I Are There Nontrivial Solutions to the Matrix Equation X C X^T = C?

[vibe3]

Hello, if I have some given vector c \in R^n, then I want to find solutions X \in R^{n\times n} to the following equation:
<br /> X C X^T = C<br />
where C = c c^T. Certainly X = I is a solution, but I'm looking for any nontrivial solutions. We can also assume X is invertible if that helps.

This equation seems related to the Sylvester equation. If the solution is nontrivial, I would appreciate if someone could point me in the direction of any work which has been done on this type of equation.

 

[fresh_42]

Hello, if I have some given vector c \in R^n, then I want to find solutions X \in R^{n\times n} to the following equation:
<br /> X C X^T = C<br />
where C = c c^T. Certainly X = I is a solution, but I'm looking for any nontrivial solutions. We can also assume X is invertible if that helps.

This equation seems related to the Sylvester equation. If the solution is nontrivial, I would appreciate if someone could point me in the direction of any work which has been done on this type of equation.

Assume you have chosen a basis such that $c=(1,0,\ldots,0)^T$. How many solutions do you get?

 

[WWGD]

Isn't C then a scalar? But then $CXX^{T}$ is a matrix and C is a scalar. Am I missing something?

 

[fresh_42]

Isn't C then a scalar? But then $CXX^{T}$ is a matrix and C is a scalar. Am I missing something?

Depends on how you write vectors, as a column or as a row: one gives a scalar and the other a rank one matrix. Either case has many solutions.

 

[WWGD]

Depends on how you write vectors, as a column or as a row: one gives a scalar and the other a rank one matrix. Either case has many solutions.

But how can an $n \times n$ matrix equal a $1 \times 1$ unless n=1? EDIT: Do you mean we extend the scalar C into a matrix (C,0,0,..,0)?

 

[fresh_42]

But how can an $n \times n$ matrix equal a $1 \times 1$ unless n=1?

You mean if $C$ is a scalar? Then $XCX^T=C=CI$ by the usual identification of the two ones in algebras with unit. Otherwise $\begin{bmatrix}c_1\\ \vdots \\c_n\end{bmatrix} \cdot \begin{bmatrix}c_1 & \ldots & c_n \end{bmatrix}$ is a $n \times n$ matrix of rank $1$.

 

[WWGD]

I mean, which case do you assume then, that C is $n \times n$ or a "scalar matrix"?

 

[vibe3]

I mean, which case do you assume then, that C is $n \times n$ or a "scalar matrix"?

C is an n-by-n matrix of rank 1

 

[WWGD]

C is an n-by-n matrix of rank 1

Thanks, kind of reminds me of least-square problems, solved by ortho projections.

 

[StoneTemplePython]

It's not clear to me what kind of "solution" OP is looking for since this problem is very much underdetermined. There are some interesting underlying relations here between singular vectors of $\mathbf X$ and the structure of $\mathbf C$, but that's about all I can add.

note: if $\mathbf c = \mathbf 0$, any $\mathbf X$ is allowed.
- - - -
Now assume $\mathbf c \neq \mathbf 0$. Thus $\mathbf C = \mathbf{cc}^T$ is a rank one matrix.

We can do singular value decomposition on $\mathbf X$ such that $\mathbf X = \mathbf{U \Sigma V}^T$

For $\mathbf{X C X}^T = \mathbf{X c} \mathbf c^T \mathbf X^T= \mathbf{cc}^T$, we know that $\mathbf c$ is given by one (or more) of the right singular vectors of $\mathbf X$ that has a singular value of 1 $^{**}$. Why? Using the trace operation we see

$trace\big(\mathbf{cc}^T\big) = \mathbf c ^T \mathbf c = trace\big(\mathbf{X c} \mathbf c^T \mathbf X^T\big)$

$\mathbf c ^T \mathbf c = trace\big(\mathbf{X c} \mathbf c^T \mathbf X^T\big) = trace\big(\mathbf c^T \mathbf X^T \mathbf{Xc}\big) = \mathbf c^T \mathbf X^T \mathbf{Xc} = \mathbf c^T \mathbf V \mathbf \Sigma^T \mathbf U^T \mathbf {U \Sigma V}^T \mathbf c = \mathbf c^T \mathbf V \mathbf{\Sigma}^2 \mathbf{V}^T \mathbf c = \mathbf y^T \mathbf{\Sigma}^2 \mathbf{y} = \mathbf y^T \mathbf y$

where $\mathbf{V}^T \mathbf c := \mathbf y$, and of course $\big \Vert \mathbf c \big \Vert_2^2 = \big \Vert \mathbf{V}^T \mathbf c \big \Vert_2^2 = \big \Vert \mathbf y \big \Vert_2^2$

- - - - - - - -
$^{**}$ to be more accurate: it is possible that that the squared singular values that non-zero elements of $\mathbf y$ gets scaled by are not one-- the exact constraint we need to observe is

$\mathbf y^T \mathbf{\Sigma}^2 \mathbf{y} = \sigma_1 ^2 y_1^2 + \sigma_2^2 y_2^2 +... + \sigma_n ^2 y_n^2 = y_1^2 + y_2^2 +... + y_n^2 =\mathbf y^T \mathbf y$
- - - - - - - -

There's also a lot of flexibility in determining $\mathbf U$. If $\mathbf X$ is full rank, you can also tease out the constraint that, $\mathbf c ^T \mathbf c = \mathbf c^T \mathbf U \mathbf{\Sigma}^{-2} \mathbf{U}^T \mathbf c = \mathbf w^T \mathbf{\Sigma}^{-2} \mathbf w = \mathbf w^T \mathbf w$, where $\mathbf{U}^T \mathbf c := \mathbf w$.

Losing some generality, here's one particularly easy approach for creating a legal $\mathbf X$: write out $\mathbf C = \mathbf {PDP}^T$, where $\mathbf D$ is the zero matrix except the top left entry is given by $trace\big(\mathbf C\big)$. From here ensure that all values of $y_k = 0$ for $k \geq 2$, and that $\sigma_1 = 1$, then set $\mathbf U := \mathbf P$.

Again, this whole thing is a very much underdetermined problem so there's a lot of different ways to approach this.