# Linear mapping/Base change/Section
## Lemma

Let $K$ denote a field, and let $V$ and $W$ denote finite-dimensional $K$-vector spaces. Let ${\mathfrak {v}}$ and ${\mathfrak {u}}$ be bases of $V$ and ${\mathfrak {w}}$ and ${\mathfrak {z}}$ bases of $W$. Let
    $\varphi \colon V\longrightarrow W$
denote a linear mapping, which is described by the matrix $M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )$ with respect to the bases ${\mathfrak {v}}$ and ${\mathfrak {w}}$. Then $\varphi$ is described with respect to the bases ${\mathfrak {u}}$ and ${\mathfrak {z}}$ by the matrix
    $M_{\mathfrak {z}}^{\mathfrak {w}}\circ (M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi ))\circ (M_{\mathfrak {u}}^{\mathfrak {v}})^{-1},$
where $M_{\mathfrak {u}}^{\mathfrak {v}}$ and $M_{\mathfrak {z}}^{\mathfrak {w}}$ are the transformation matrices, which describe the change of basis from ${\mathfrak {v}}$ to ${\mathfrak {u}}$ and from
${\mathfrak {w}}$
${\mathfrak {z}}$

### Proof

The linear standard mappings $K^{n}\rightarrow V$ and $K^{m}\rightarrow W$ for the various bases are denoted by $\Psi _{\mathfrak {v}},\,\Psi _{\mathfrak {u}},\,\Psi _{\mathfrak {w}},\,\Psi _{\mathfrak {z}}$. We consider the commutative diagram
    ${\begin{matrix}K^{n}&&&{\stackrel {M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )}{\longrightarrow }}&&&K^{m}\\&\searrow \Psi _{\mathfrak {v}}\!\!\!\!\!&&&&\Psi _{\mathfrak {w}}\swarrow \!\!\!\!\!&\\\!\!\!\!\!M_{\mathfrak {u}}^{\mathfrak {v}}\downarrow &&V&{\stackrel {\varphi }{\longrightarrow }}&W&&\,\,\,\,\downarrow M_{\mathfrak {z}}^{\mathfrak {w}}\\&\nearrow \Psi _{\mathfrak {u}}\!\!\!\!\!&&&&\Psi _{\mathfrak {z}}\nwarrow \!\!\!\!\!&\\K^{n}&&&{\stackrel {M_{\mathfrak {z}}^{\mathfrak {u}}(\varphi )}{\longrightarrow }}&&&K^{m},\!\!\!\!\!\end{matrix}}$
where the commutativity rests on fact and fact. In this situation, we have altogether
    ${\begin{aligned}M_{\mathfrak {z}}^{\mathfrak {u}}(\varphi )&=\Psi _{\mathfrak {z}}^{-1}\circ \varphi \circ \Psi _{\mathfrak {u}}\\&=\Psi _{\mathfrak {z}}^{-1}\circ (\Psi _{\mathfrak {w}}\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\circ \Psi _{\mathfrak {v}}^{-1})\circ \Psi _{\mathfrak {u}}\\&=(\Psi _{\mathfrak {z}}^{-1}\circ \Psi _{\mathfrak {w}})\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\circ (\Psi _{\mathfrak {v}}^{-1}\circ \Psi _{\mathfrak {u}})\\&=(\Psi _{\mathfrak {z}}^{-1}\circ \Psi _{\mathfrak {w}})\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\circ (\Psi _{\mathfrak {u}}^{-1}\circ \Psi _{\mathfrak {v}})^{-1}\\&=M_{\mathfrak {z}}^{\mathfrak {w}}\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\circ (M_{\mathfrak {u}}^{\mathfrak {v}})^{-1}.\end{aligned}}$
$\Box$

## Corollary

Let $K$ denote a field, and let $V$ denote a $K$-vector space of finite dimension. Let
    $\varphi \colon V\longrightarrow V$
be a linear mapping. Let ${\mathfrak {u}}$ and ${\mathfrak {v}}$ denote bases of $V$. Then the matrices which describe the linear mapping with respect to ${\mathfrak {u}}$ and ${\mathfrak {v}}$ respectively (on both sides), fulfil the relation
    $M_{\mathfrak {u}}^{\mathfrak {u}}(\varphi )=M_{\mathfrak {u}}^{\mathfrak {v}}\circ M_{\mathfrak {v}}^{\mathfrak {v}}(\varphi )\circ (M_{\mathfrak {u}}^{\mathfrak {v}})^{-1}\,.$

### Proof

This follows directly from fact.
$\Box$
It is an important goal of linear algebra to find, for a given linear mapping $\varphi \colon V\rightarrow V$, a basis ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$ such that the describing matrix $M_{\mathfrak {v}}^{\mathfrak {v}}(\varphi )$ becomes "quite simple“.