Solve $A\mathbf{x}=\mathbf{v}(\mathbf{x}^TD\mathbf{v})$ for $A$

Question

Is there a way to simplify $$f(\mathbf{x})=\mathbf{x}-\mathbf{v}(\mathbf{x}^TD\mathbf{v})$$

to the form

$$f(\mathbf{x})=(I-A)\mathbf{x}$$

where $\mathbf{x}$ and $\mathbf{v}$ are column-vectors and $D$ is a matrix?

In particular, I seek to extract the variable $\mathbf{x}$ as a common factor so that it appears only once in the expression for $f(\mathbf{x})$.

I could write $\mathbf{x}-\mathbf{v}(\mathbf{v}^TD^T\mathbf{x})$ but then I still don't know how to get that into the form $(...\text{constant matrix expression }...)\mathbf{x}$.

Answer 1

This expression is linear in $x$ so it can be reduced to such form:

$f(\mathbf{x})=M \mathbf{x}$

where explicitely the components of the matrix $M$ are $M_{a,b}=\delta_{a,b}-v_a (D\mathbf{v})_b$

We can put it in a more vector form, if needed:

$v_a (D\mathbf{v})_b=\sum_c v_a D_{b,c}v_c=\sum_c (\mathbf{v}\mathbf{v}^+)_{a,c}D_{b,c}=[(\mathbf{v}\mathbf{v}^+) D^+]_{a,b}$

Answer 2

Checking with 2 by 2 matrix, $$f(x)=(I-vv^TD^T)x$$