First I am showing that one proof which will help to prove $I+A$ is an invertible matrix.
Statement :- Show that if $\lambda$ is an eigen value of A , then $(1+\lambda)$ is an eigen value of $I+A$ .
Proof :- Say $x$ is an eigen vector of A corresponding to eigen value $\lambda$ ,
$(I+A)x=Ix+Ax=1x+\lambda x=(1+\lambda)x$
So,$x$ is an eigen vector of $(A+I) $ corresponding to eigen value $(1+\lambda)$.
Now given matrix A has the property $A^{p}=0$ for some positive integer $p$. We call this matrix nilpotent matrix as well .
Statement :- If A is nilpotent matrix ,then All its eigen values are zero .
Proof:- Given $A^{p}=0$ . Let $\lambda$ is an eigen value of $A$ with eigen vector $x$.
So , $Ax=\lambda x$
or, $A^{p}x=A^{p-1}Ax=A^{p-1}\lambda x=\lambda A^{p-1}x=\lambda A^{p-2}Ax=\lambda A^{p-2}\lambda x=\lambda^{2} A^{p-2}x$$=……….=\lambda^{p}x$
So, $A^{p}x=\lambda^{p}x$
Since $A^{p}=0$ we can conclude either $\lambda^{p}=0$ or $x=0$ now as $x$ is eigen vector and eigen vector cannot be zero so ,
$\lambda^{p}=0$ so, $\lambda=0$ .
Now A has all $n$ eigen value as $0,0,0,0,0,…0$ .So A+I has all eigen value as n 1’s $1,1,1,1,1,...1.$
Now we know determinant of a matrix is product of all its eigen values .So, determinant of $\left | A+I \right |=1*1*1*1*1*.......*1=1$
Now We know that a matrix is invertible if its determinant is non-zero , so $A+I$ is inverible .
Part 2:- We know that ,the roots of the characteristic polynomials are eigen values of that matrix . If the n eigen values of the matrix is $\lambda _{1},\lambda _{2},\lambda _{3},...........,\lambda _{n}$ then characteristic polynomial is ,
$f(x)=(x-\lambda _{1})(x-\lambda _{2})(x-\lambda _{3}).......(x-\lambda _{n})$
Now Given Matrix $A$ is nilpotent matrix whose eigen values are all zero (already proved) .
So characteristic polynomial of A is ,
$f(x)=(x-0)(x-0)(x-0)........(x-0)=x^{n}$
$\large f(x)=x^{n}$
