# Vector space/Base change/Diagram/Introduction/Section

We know, due to fact, that in a finite-dimensional vector space, any two bases have the same length, the same number of vectors. Every vector has, with respect to every basis, unique coordinates (the coefficient tuple). How do these coordinates behave when we change the bases? This is answered by the following statement.

## Lemma

Let $K$ be a field, and let $V$ be a $K$-vector space of dimension $n$. Let ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$ and ${\mathfrak {w}}=w_{1},\ldots ,w_{n}$ denote bases of $V$. Suppose that
    $v_{j}=\sum _{i=1}^{n}c_{ij}w_{i}\,$
with coefficients $c_{ij}\in K$, which we collect into the $n\times n$-matrix
    $M_{\mathfrak {w}}^{\mathfrak {v}}={\left(c_{ij}\right)}_{ij}\,.$
$u$
${\begin{pmatrix}s_{1}\\\vdots \\s_{n}\end{pmatrix}}$
${\mathfrak {v}}$
    ${\begin{pmatrix}t_{1}\\\vdots \\t_{n}\end{pmatrix}}=M_{\mathfrak {w}}^{\mathfrak {v}}{\begin{pmatrix}s_{1}\\\vdots \\s_{n}\end{pmatrix}}={\begin{pmatrix}c_{11}&c_{12}&\ldots &c_{1n}\\c_{21}&c_{22}&\ldots &c_{2n}\\\vdots &\vdots &\ddots &\vdots \\c_{n1}&c_{n2}&\ldots &c_{nn}\end{pmatrix}}{\begin{pmatrix}s_{1}\\\vdots \\s_{n}\end{pmatrix}}\,$
${\mathfrak {w}}$

### Proof

This follows directly from
    $u=\sum _{j=1}^{n}s_{j}v_{j}=\sum _{j=1}^{n}s_{j}{\left(\sum _{i=1}^{n}c_{ij}w_{i}\right)}=\sum _{i=1}^{n}{\left(\sum _{j=1}^{n}s_{j}c_{ij}\right)}w_{i}\,,$
and the definition of matrix multiplication.
$\Box$
If for a basis ${\mathfrak {v}}$, we consider the corresponding bijective mapping (see remark)
    $\Psi _{\mathfrak {v}}\colon K^{n}\longrightarrow V,$
then we can express the preceding statement as saying that the triangle
    ${\begin{matrix}K^{n}&{\stackrel {M_{\mathfrak {w}}^{\mathfrak {v}}}{\longrightarrow }}&K^{n}&\\&\!\!\!\!\!\Psi _{\mathfrak {v}}\searrow &\downarrow \Psi _{\mathfrak {w}}\!\!\!\!\!&\\&&V&\!\!\!\!\!\!\!\!\\\end{matrix}}$
commutes.

## Definition

Let $K$ denote a field and let $V$ denote a $K$-vector space of dimension $n$. Let ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$ and ${\mathfrak {w}}=w_{1},\ldots ,w_{n}$ denote two bases of $V$. Let
    $v_{j}=\sum _{i=1}^{n}c_{ij}w_{i}\,$
with coefficients $c_{ij}\in K$. Then the $n\times n$-matrix
    $M_{\mathfrak {w}}^{\mathfrak {v}}=(c_{ij})_{ij}\,$
${\mathfrak {v}}$
${\mathfrak {w}}$

## Remark

The $j$-th column of a transformation matrix $M_{\mathfrak {w}}^{\mathfrak {v}}$ consists of the coordinates of $v_{j}$ with respect to the basis ${\mathfrak {w}}$. The vector $v_{j}$ has the coordinates $e_{j}$ with respect to the basis ${\mathfrak {v}}$, and when we apply the matrix to $e_{j}$ we get the $j$-th column of the Matrix, and this is just the coordinate tuple of $v_{j}$ with respect to the basis ${\mathfrak {w}}$.
For a one-dimensional space and
    $v=cw\,,$
we have $M_{\mathfrak {w}}^{\mathfrak {v}}=c={\frac {v}{w}}$, where the fraction is well-defined. This might help in memorizing the order of the bases in this notation.
Another important relation is
    ${\mathfrak {v}}={{\left(M_{\mathfrak {w}}^{\mathfrak {v}}\right)}^{\text{tr}}}{\mathfrak {w}}\,.$
Note that here, the matrix is not applied to an $n$-tuple of $K$, but to an $n$-tuple of $V$, yielding a new $n$-tuple of $V$. This equation be an argument to define the transformation matrix the other way round, however, we consider the behavior in fact as decisive.
In case
    $V=K^{n}\,,$
if ${\mathfrak {e}}$ is the standard basis and ${\mathfrak {v}}$ some further basis, we obtain the transformation matrix $M_{\mathfrak {v}}^{\mathfrak {e}}$ of the base change from ${\mathfrak {e}}$ to ${\mathfrak {v}}$ by expressing $e_{j}$ as a linear combination of the basis vectors $v_{1},\ldots ,v_{n}$, and write down the corresponding tuple as column. The inverse transformation matrix, $M_{\mathfrak {e}}^{\mathfrak {v}}$, consists simply in $v_{1},\ldots ,v_{n}$, written as columns.

