Find the determinant of $I+A$ Let $A$ be a $2\times2$ matrix with real entries such that $A^2=0$.Find the determinant of $I+A$ where $I$ denotes the identity matrix. I proceed in this way :Note that $(I+A)A=A+A^2 \Longrightarrow (I+A)A=A$ (Since $A^2=0$).
Now taking determinant both side we get 
$|(I+A)A| = |A| \Longrightarrow |(I+A)|\cdot|A|=|A|\Longrightarrow |I+A|=1$.
Am I proceed in correct way? In the last line am I need to consider $|A|=0$?
 A: The matrix $A$ is nilpotent with characteristic polynomial $x^2.$ But by definition $x^2 = \det(A - xI).$ If we let $x = -1,$ then we have $\det(A + I) = 1.$
A: Let $x = \det (I+A)$
$$(I+A).(I+nA)=I+(n+1)A$$
Hence $\det(I+nA)=x^n$
But $\det(I+nA)=P(n)$ with P a polynomial (of maximum degree the size of matrix A).
So the only solutions are $x=1$ or $x=0$. But If $x=0$, there is a $v\neq 0$ such that $v+Av=0$, hence, $Av=-v$, and $A^2v=v$, that is not possible.
So $\det(I+A)=1$
Note that this proof is true for any size of matrix.
A: I like this one:
The characteristic polynomial of $A$ is $x^2$.
By substitution we get
$$(x-1)^2=x^2-2x+1$$
as characteristic polynomial of $A+I$. Now we see trace and determinant.
A: The trace of $A$ is $0$, using $$A^2-\operatorname{tr}(A)\cdot A+(\det A)\cdot I=0$$
and the fact that the determinant of $A$ is $0$. Applying this to $A+I$, we get 
$$\tag{1} A+2I-\operatorname{tr}(A+I)\cdot(A+I)+\det(A+I)\cdot I=0,$$
hence 
$$2A+I-2(A+I)+\det(A+I)\cdot I=0,$$
and $$\det(A+I)\cdot I=I,$$
hence $\det (A+I)$. 
The formula $(1)$ can we deduce either by Cayley Hamilton theorem (sledgehammer), or by hand.
A: $A$ is similar to (and hence has the same determinant as) a matrix of the form $\begin{pmatrix} 0 & * \\ 0 & 0 \end{pmatrix}$. That is, there is an invertible matrix $P$ such that
$$A':=P^{-1}AP = \begin{pmatrix} 0 & * \\ 0 & 0 \end{pmatrix}$$
and $\det A = \det A'$. But $I':=P^{-1}IP=I$, and so $P^{-1}(I+A)P=I+A'$, and so $\det (I+A) = \det (I+A')$.
So what is $\det (I+A)$?
