The changing basis matrix is invertible I'm trying to prove the changing basis matrix is invertible. I would like to know if my reasoning is correct.
Let $P_1$ and $P_2$ be changing basis matrices from the basis $\mathfrak B$ to the $\mathfrak B'$ and from $\mathfrak B'$ to $\mathfrak B$ respectively.
Then
$[\alpha]_{\mathfrak B'}=P_1[\alpha]_{\mathfrak B}$
$[\alpha]_{\mathfrak B}=P_2[\alpha]_{\mathfrak B'}$
Therefore:
$[\alpha]_{\mathfrak B}=P_2P_1[\alpha]_{\mathfrak B}$, for every $[\alpha]_{\mathfrak B}$.
So using the same reasoning of this question, we get $P_2P_1=I_n$, am I right?
 A: $P_1$ is defined by $[\alpha]_{\mathfrak B'}=P_1[\alpha]_{\mathfrak B}$,
where ${\mathfrak B},{\mathfrak B'}$ are bases of the same underlying space.
Also, note that when ${ \beta}$ is a basis, then $[\alpha]_\beta = 0$ iff $\alpha = 0$.
Since $\dim {\mathfrak B} = \dim {\mathfrak B'}$. it is sufficient to show
that $\ker P_1 = \{0\}$.
If $P_1[\alpha]_{\mathfrak B} = 0$, then $[\alpha]_{\mathfrak B'} = 0$
from which we get $\alpha = 0$ and hence $[\alpha]_{\mathfrak B} = 0$.
Hence $P_1$ is invertible.
A: $ {\mathfrak B'} $ and $  {\mathfrak B}  $ are   varying  so $P_1 $and  $ P _2 $ will be also be varying
And also you must be using the fact that you are using a basis right. That information is the most important otherwise this will not happen.So 


*

*Assuming that the dimension of the Vector Space is n.
The basis matrix has dimension n.
Now using the inequality  that rank (AB)<= rank(A),rank (B) 
And rank (A (nxn))<=n we get the equality . Please check


If yoi think i have a logic gap then tell me i will explain it further.
