Given a unitary vector $(x_1,\ldots,x_n)^T$ and a vector $y=(x_2,\ldots, x_n)^T$, prove the columns of a created matrix form an orthonormal basis 
Let $X=\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}\in M_{n1}(\Bbb F),x_1\ne-1$ be a unit vector wrt the standard scalar product and let $Y=\begin{bmatrix}x_2\\x_3\\\vdots\\x_n\end{bmatrix}\in M_{n-1,1}(\Bbb F).$
Prove that the columns of the matrix $$M=\begin{bmatrix}x_1&y^*\\y&\frac1{1+x_1}yy^*-I\end{bmatrix}$$ form an orthonormal basis for $M_{n1}(\Bbb F)$
Note: $y^*$ denotes the conjugate transpose of $y$.

This is what I tried:
I would like to prove the given
matrix $M$ is unitary by showing $MM^*=I$.
First, I wanted to analyse the Hermitian matrix $$yy^*=\begin{bmatrix}x_2\bar x_2&x_2\bar x_3&\ldots&x_2\bar x_n\\x_3\bar x_2&x_3\bar x_3&\ldots &x_3\bar x_n\\\vdots&\vdots&\ddots&\vdots\\x_n\bar x_2&x_n\bar x_3&\ldots&x_n\bar x_n\end{bmatrix}$$
We can see that $\operatorname{rank}(yy^*)=1\implies\operatorname{Ker}(yy^*)=n-1$.
Furthermore, $$yy^*y=y(y^*y)=\|y\|^2y,$$
which means $y$ is an eigenvector of the matrix $yy^*$ corresponding to the eigenvalue $\|y\|^2=1-|x_1|^2=1-x_1\bar x_1$ since $\|x\|=\sqrt{\sum\limits_{i=1}^n|x_i|^2}=1$, and we conclude $\sigma(yy^*)=\{0,1-x_1\bar x_1\}$.
If $x_1\in\Bbb R, M=M^*,$ but it can happen that $x_1\not\in\Bbb R$.
I tried writing: $$\left(\frac1{1+x_1}yy^*-I\right)y=\begin{bmatrix}\left(\frac{1-x_1\bar x_1}{1+x_1}-1\right)x_2\\\left(\frac{1-x_1\bar x_1}{1+x_1}-1\right)x_3\\\vdots\\\left(\frac{1-x_1\bar x_1}{1+x_1}-1\right)x_n\end{bmatrix}=\begin{bmatrix}-\frac{1+\bar x_1}{1+x_1}x_1x_2\\-\frac{1+\bar x_1}{1+x_1}x_1x_3\\\vdots\\-\frac{1+\bar x_1}{1+x_1}x_1x_n\end{bmatrix}$$
but it wasn't quite helpful.
How should I proceede?
Thank you very much for your time and patience!

As given, $\ M\ $ is not always unitary.  Take
$$
X=\begin{bmatrix}\frac{i}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}
\end{bmatrix}\ .
$$
then
$$
M=\begin{bmatrix}\frac{i}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}&\frac{1}{2+i\sqrt{2}}-1
\end{bmatrix}=\begin{bmatrix}\frac{i}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}&-\frac{1+i\sqrt{2}}{2+i\sqrt{2}}
\end{bmatrix}
$$
\begin{align}
MM^*&=\begin{bmatrix}\frac{i}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}&-\frac{1+i\sqrt{2}}{2+i\sqrt{2}}
\end{bmatrix}\begin{bmatrix}-\frac{i}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}&\frac{i\sqrt{2}-1}{2-i\sqrt{2}}
\end{bmatrix}\\
&=\begin{bmatrix}1&\frac{i}{2}+\frac{i\sqrt{2}-1}{2\sqrt{2}-2i}\\
\frac{-i}{2}-\frac{1+i\sqrt{2}}{2\sqrt{2}+2i}&1
\end{bmatrix}\\
&=\begin{bmatrix}
1&\frac{2\sqrt{2}i}{2\sqrt{2}-2i}\\
\frac{-2\sqrt{2}i}{2\sqrt{2}+2i}&1
\end{bmatrix}\\
&=\begin{bmatrix}
1&\frac{-\sqrt{2}+2i}{3}\\
\frac{-\sqrt{2}-2i}{3}&1
\end{bmatrix}\ .
\end{align}
