# Linear mapping/Change of basis/No proof/Section

## Lemma

Let ${\displaystyle {}K}$ denote a field, and let ${\displaystyle {}V}$ and ${\displaystyle {}W}$ denote finite-dimensional ${\displaystyle {}K}$-vector spaces. Let ${\displaystyle {}{\mathfrak {v}}}$ and ${\displaystyle {}{\mathfrak {u}}}$ be bases of ${\displaystyle {}V}$ and ${\displaystyle {}{\mathfrak {w}}}$ and ${\displaystyle {}{\mathfrak {z}}}$ bases of ${\displaystyle {}W}$. Let

${\displaystyle \varphi \colon V\longrightarrow W}$

denote a linear mapping, which is described by the matrix ${\displaystyle {}M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )}$ with respect to the bases ${\displaystyle {}{\mathfrak {v}}}$ and ${\displaystyle {}{\mathfrak {w}}}$. Then ${\displaystyle {}\varphi }$ is described with respect to the bases ${\displaystyle {}{\mathfrak {u}}}$ and ${\displaystyle {}{\mathfrak {z}}}$ by the matrix

${\displaystyle M_{\mathfrak {z}}^{\mathfrak {w}}\circ (M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi ))\circ (M_{\mathfrak {u}}^{\mathfrak {v}})^{-1},}$

where ${\displaystyle {}M_{\mathfrak {u}}^{\mathfrak {v}}}$ and ${\displaystyle {}M_{\mathfrak {z}}^{\mathfrak {w}}}$ are the transformation matrices, which describe the change of basis from ${\displaystyle {}{\mathfrak {v}}}$ to ${\displaystyle {}{\mathfrak {u}}}$ and from ${\displaystyle {}{\mathfrak {w}}}$ to ${\displaystyle {}{\mathfrak {z}}}$.

### Proof

This proof was not presented in the lecture.
${\displaystyle \Box }$

## Corollary

Let ${\displaystyle {}K}$ denote a field, and let ${\displaystyle {}V}$ denote a ${\displaystyle {}K}$-vector space of finite dimension. Let

${\displaystyle \varphi \colon V\longrightarrow V}$

be a linear mapping. Let ${\displaystyle {}{\mathfrak {u}}}$ and ${\displaystyle {}{\mathfrak {v}}}$ denote bases of ${\displaystyle {}V}$. Then the matrices which describe the linear mapping with respect to ${\displaystyle {}{\mathfrak {u}}}$ and ${\displaystyle {}{\mathfrak {v}}}$ respectively (on both sides), fulfil the relation

${\displaystyle {}M_{\mathfrak {u}}^{\mathfrak {u}}(\varphi )=M_{\mathfrak {u}}^{\mathfrak {v}}\circ M_{\mathfrak {v}}^{\mathfrak {v}}(\varphi )\circ (M_{\mathfrak {u}}^{\mathfrak {v}})^{-1}\,.}$

### Proof

This follows directly from fact.

${\displaystyle \Box }$

## Definition

Two square matrices ${\displaystyle {}M,N\in \operatorname {Mat} _{n}(K)}$ are called similar, if there exists an invertible matrix ${\displaystyle {}B}$ with

${\displaystyle {}M=BNB^{-1}}$.

Due to fact, for a linear mapping ${\displaystyle {}\varphi \colon V\rightarrow V}$, the describing matrices with respect to several bases are similar.