Power operation on a quadratic form

Question

I have the following question:

Assume that you have a vector $x\in\mathbb{R}^{n\times 1}$ and a matrix $A\in\mathbb{R}^{n\times n}$. I'm interested in the following operation: $$ \left(x^TAx\right)^{k} = \hspace{5pt} ? \qquad k\in\mathbb{Z}_+$$

Is there any nice property to apply in this case? It would be ideal that, for some cases (I would like to know which ones, if any) the following relation $$ \left(x^TAx\right)^{k} = x^TA^{k}x$$ holds.

Thanks in advance!!!

Answer 1

To be clear: Are you looking for matrices $A$ such that for every $x$ you have $(x^T A x)^k = x^T A^k x$? This is clearly not possible.

We can see this by contradiction. Let's assume we found a matrix such that the above relation is true for all vectors. Now take a vector $x$ such that $A x \neq 0$ (any non-zero matrix should have at least one). Then $(x^T A x)^k = x^T A^k x$ by assumption. Now consider $y = cx$ for $c \in \mathbb{R}_{\neq 0}$. Note that $$(y^T A y)^k = c^{2k} (x^T Ax)^k = c^{2k} x^T A^k x \neq c^2 x^T A^k x = (cx)^T A^k (cx) = y^T A^k y$$ unless $k = 1$ or $c=1$. Therefore, the relation it is not true for any $y \neq x$ and hence not true for all vectors, which is a contradiction.

From this we can see that the only matrix that satisfies the above condition is the zero matrix.

If you are looking for specific vectors $x$, the problem is different. Your relation is for instance true whenever

• $x$ is an eigenvector of $A$ scaled to unit norm, since then $A x = \lambda x$ means $x^T A x = \lambda$ and $x^T A^k x = \lambda^k$.
• $x$ is in the null space of $A$ since $A x = 0$ implies $A^k x = 0$ and hence $(x^T A x)^k = x^T A^k x = 0$.

Answer 2

We do have $$ (x^TAx)^k = x^TAxx^TAx\cdots x^TAx\\ = x^T(Axx^T)^{k-1}Ax \\ = x^TA(xx^TA)^{k-1}x $$ but apart from that I don't think there is much that can be done.