Injective linear map L

Question

Let $n \in \mathbb{N}_{0}$ and let $L: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ be an injective linear map. Let $A \in \mathbb{R}^{n \times n}$ be an invertible matrix. Prove that for every base $\alpha$ of $\mathbb{R}^{n}$ there exists a base $\beta$ of $\mathbb{R}^{n}$ so that $A=L^{\beta}_{\alpha}$.

Answer 1

Since $L$ is injective, we know that $\ker(L)=\{0\}$, hence $\dim(Im(L))=\dim(\mathbb{R}^{n}) = n$ because of the rank–nullity theorem. Because of this we know that the matrix for $L$ is invertible, which we will need later.

We will now try to find the vectors in $\beta$ so that $A=L^{\beta}_{\alpha}$ holds.

Consider $\varepsilon$ the standard basis of $\mathbb{R}^{n}$. We can then see that

\begin{equation} \begin{split} A & = L^{\beta}_{\alpha} \\ & = Id^{\beta}_{\varepsilon} \cdot L \cdot Id^{\varepsilon}_{\alpha} \\ \end{split} \end{equation}

(Where $Id^{b}_{a}$ is the matrix for change of basis from basis $a$ to basis $b$. Also note that $(Id^{b}_{a})^{-1}=Id^{a}_{b}$ )

From this we can find that

$$ A \cdot Id^{\alpha}_{\varepsilon} \cdot L^{-1} = Id^{\beta}_{\varepsilon} \\ $$

If we take the inverse of this, we get $(Id^{\beta}_{\varepsilon})^{-1}=Id^{\varepsilon}_{\beta}$. From this matrix we can get the elements we need to construct $\beta$. Consider $\beta = \{\beta_{1},\beta_{2},...,\beta_{n}\}$. Then $\beta_{i}$ is equal to the elements in the $i$-th column of $Id^{\varepsilon}_{\beta}$.