Find an orthogonal matrix $V$ such that $V^{T}B(\gamma)V=diag(1+\gamma n,1,1,\cdots,1)$

Question

Let $B(\gamma)=I+\gamma ee^{T}$ where $e=(1,1,\cdots,1)^{T}$ and $\gamma$ a strictly positive real number.

I would like to find an orthogonal matrix $V$ such that $V^{T}B(\gamma)V=diag(1+\gamma n,1,1,\cdots,1)$

It seems closely related to find a 'base change matrix'.

Moreover $e$ is eigenvector of $B(\gamma)$, indeed we have $B(\gamma)e=Ie+\gamma ee^{T}e=e+\gamma e n=(1+\gamma n)e$. So it seems that I have to find a basis of eigenvector but how can I do this here?

Answer 1

Take an orthogonal matrix $V$ such that $V^Te=\|e\|_2e_1=\sqrt{n}e_1$, where $e_1=[1,0,\ldots,0]^T$ (take any orthogonal matrix $V$ with the first column equal to $e/\sqrt{n}$). Then $$ V^T(I+\gamma e e^T)V=V^TV+\gamma (V^Te)(V^Te)^T=I+\gamma n e_1e_1^T $$ is your diagonal matrix.