## Example

We consider in $\mathbb {R} ^{2}$ the standard basis,
    ${\mathfrak {u}}={\begin{pmatrix}1\\0\end{pmatrix}},\,{\begin{pmatrix}0\\1\end{pmatrix}}\,,$
and the basis
    ${\mathfrak {v}}={\begin{pmatrix}1\\2\end{pmatrix}},\,{\begin{pmatrix}-2\\3\end{pmatrix}}\,.$
The basis vectors of ${\mathfrak {v}}$ can be expressed directly with the standard basis, namely
    $v_{1}={\begin{pmatrix}1\\2\end{pmatrix}}=1{\begin{pmatrix}1\\0\end{pmatrix}}+2{\begin{pmatrix}0\\1\end{pmatrix}}\,\,{\text{ and }}\,\,v_{2}={\begin{pmatrix}-2\\3\end{pmatrix}}=-2{\begin{pmatrix}1\\0\end{pmatrix}}+3{\begin{pmatrix}0\\1\end{pmatrix}}.$
Therefore, we get immediately
    $M_{\mathfrak {u}}^{\mathfrak {v}}={\begin{pmatrix}1&-2\\2&3\end{pmatrix}}\,.$
For example, the vector which has with respect to ${\mathfrak {v}}$ the coordinates $(4,-3)$, has the coordinates
    $M_{\mathfrak {u}}^{\mathfrak {v}}{\begin{pmatrix}4\\-3\end{pmatrix}}={\begin{pmatrix}1&-2\\2&3\end{pmatrix}}{\begin{pmatrix}4\\-3\end{pmatrix}}={\begin{pmatrix}10\\-1\end{pmatrix}}\,$
with respect to the standard basis ${\mathfrak {u}}$. The transformation matrix $M_{\mathfrak {v}}^{\mathfrak {u}}$ is more difficult to compute: We have to write the standard vectors as linear combinations of $v_{1}$ and $v_{2}$. A direct computation (solving two linear systems) yields
    ${\begin{pmatrix}1\\0\end{pmatrix}}={\frac {3}{7}}{\begin{pmatrix}1\\2\end{pmatrix}}-{\frac {2}{7}}{\begin{pmatrix}-2\\3\end{pmatrix}}\,$
and
    ${\begin{pmatrix}0\\1\end{pmatrix}}={\frac {2}{7}}{\begin{pmatrix}1\\2\end{pmatrix}}+{\frac {1}{7}}{\begin{pmatrix}-2\\3\end{pmatrix}}\,.$
Hence,
    $M_{\mathfrak {v}}^{\mathfrak {u}}={\begin{pmatrix}{\frac {3}{7}}&{\frac {2}{7}}\\-{\frac {2}{7}}&{\frac {1}{7}}\end{pmatrix}}\,.$

## Lemma

Let $K$ be a field, and let $V$ be a $K$-vector space of dimension $n$. Let ${\mathfrak {u}}=u_{1},\ldots ,u_{n},\,{\mathfrak {v}}=v_{1},\ldots ,v_{n}$ and ${\mathfrak {w}}=w_{1},\ldots ,w_{n}$ denote bases of $V$. Then the three transformation matrices fulfill the relation
    $M_{\mathfrak {w}}^{\mathfrak {u}}=M_{\mathfrak {w}}^{\mathfrak {v}}\circ M_{\mathfrak {v}}^{\mathfrak {u}}\,.$
    $M_{\mathfrak {v}}^{\mathfrak {u}}\circ M_{\mathfrak {u}}^{\mathfrak {v}}=E_{n}\,.$

### Proof

$\Box$
* ↑ The commutativity of such a diagram of arrows and mappings means that all composed mappings coincide as long as their domain and codomain coincide. In this case, it simply means that $\Psi _{\mathfrak {v}}=\Psi _{\mathfrak {w}}\circ M_{\mathfrak {w}}^{\mathfrak {v}}$ holds.