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I have a vector, $u := [u_1, \ldots, u_n]^\mathrm{T}$. I am trying to find a coordinate transformation matrix, $Q \in \mathbb{R}^{n \times n}$, which is nonsingular, satisfying: \begin{align*} \begin{bmatrix} 0 \\ \vdots \\ 0 \\ ||u|| \end{bmatrix} = Q u. \end{align*} I would appreciate any idea to find this matrix, $Q$.

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Recall that if $W$ is a subespace of $\mathbb{R}^{n}$, then $$\dim(W)+\dim(W^{\perp})=n$$ Take $W$ as the span of $u$, so $$\dim(<u>)=n-1.$$ 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 $u$. $$Q=\left(\begin{array}{cccc} v_{11}&v_{12}&\cdots&v_{1n}\\ v_{21}&v_{22}&\cdots &v_{2n}\\ v_{31}&v_{32}&\cdots&v_{3n}\\ \vdots&\vdots&\vdots&\vdots\\ v_{n-1,1}&v_{n-1,2}&\cdots&v_{n-1,n}\\ \displaystyle{\frac{u_1}{\vert\vert{u}\vert\vert}}&\displaystyle{\frac{u_2}{\vert\vert{u}\vert\vert}}&\cdots&\displaystyle{\frac{u_n}{\vert\vert{u}\vert\vert}}\\ \end{array}\right)$$ where $v_{i}=(v_{i1},\ldots,v_{in})^{T}$ are vectors in a basis $\{v_{1},\ldots,v_{n-1}\}$ of the ortogonal complement of the linear span of $u$, for all $i$ with $1\leq i\leq n-1$.

Remark: This matrix $Q$ is non singular because it's rows are linearly independent because the following set is a basis of $\mathbb{R}^{n}$: $$\mathcal{B}=\{u,v_1,\ldots,v_{n-1}\}$$

Also: You can use Gram Schmidt process to get an ortogonal basis of $<u>^{\perp}$ starting from $u$. In this way you will find the rights vectors $v_i$ such that $Q$ is inversible.

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  • $\begingroup$ Thank you for your comment. I omitted an info that Q should be nonsingular. $\endgroup$ – H.C. Nov 19 '17 at 4:12
  • $\begingroup$ Oh ! Yes sure, you want a matrix to do change of basis :) $\endgroup$ – Hector Blandin Nov 19 '17 at 4:13
  • $\begingroup$ You said it when you refer to coordinate transformation matrix, this matrix should be inversible. So, is my mistake :) $\endgroup$ – Hector Blandin Nov 19 '17 at 4:14
  • $\begingroup$ @H.C.: You can use my matrix $Q_1$ and Gram Schmidt algorithm see here: en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process $\endgroup$ – Hector Blandin Nov 19 '17 at 4:18
  • $\begingroup$ @H.C.: Look to this answer also: math.stackexchange.com/questions/1766613/… $\endgroup$ – Hector Blandin Nov 19 '17 at 4:34

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