard Basis]]. For example, consider $\mathbb{R}^3$ with the two bases:
$
\begin{align}
\mathcal{B} &= \{\pmb{e}_1 = (1,0,0), \pmb{e}_2 = (0,1,0), \pmb{e}_3 = (0,0,1)\}, \\
\mathcal{B'} &= \{\pmb{e}'_1 = (1,1,0), \pmb{e}'_2 = (0,1,1), \pmb{e}'_3 = (1,0,1)\}
\end{align}
$
Observe that:
$
[\pmb{e}_1]_{\mathcal{B}} =
\begin{pmatrix}
1 \\
0 \\
0
\end{pmatrix}, \quad
[\pmb{e}_2]_{\mathcal{B}} =
\begin{pmatrix}
0 \\
1 \\
0
\end{pmatrix}, \quad
[\pmb{e}_3]_{\mathcal{B}} =
\begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix},
$
and
$
[\pmb{e}'_1]_{\mathcal{B}'} =
\begin{pmatrix}
1 \\
0 \\
0
\end{pmatrix}, \quad
[\pmb{e}'_2]_{\mathcal{B}'} =
\begin{pmatrix}
0 \\
1 \\
0
\end{pmatrix}, \quad
[\pmb{e}'_3]_{\mathcal{B}'} =
\begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix},
$
# One-Forms
The construction for [[one-form|one-forms]] becomes:
$
\tilde{q} = \sum_{j=1}^n q_j \tilde{e}^j
$
where $q_j$ are the components and $\tilde{e}^j$ are the [[dual basis]].
# Tensors
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Just like the components of a one-form, the components of a [[tensor]] $\overline{S}$ (e.g. of type $(3,2)$) are its values when it takes [[basis]] vectors and [[basis]] one-form as its arguments.
$S^{ijk}_{\,\,\,\,\,lm} = \overline{S}(\tilde{e}^i, \tilde{e}^j, \tilde{e}^k; \pmb{e}_l, \pmb{e}_m)$
Note that the components of a [[tensor]] (and thus, a [[vector]] and a [[one-form]]) in a certain [[basis]] are unique, i.e., no two tensors have the same components in the same basis.
# Linear Operators
Consider a [[Linear Operator]] $T$ on a finite dimensional space $V$, $\pmb{v} \in V$, and choose a [[basis]] $\mathcal{B} = \{e_i\}_{\{i = 1, \ldots, n\}}$. The action of $T$ is determined by its action on the basis [[vectors]] as follows:
$
T(\pmb{v}) = T( \sum_{i = 1}^n v^i e_i) = \sum_{i = 1}^n v^i T(e_i) = \sum_{i, j = 1} v^i T_{i}^{\,j} e_j
$
where
$
T(e_i) = \sum_{j = 1}^n T_{i}^{\,j} e_j
$
the numbers $T_{i}^{\,j}$ are the components of $T$ in the basis $\mathcal{B}$.
The [[matrix]] of $T$ in the basis $\mathcal{B}$, $[T]_\mathcal{B}$, just has components $T_{i}^{\,j}$ where the lower index refers to columns and the upper index refers to rows.The action of the operator in a basis takes the form:
$
[T(\pmb{v})]_\mathcal{B} = [T]_\mathcal{B} [\pmb{v}]_\mathcal{B}
$
Thus, when we pick a basis, we can use the operator to act on vectors via matrix multiplication.
# Linear Maps
This case of [[linear map|linear maps]] is similar to the linear operator discussed above but the map is no longer an [[endomorphism]]. Let our [[vector space|vector spaces]] be $E$ and $F$ and their [[basis|bases]] $\{e_i\}$ and $\{f_i\}$.
We have four cases:
1.
$
\begin{align}
T: E &\rightarrow F\\
T(e_i) &= T_{\,\,i}^{j} f_j
\end{align}
$
2.
$
\begin{align}
T: E^* &\rightarrow F\\
T(e^i) &= T^{ji} f_j
\end{align}
$
3.
$
\begin{align}
T: E &\rightarrow F^*\\
T(e_i) &= T_{ji} f^j
\end{align}
$
4.
$
\begin{align}
T: E^* &\rightarrow F^*\\
T(e^i) &= T^{\,\,i}_{j} f^j
\end{align}
$