Callum Dudley

2022-07-01

Find the solutions of this linear equation:

$\begin{array}{}\text{(}\star \text{)}& X+({X}^{\mathrm{\top}}C)C=D\end{array}$

where $C=({c}_{1},\dots ,{c}_{n}{)}^{\mathrm{\top}}\in {\mathbb{R}}^{n}$ and $D=({d}_{1},\dots ,{d}_{n}{)}^{\mathrm{\top}}\in {\mathbb{R}}^{n}$ are given and non-zero.

I noticed that the equation $(\star )$ is equivalent to:

$\begin{array}{}\text{(}\star \star \text{)}& \stackrel{~}{C}X=D\end{array}$

where $\stackrel{~}{C}\in {\mathrm{M}\mathrm{a}\mathrm{t}}_{n}(\mathbb{R})$ such that: $\stackrel{~}{C}}_{i,j}=\{\begin{array}{ll}{c}_{j}{c}_{i}& \text{if}i\ne j\\ 1+{c}_{i}^{2}& \text{if}i=j\end{array$. To do so, I just said that the $i$-th component of the LHS of $(\star )$ is $(1+{c}_{i}^{2}){x}_{i}+\sum _{j\ne i}{x}_{j}{c}_{j}{c}_{i}$

Is there another method to prove that $(\star )$ is equivalent to $(\star \star )$?

$\begin{array}{}\text{(}\star \text{)}& X+({X}^{\mathrm{\top}}C)C=D\end{array}$

where $C=({c}_{1},\dots ,{c}_{n}{)}^{\mathrm{\top}}\in {\mathbb{R}}^{n}$ and $D=({d}_{1},\dots ,{d}_{n}{)}^{\mathrm{\top}}\in {\mathbb{R}}^{n}$ are given and non-zero.

I noticed that the equation $(\star )$ is equivalent to:

$\begin{array}{}\text{(}\star \star \text{)}& \stackrel{~}{C}X=D\end{array}$

where $\stackrel{~}{C}\in {\mathrm{M}\mathrm{a}\mathrm{t}}_{n}(\mathbb{R})$ such that: $\stackrel{~}{C}}_{i,j}=\{\begin{array}{ll}{c}_{j}{c}_{i}& \text{if}i\ne j\\ 1+{c}_{i}^{2}& \text{if}i=j\end{array$. To do so, I just said that the $i$-th component of the LHS of $(\star )$ is $(1+{c}_{i}^{2}){x}_{i}+\sum _{j\ne i}{x}_{j}{c}_{j}{c}_{i}$

Is there another method to prove that $(\star )$ is equivalent to $(\star \star )$?

Camron Herrera

Beginner2022-07-02Added 16 answers

As $X$ and $C$ are real $n$-vectors,

$\begin{array}{rl}\text{(1)}& X+({X}^{\mathrm{\top}}C)C& =X+C({X}^{\mathrm{\top}}C)\text{(2)}& & =X+C({X}^{\mathrm{\top}}C{)}^{\mathrm{\top}}& =X+C({C}^{\mathrm{\top}}X)\\ & =(I+C{C}^{\mathrm{\top}})X.\end{array}$

Explanations:

$(1)$: if $u$ is a vector and $s$ is a scalar, we have $[s]u=u[s]$, where on the LHS we identify the $1\times 1$ matrix $[s]$ with the scalar $s$, so that $[s]u$ is the scalar multiplication $su$.

$(2)$: the transpose of a real $1\times 1$ matrix is the matrix itself.

$\begin{array}{rl}\text{(1)}& X+({X}^{\mathrm{\top}}C)C& =X+C({X}^{\mathrm{\top}}C)\text{(2)}& & =X+C({X}^{\mathrm{\top}}C{)}^{\mathrm{\top}}& =X+C({C}^{\mathrm{\top}}X)\\ & =(I+C{C}^{\mathrm{\top}})X.\end{array}$

Explanations:

$(1)$: if $u$ is a vector and $s$ is a scalar, we have $[s]u=u[s]$, where on the LHS we identify the $1\times 1$ matrix $[s]$ with the scalar $s$, so that $[s]u$ is the scalar multiplication $su$.

$(2)$: the transpose of a real $1\times 1$ matrix is the matrix itself.

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