linear algebra – 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 $X=begin{bmatrix}x_1\x_2\vdots\x_nend{bmatrix}in M_{n1}(Bbb F),x_1ne-1$ be a unit vector wrt the standard scalar product and let $Y=begin{bmatrix}x_2\x_3\vdots\x_nend{bmatrix}in M_{n-1,1}(Bbb F).$

Prove that the columns of the matrix $$M=begin{bmatrix}x_1&y^*\y&frac1{1+x_1}yy^*-Iend{bmatrix}$$ 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 $$yy^*=begin{bmatrix}x_2bar x_2&x_2bar x_3&ldots&x_2bar x_n\x_3bar x_2&x_3bar x_3&ldots &x_3bar x_n\vdots&vdots&ddots&vdots\x_nbar x_n&x_1bar x_2&ldots&x_nbar x_nend{bmatrix}$$
We can see that $operatorname{rank}(yy^*)=1impliesoperatorname{Ker}(yy^*)=1$.

Furthermore, $$yy^*y=y(y^*y)=|y|^2y,$$

which means $y$ is an eigenvector of the matrix $yy^*$ corresponding to the eigenvalue $|y|^2=1-|x_1|^2=1-x_1bar x_1$ and we conclude $sigma(yy^*)={0,1-x_1bar x_1}$.

If $x_1inBbb R, M=M^*,$ but it can happen that $x_1notinBbb R$.
I tried writing: $$left(frac1{1+x_1}yy^*-Iright)y=begin{bmatrix}left(frac{1-x_1bar x_1}{1+x_1}-1right)x_2\left(frac{1-x_1bar x_1}{1+x_1}-1right)x_3\vdots\left(frac{1-x_1bar x_1}{1+x_1}-1right)x_nend{bmatrix}=begin{bmatrix}-frac{1+bar x_1}{1+x_1}x_1x_2\-frac{1+bar x_1}{1+x_1}x_1x_3\vdots\-frac{1+bar x_1}{1+x_1}x_1x_nend{bmatrix}$$
but it wasn’t quite helpful.

How should I proceede?

Thank you very much for your time and patience!