Linear equation

Question

I would like to find the solutions of this linear equation :

$$ X + (X^\top C)C = D \tag{$\star$}$$

where $C = (c_{1},\ldots,c_{n})^\top \in \mathbb{R}^{n}$ and $D = (d_{1},\ldots,d_{n})^\top \in \mathbb{R}^{n}$ are given and non-zero.

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

$$ \tilde{C}X = D \tag{$\star\star$}$$

where $\tilde{C} \in \mathrm{Mat}_{n}(\mathbb{R})$ such that : $\displaystyle \tilde{C}_{i,j} = \begin{cases} c_{j}c_{i} & \text{if } i \neq j \\ 1+c_{i}^{2} & \text{if } i=j \end{cases}$. To do so, I just said that the $i$-th component of the LHS of $(\star)$ is $\displaystyle (1+c_{i}^{2})x_{i} + \sum_{j \neq i}x_{j}c_{j}c_{i}$.

My question is this one : is there another method to prove that $(\star)$ is equivalent to $(\star\star)$ ?

Answer 1

Since $x^Tc$ is a scalar, you can move it to anywhere. Say $$(x^Tc)c=(c^Tx)c=c(c^Tx)=(c\otimes c^T)x$$ So $$x+(x^Tc)c=(I+c\otimes c^T)x=\tilde Cx=d$$ where $$\tilde C=I+c\otimes c^T$$

Answer 2

As $X$ and $C$ are real $n$-vectors, \begin{align*} X+(X^\top C)C &=X+C(X^\top C)\tag{1}\\ &=X+C(X^\top C)^\top\tag{2}\\ &=X+C(C^\top X)\\ &=(I+CC^\top)X. \end{align*}

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 $1\times1$ matrix $[s]$ with the scalar $s$, so that $[s]u$ is the scalar multiplication $su$.

$(2)$: the transpose of a real $1\times1$ matrix is the matrix itself.