# 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: $$left(frac1{1+x_1}yy^*-Iright)y=begin{bmatrix}left(frac{1-x_1bar x_1}{1+x_1}-1right)x_2\left(frac{1-x_1bar x_1}{1+x_1}-1right)x_3\vdots\left(frac{1-x_1bar x_1}{1+x_1}-1right)x_nend{bmatrix}=begin{bmatrix}-frac{1+bar x_1}{1+x_1}x_1x_2\-frac{1+bar x_1}{1+x_1}x_1x_3\vdots\-frac{1+bar x_1}{1+x_1}x_1x_nend{bmatrix}$$