Problem related to a square matrix Let $A$ be an $n\times n$ matrix with real entries such that $A^{2}+I=\mathbf{0}$. Then: 
(A) $n$ is an odd integer. 
(B) $n$ is an even integer. 
(C) $n$ has to be $2$ 
(D) $n$ could be any positive integer.
I was thinking about the problem.I noticed for a  $2\times 2$ matrix $A$ of the form 
$$\begin{pmatrix}
1 &-2 \\ 
 1& -1
\end{pmatrix},$$ the given condition holds good.So option (C) is a possibility.But I am not sure about other options. Is there any convenient way to tackle it?With regards..
 A: As $A$ is real, so is $\det A$. Yet $A^2+I=0$ implies that $(\det A)^2=\det(-I)=(-1)^n$. Therefore $n$ has to be an even integer and (A), (D) are incorrect. Now, if there is a $2\times 2$ matrix $A$ such that $A^2+I=0$, then $\tilde{A}=\begin{pmatrix}A\\&A\end{pmatrix}$ would be a $4\times4$ matrix such that $\tilde{A}^2+I=0$. Therefore (C) is incorrect too. To verify that (B) is indeed the correct answer, consider the aforementioned $\tilde{A}$ with $A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$.
A: Note: 
$n=1\,$ is ruled out, since e.g., $A$ consists of the single scalar entry $1$: $A = [1],\; I = I_1,\;, A^2 + I = 2.\;$ 
Indeed there is no real scalar $\,k\,\neq 0\,$ (in the case $n = 1, A = k\,$) such that $\,k^2 = -1.\,$ 
So option (A) is ruled out, since $\,n = 1\,$ is odd, and option (D) is ruled out, since $\,n = 1 >0\, n \in \mathbb{Z}^+$. 
What remains is to decide between (B) and (C). 
You know for $n = 2\,$, the equality is satisfied (hence "(C)" is in the "running") but $n = 2\,$ is also even: so "(B)" has a chance. You can rule out (C) if there exists any $\,n= 2k,\; k\in \mathbb{Z}^+$, $n > 2,\,$ such that $\,A_{n\times n}^2 + I_n\, = 0$. 


*

*Hint: try $n = 4$: construct a $4\times 4$ matrix made of $4$-square block $2 \times 2$ matrices, using your matrix for each of the two block entries on the diagonal, and zero blocks off the diagonal.)


Let $A = 
\begin{pmatrix}
1 &-2 \\ 
 1& -1
\end{pmatrix}.
\quad$ So using $A^2 + I = 0$, construct
$A_{4 \times 4} = 
\begin{pmatrix}
A & 0 \\
0 & A
\end{pmatrix}
$
$$
A_{4\times 4}^2 + I_4 = 
\begin{pmatrix}
A &0 \\ 
0& A
\end{pmatrix}
\cdot 
\begin{pmatrix}
A & 0 \\ 
0 & A
\end{pmatrix}
+ I_4 =
\begin{pmatrix}
A^2 & 0 \\ 
 0& A^2
\end{pmatrix} +
\begin{pmatrix}
I_2 & 0\\
0 & I_2
\end{pmatrix}
$$
$$
=
\begin{pmatrix} 
A^2 + I_2 & 0\\
0 & A^2 + I_2
\end{pmatrix}= 0
$$
A: Hint: Let's $n>1$ an interger positive and even. Set the $n\times n$ matrix
$$
J_{n\times n}=\begin{pmatrix} 
 0 & \ldots & 0 & \ldots  &-1 \\
 \vdots &  & \vdots  &  & \vdots \\ 
 0 & \ldots & -1 & \ldots & 0 \\
 \vdots &  & \vdots  &  & \vdots \\ 
 -1 & \ldots & 0 & \ldots  &0 \\  \end{pmatrix}.
$$ 
Note that $I_{n\times n}=J_{n\times n}^2$ for if $n$ is even and $$A^2+I_{n\times n}= ( A+J_{n\times n})(A-J_{n\times n})=0.
$$
Then we have $A=\pm J$. If for n=1 we are a conter exemple. It's exclude (D). Then the answer is (C).
