$$\begin{array}{r}\left[\begin{array}{c}0\\ \vdots \\ 0\\ ||u||\end{array}\right]=Qu.\end{array}$$

Celinamg8
2022-09-20
Answered

Vector, $u:=[{u}_{1},\dots ,{u}_{n}{]}^{\mathrm{T}}$. I am trying to find a coordinate transformation matrix, $Q\in {\mathbb{R}}^{n\times n}$, which is nonsingular, satisfying:

$$\begin{array}{r}\left[\begin{array}{c}0\\ \vdots \\ 0\\ ||u||\end{array}\right]=Qu.\end{array}$$

$$\begin{array}{r}\left[\begin{array}{c}0\\ \vdots \\ 0\\ ||u||\end{array}\right]=Qu.\end{array}$$

You can still ask an expert for help

Abagail Stephenson

Answered 2022-09-21
Author has **9** answers

Recall that if $W$ is a subespace of ${\mathbb{R}}^{n}$, then

$$\mathrm{dim}(W)+\mathrm{dim}({W}^{\perp})=n$$

Take $W$ as the span of $u$, so

$$\mathrm{dim}(<u>)=n-1.$$

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 $u$.

$$Q=\left(\begin{array}{cccc}{v}_{11}& {v}_{12}& \cdots & {v}_{1n}\\ {v}_{21}& {v}_{22}& \cdots & {v}_{2n}\\ {v}_{31}& {v}_{32}& \cdots & {v}_{3n}\\ \vdots & \vdots & \vdots & \vdots \\ {v}_{n-1,1}& {v}_{n-1,2}& \cdots & {v}_{n-1,n}\\ {\displaystyle \frac{{u}_{1}}{||u||}}& {\displaystyle \frac{{u}_{2}}{||u||}}& \cdots & {\displaystyle \frac{{u}_{n}}{||u||}}\end{array}\right)$$

where ${v}_{i}=({v}_{i1},\dots ,{v}_{in}{)}^{T}$ are vectors in a basis $\{{v}_{1},\dots ,{v}_{n-1}\}$ of the ortogonal complement of the linear span of $\u0433$, for all $i$ with $1\le i\le n-1$.

$$\mathrm{dim}(W)+\mathrm{dim}({W}^{\perp})=n$$

Take $W$ as the span of $u$, so

$$\mathrm{dim}(<u>)=n-1.$$

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 $u$.

$$Q=\left(\begin{array}{cccc}{v}_{11}& {v}_{12}& \cdots & {v}_{1n}\\ {v}_{21}& {v}_{22}& \cdots & {v}_{2n}\\ {v}_{31}& {v}_{32}& \cdots & {v}_{3n}\\ \vdots & \vdots & \vdots & \vdots \\ {v}_{n-1,1}& {v}_{n-1,2}& \cdots & {v}_{n-1,n}\\ {\displaystyle \frac{{u}_{1}}{||u||}}& {\displaystyle \frac{{u}_{2}}{||u||}}& \cdots & {\displaystyle \frac{{u}_{n}}{||u||}}\end{array}\right)$$

where ${v}_{i}=({v}_{i1},\dots ,{v}_{in}{)}^{T}$ are vectors in a basis $\{{v}_{1},\dots ,{v}_{n-1}\}$ of the ortogonal complement of the linear span of $\u0433$, for all $i$ with $1\le i\le n-1$.

Darius Miles

Answered 2022-09-22
Author has **3** answers

This matrix $Q$ is non singular because it's rows are linearly independent because the following set is a basis of ${\mathbb{R}}^{n}$:

$$\mathcal{B}=\{u,{v}_{1},\dots ,{v}_{n-1}\}$$

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.

$$\mathcal{B}=\{u,{v}_{1},\dots ,{v}_{n-1}\}$$

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.

asked 2021-09-13

Suppose that A is row equivalent to B. Find bases for the null space of A and the column space of A.

$A=\left[\begin{array}{ccccc}1& 2& -5& 11& -3\\ 2& 4& -5& 15& 2\\ 1& 2& 0& 4& 5\\ 3& 6& -5& 19& -2\end{array}\right]$

$B=\left[\begin{array}{ccccc}1& 2& 0& 4& 5\\ 0& 0& 5& -7& 8\\ 0& 0& 0& 0& -9\\ 0& 0& 0& 0& 0\end{array}\right]$

asked 2021-09-18

I need to find a unique description of Nul A, namely by listing the vectors that measure the null space.

