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?
$P_1$ 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.
$ {\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)&lt;= rank(A),rank (B) 
And rank (A (nxn))&lt;=n we get the equality . Please check


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