<s>[INST] Show that $B$ is invertible if $B=A^2-2A+2I$ and $A^3=2I$ If $A$ is $40\times 40$ matrix such that $A^3=2I$ show that $B$ is invertible where $B=A^2-2A+2I$. 
I tried to evaluate $B(A-I)$ , $B(A+I)$ , $B(A-2I)$ ...  but I couldn't find anything.
 [/INST] One of the easiest ways to prove $B$ is invertible is the following;
$$B=A^2-2A+2I$$
Using that $A^3=2I$, we can write 
$$B=A^2-2A+2I= A^2-2A + A^3 = A(A^2+A-2I)=A(A+2I)(A-I)$$
Now let us prove $A$, $A+2I$ and $A-I$ are invertible. 


*

*Supose there is $v$ such that $Av=0$, then we have $0=A^20=A^3v=2Iv=2v$. So we have $v=0$. So $A$ is invertible. 

*Supose there is $v$ such that $(A+2I)v=0$, then we have $Av=-2v$ and we have $A^3v=-8v$. So we get $-8v=A^3v=2Iv=2v$. So we have $v=0$. So $(A+2I)$ is invertible.   

*Supose there is $v$ such that $(A-I)v=0$, then we have $Av=v$ and we have $A^3v=v$. So we get $v=A^3v=2Iv=2v$. So we have $v=0$. So $(A-I)$ is invertible.
Since B is the product of three invertible matrices, B itself is invertible. 
Remark: For item 1 above, we also simply present the inverse of $A$. From $A^3=2I$, we have $A^{-1}=\frac{1}{2}A^2$.
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