$A=\left[\begin{array}{ccccc}1& 5& -4& -3& 1\\ 0& 1& -2& 1& 0\\ 0& 0& 0& 0& 0\end{array}\right]$

asked 2021-06-13

For the matrix A below, find a nonzero vector in the null space of A and a nonzero vector in the column space of A

$A=\left[\begin{array}{cccc}2& 3& 5& -9\\ -8& -9& -11& 21\\ 4& -3& -17& 27\end{array}\right]$

Find a vector in the null space of A that is not the zero vector

$A=\left[\begin{array}{c}-3\\ 2\\ 0\\ 1\end{array}\right]$

asked 2022-06-26

A linear transformation $T:{\mathbb{R}}^{3}\to {\mathbb{R}}^{2}$ whose matrix is

$\left(\begin{array}{ccc}1& 3& 3\\ 2& 6& -3.5+k\end{array}\right)$

is onto if and only if $k\ne $

$\left(\begin{array}{ccc}1& 3& 3\\ 2& 6& -3.5+k\end{array}\right)$

is onto if and only if $k\ne $

asked 2022-01-31

Can a triangle be represented as a matrix?

Triangle1 (1, 1), (1, 2), (3, 1)

Triangle2 (1, 3), (1, 6), (3, 3)

Find the matrix which represents the stretch that maps triangle T1 onto triangle T2.

Triangle1 (1, 1), (1, 2), (3, 1)

Triangle2 (1, 3), (1, 6), (3, 3)

Find the matrix which represents the stretch that maps triangle T1 onto triangle T2.

asked 2022-05-19

Finding alternate transformation matrix for similarity transformation

A pair of square matrices $X$ and $Y$ are called similar if there exists a nonsingular matrix $T$ such that ${T}^{-1}XT=Y$ holds. It is known that the transformation matrix $T$ is not unique for given $X$ and $Y$. I'm just wondering whether those non-unique transformation matrices would have any relation among themselves, like having column vectors with same directions...

What I want to mean is: Given $X$ and $Y$ a pair of similar matrices, if $S$ and $T$ are two possible transformation matrices satisfying ${S}^{-1}XS={T}^{-1}XT=Y$, is there any generic (apart from scaling) relation between $T$ and $S$ (e.g., direction of column vectors)?

For a specific example, consider $X=\left[\begin{array}{cc}A& BK\\ C& 0\end{array}\right]$ and $Y=\left[\begin{array}{cc}A+{A}^{-1}BKC& -{A}^{-1}BKC{A}^{-1}B\\ KC& -KC{A}^{-1}B\end{array}\right]$. Assuming $K$ to be invertible it can be shown that $X$ and $Y$ are similar with transformation matrix $T=\left[\begin{array}{cc}I& -{A}^{-1}B\\ 0& {K}^{-1}\end{array}\right]$. Can there be any other matrix $S$ which will be independent of $K$, and would result ${S}^{-1}XS=Y$?

A pair of square matrices $X$ and $Y$ are called similar if there exists a nonsingular matrix $T$ such that ${T}^{-1}XT=Y$ holds. It is known that the transformation matrix $T$ is not unique for given $X$ and $Y$. I'm just wondering whether those non-unique transformation matrices would have any relation among themselves, like having column vectors with same directions...

What I want to mean is: Given $X$ and $Y$ a pair of similar matrices, if $S$ and $T$ are two possible transformation matrices satisfying ${S}^{-1}XS={T}^{-1}XT=Y$, is there any generic (apart from scaling) relation between $T$ and $S$ (e.g., direction of column vectors)?

For a specific example, consider $X=\left[\begin{array}{cc}A& BK\\ C& 0\end{array}\right]$ and $Y=\left[\begin{array}{cc}A+{A}^{-1}BKC& -{A}^{-1}BKC{A}^{-1}B\\ KC& -KC{A}^{-1}B\end{array}\right]$. Assuming $K$ to be invertible it can be shown that $X$ and $Y$ are similar with transformation matrix $T=\left[\begin{array}{cc}I& -{A}^{-1}B\\ 0& {K}^{-1}\end{array}\right]$. Can there be any other matrix $S$ which will be independent of $K$, and would result ${S}^{-1}XS=Y$?

asked 2020-12-25

a) Let A and B be symmetric matrices of the same size.

Prove that AB is symmetric if and only$AB=BA.$

b) Find symmetric$2\cdot 2$

matrices A and B such that$AB=BA.$

Prove that AB is symmetric if and only

b) Find symmetric

matrices A and B such that