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

${}{\mathfrak {w}}$ to ${}{\mathfrak {z}}$.

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 that 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“.