# Vector, u:=[u_1,…,u_n]T^. Find a coordinate transformation matrix, Q in R^(nxn), which is nonsingular, satisfying: [0,...,0,||u||]=Qu

Vector, $u:=\left[{u}_{1},\dots ,{u}_{n}{\right]}^{\mathrm{T}}$. I am trying to find a coordinate transformation matrix, $Q\in {\mathbb{R}}^{n×n}$, which is nonsingular, satisfying:
$\begin{array}{r}\left[\begin{array}{c}0\\ ⋮\\ 0\\ ||u||\end{array}\right]=Qu.\end{array}$
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Abagail Stephenson
Recall that if $W$ is a subespace of ${\mathbb{R}}^{n}$, then
$\mathrm{dim}\left(W\right)+\mathrm{dim}\left({W}^{\perp }\right)=n$
Take $W$ as the span of $u$, so
$\mathrm{dim}\left(\right)=n-1.$
You can construct a matrix $Q$ that will satisfy the same condition taking any set of $n-1$ vectors ${v}_{i}$ from any basis of the ortogonal complement of the span of $u$.
$Q=\left(\begin{array}{cccc}{v}_{11}& {v}_{12}& \cdots & {v}_{1n}\\ {v}_{21}& {v}_{22}& \cdots & {v}_{2n}\\ {v}_{31}& {v}_{32}& \cdots & {v}_{3n}\\ ⋮& ⋮& ⋮& ⋮\\ {v}_{n-1,1}& {v}_{n-1,2}& \cdots & {v}_{n-1,n}\\ \frac{{u}_{1}}{||u||}& \frac{{u}_{2}}{||u||}& \cdots & \frac{{u}_{n}}{||u||}\end{array}\right)$
where ${v}_{i}=\left({v}_{i1},\dots ,{v}_{in}{\right)}^{T}$ are vectors in a basis $\left\{{v}_{1},\dots ,{v}_{n-1}\right\}$ of the ortogonal complement of the linear span of $г$, for all $i$ with $1\le i\le n-1$.
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Darius Miles
This matrix $Q$ is non singular because it's rows are linearly independent because the following set is a basis of ${\mathbb{R}}^{n}$:
$\mathcal{B}=\left\{u,{v}_{1},\dots ,{v}_{n-1}\right\}$
Also: You can use Gram Schmidt process to get an ortogonal basis of $}^{\perp }$ starting from $u$. In this way you will find the rights vectors ${v}_{i}$ such that $Q$ is inversible.