Let $A$ be an $n \times n$ matrix with real entries such that $A^2 + I = 0$ then $n$ is even. Let $A$ be an $n \times n$ matrix with real entries such that $A^2 + I = 0$ then $n$ is even. And if $n = 2k$, then $A$ is similar over the field of real numbers to a matrix of the block form 

$$\begin{bmatrix}
    0      &amp; -I \\
    I &amp; 0 \\
\end{bmatrix}$$ 
where $I$ is the $k \times k$ identity matrix. 

I have done the first part. Since $\det(A^2) = \det (-I)$ we must have $\det (-I)$ non negative and hence $n$ must be even.

Need Help in the second part.

Thank You.
You can solve both parts of the question at the same time as follows: Given $A$, define a map $\mathbb C\times\mathbb R^n\to\mathbb R^n$ by $(u+iv,x)\mapsto (u+iv)\cdot x:=ux+vAx$. Evidently, this is linear over $\mathbb R$ in both components and satisfies $(1,x)\mapsto x$. Using that $A^2=-I$, one easily verifies that it is multiplicative, i.e. $z\cdot (w\cdot x)=(zw)\cdot x$. Hence is satisfies all properties required from scalar multiplication, and hence makes $\mathbb R^n$ into a complex vector space, so $n=2m$ for some $m\in\mathbb N$. Moreover, choosing a complex basis $\{x_1,\dots,x_m\}$ for this vector space, $\{x_1,\dots,x_m,i\cdot x_1,\dots,i\cdot x_m\}$ is a real basis for $\mathbb R^n$. By definition the matrix representation of $A$ with respect to this real basis is the block matrix you have indicated.
I solve it another method : $det(A^2)=det(-I)=(-1)^n ....(1)$.

Now $\lambda$ is an eigenvalue of $A$ then ${\lambda}^2+I^2=0\Rightarrow \lambda =i,-i.$ So $detA=i(-i)=1\Rightarrow det(A^2)=(detA)^2=1 ....(2)$

So from $(1)$ and $(2)$ we get $n=2m$ for dome $m\in \mathbb{N} $