Linear mapping/Change of basis/No proof/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 that describe the change of basis from ${}{\mathfrak {v}}$ to ${}{\mathfrak {u}}$ and from

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

Proof

This proof was not presented in the lecture.

$\Box$

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) fulfill 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.

$\Box$

Definition

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

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

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