Show that $a\gamma_B(v)=v$ for all $v\in \mathbb{R}^n$ Let $1\leq n\in \mathbb{N}$, $E=\left (e_1, \ldots , e_n\right )$ the standardbasis of $\mathbb{R}^n$, $B=\left (b_1, \ldots , b_n\right )$ a basis of $\mathbb{R}^n$ and $a:=\left (b_1\mid b_2\mid \ldots \mid b_n\right )\in M_n(\mathbb{R}^n)$.
Show the following:

*

*$a\gamma_B(v)=v$ for all $v\in \mathbb{R}^n$.


*$a$ is invertible and it holds that $a^{-1}=\left (\gamma_B(e_1)\mid \gamma_B(e_2)\mid \ldots \mid \gamma_B(e_n)\right )$.
I have done an example and I have seen that it holds that $\left (b_1\mid b_2\mid b_3\right )\left (\gamma_B(e_1)\mid \gamma_B(e_2)\mid \gamma_B(e_3)\right )=I_{3\times 3}$.
But how can we show that it holds?
We have that $$\gamma_B(v)=v \Rightarrow c_1b_1+c_2b_2+\ldots c_nb_n=v$$ then multiplying with the matrix $a$ and knowing that $B$ is a basis we get the result, but how exactly can we show that?
 A: Let $v=c_1 b_1+\dots+c_n b_n$ so that $\gamma_B(v) = (c_1,\dots,c_n)^T$. We then have
$$
a\,\gamma_B(v) = (b_1\mid \dots\mid b_n)(c_1,\dots,c_n)^T = c_1 b_1 + \dots + c_n b_n = v.
$$
Since the columns of $a$ form a basis of $\mathbb R^n$, the matrix has full rank and hence is invertible.
Now note that $a\,\gamma_B(v) = v$ is equivalent to $\gamma_B(v) = a^{-1} v$. Consider $v=e_i$ to obtain $\gamma_B(e_i) = a^{-1} e_i$. Since in general $Me_i$ is the $i$-th column of the matrix $M$, we conclude that
$$
a^{-1} = (\gamma_B(e_1)\mid\dots\mid\gamma_B(e_n)).
$$
A: Ok, we can temporarily suspend disbelief and assume that $a$ is a vector and not a matrix. If $v = c_1b_1+\cdots + c_nb_n$, we have
$$a\gamma_B(v) = \begin{bmatrix} b_1 & b_2 & \cdots & b_n\end{bmatrix}\begin{bmatrix}c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix}  = c_1b_1 + c_2b_2 +\cdots c_nb_n = v.$$
Now, using this we get that $$a\gamma_B(e_i) = e_i, \quad 1 \le i \le n$$
so
$$a\begin{bmatrix}\gamma_B(e_1)  & \cdots & \gamma_B(e_n)\end{bmatrix} = \begin{bmatrix}a\gamma_B(e_1)  & \cdots & a\gamma_B(e_n)\end{bmatrix} = \begin{bmatrix} e_1 & \cdots & e_n\end{bmatrix} = I$$
so $a$ is invertible with inverse $a^{-1} = \begin{bmatrix}\gamma_B(e_1)  & \cdots & \gamma_B(e_n)\end{bmatrix}$.
A: Assuming $(i)$ we get $(ii)$ by direct computation:
$$
a(\gamma_B(e_q)\mid\cdots\mid\gamma_B(e_n)) = (a\gamma_B(e_q)\mid \cdots \mid a\gamma_B(e_n)) = (e_1 \mid \cdots \mid e_n) = I 
$$
so $a^{-1} = (\gamma_B(e_q)\mid\cdots\mid\gamma_B(e_n))$. Here we use that for square matrices $ab = 1$ implies $ba = 1$.
Now, for $(i)$, if $\gamma_B(v) = (\lambda_1,\ldots,\lambda_n)$ then this precisely means that $v = \sum_i \lambda_i b_i$, which is the same as
$$
v_j = \sum_i \lambda_i (b_i)_j = \sum_i \lambda_i a_{ij} = \sum_i a_{ij}\lambda_i = [a\cdot (\lambda_1, \ldots, \lambda_n)]_j = [a \gamma_B(v)]_j
$$
for all $j$, and so $v = a\gamma_B(v)$.
