Prove $A^2 = A$ where $A = I_n − \alpha\alpha^T$ and $\alpha$ is an $n × 1$ vector with $\alpha^T\alpha=1$ Consider a $n × n$ matrix $A = I_n − \alpha\alpha^T$, where $I_n$ is the $n × n$
identity matrix and $α$ is an $n × 1$ column vector such that $\alpha^T\alpha = 1$. Show that $A^2 = A$.

My proof:
$\alpha\alpha^T$ is a $n × n$ matrix. It is given that $\alpha^T\alpha = 1$. Multiplying both sides of this by the inverse of $\alpha^T$ gives, ${{\alpha^T}^{-1}}\alpha^T={\alpha^T}^{-1}I$.
$I\alpha={\alpha^T}^{-1}$ which means $\alpha={\alpha^T}^{-1}$. 
Now, $A = I_n − \alpha\alpha^T$,
=$I_n - {\alpha^T}^{-1}{\alpha^T}$ which means $A=I_n-I_n=\mathsf{O}$.
Therefore, $\forall n \in \mathbb{N}, A^n =\mathsf{O}$.
Is it correct?

$(I_n-\alpha\alpha^T)^2 = (I_n-\alpha\alpha^T)(I_n-\alpha\alpha^T) = I_n-2\alpha\alpha^T+\alpha\alpha^T\alpha\alpha^T = I_n-2\alpha\alpha^T+\alpha1\alpha^T = I_n-\alpha\alpha^T$
