 # Find the solutions of this linear equation: <mtable displaystyle="true"> <mlabeledtr> Callum Dudley 2022-07-01 Answered
Find the solutions of this linear equation:
$\begin{array}{}\text{(}\star \text{)}& X+\left({X}^{\mathrm{\top }}C\right)C=D\end{array}$
where $C=\left({c}_{1},\dots ,{c}_{n}{\right)}^{\mathrm{\top }}\in {\mathbb{R}}^{n}$ and $D=\left({d}_{1},\dots ,{d}_{n}{\right)}^{\mathrm{\top }}\in {\mathbb{R}}^{n}$ are given and non-zero.
I noticed that the equation $\left(\star \right)$ is equivalent to:
$\begin{array}{}\text{(}\star \star \text{)}& \stackrel{~}{C}X=D\end{array}$
where $\stackrel{~}{C}\in {\mathrm{M}\mathrm{a}\mathrm{t}}_{n}\left(\mathbb{R}\right)$ such that: . To do so, I just said that the $i$-th component of the LHS of $\left(\star \right)$ is $\left(1+{c}_{i}^{2}\right){x}_{i}+\sum _{j\ne i}{x}_{j}{c}_{j}{c}_{i}$
Is there another method to prove that $\left(\star \right)$ is equivalent to $\left(\star \star \right)$?
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As $X$ and $C$ are real $n$-vectors,
$\begin{array}{rl}\text{(1)}& X+\left({X}^{\mathrm{\top }}C\right)C& =X+C\left({X}^{\mathrm{\top }}C\right)\text{(2)}& & =X+C\left({X}^{\mathrm{\top }}C{\right)}^{\mathrm{\top }}& =X+C\left({C}^{\mathrm{\top }}X\right)\\ & =\left(I+C{C}^{\mathrm{\top }}\right)X.\end{array}$
Explanations:
$\left(1\right)$: if $u$ is a vector and $s$ is a scalar, we have $\left[s\right]u=u\left[s\right]$, where on the LHS we identify the $1×1$ matrix $\left[s\right]$ with the scalar $s$, so that $\left[s\right]u$ is the scalar multiplication $su$.
$\left(2\right)$: the transpose of a real $1×1$ matrix is the matrix itself.

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