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Adrien Hardy
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Consider a fixed $N\times N$ positive definite symmetric matrix $A$. We assumeAssume $N=d^r$ for some $d,r\geq 1$.

I wonder if one can find a closed formula for the maximizer/maximum of the function $$f(x):=\langle x^{\otimes r}, A x^{\otimes r}\rangle$$ over $x\in\mathbb R^d$ under the constraint $\||x\||=1$$\|x\|=1$ (unit sphere).

Of course when $r=1$ the maximum is reached at the eigenvector associated with the maximal eigenvalue of $A$, which is the maximal value of $f$. But what if $r\geq 2$?

When $r\geq 2$, this is still true when $A$ is the $r$-fold product of a $d\times d$ matrix $B$, namely $A=B^{\otimes r}$.

So my question is: What can we say when $r\geq 2$ and $A$ not the $r$-fold product of a $d\times d$ matrix? Somehow we are looking at eigenvectors of $A$ that are tensor products, hence the title.

Consider a fixed $N\times N$ positive definite symmetric matrix $A$. We assume $N=d^r$ for some $d,r\geq 1$.

I wonder if one can find a closed formula for the maximizer/maximum of the function $$f(x):=\langle x^{\otimes r}, A x^{\otimes r}\rangle$$ over $x\in\mathbb R^d$ under the constraint $\||x\||=1$ (unit sphere).

Of course when $r=1$ the maximum is reached at the eigenvector associated with the maximal eigenvalue of $A$, which is the maximal value of $f$. But what if $r\geq 2$?

When $r\geq 2$, this is still true when $A$ is the $r$-fold product of a $d\times d$ matrix $B$, namely $A=B^{\otimes r}$.

So my question is: What can we say when $r\geq 2$ and $A$ not the $r$-fold product of a $d\times d$ matrix? Somehow we are looking at eigenvectors of $A$ that are tensor products, hence the title.

Consider a fixed $N\times N$ positive definite symmetric matrix $A$. Assume $N=d^r$ for some $d,r\geq 1$.

I wonder if one can find a closed formula for the maximizer/maximum of the function $$f(x):=\langle x^{\otimes r}, A x^{\otimes r}\rangle$$ over $x\in\mathbb R^d$ under the constraint $\|x\|=1$ (unit sphere).

Of course when $r=1$ the maximum is reached at the eigenvector associated with the maximal eigenvalue of $A$, which is the maximal value of $f$. But what if $r\geq 2$?

When $r\geq 2$, this is still true when $A$ is the $r$-fold product of a $d\times d$ matrix $B$, namely $A=B^{\otimes r}$.

So my question is: What can we say when $r\geq 2$ and $A$ not the $r$-fold product of a $d\times d$ matrix? Somehow we are looking at eigenvectors of $A$ that are tensor products, hence the title.

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Adrien Hardy
  • 2.1k
  • 17
  • 21

Eigenvectors that are tensor products?

Consider a fixed $N\times N$ positive definite symmetric matrix $A$. We assume $N=d^r$ for some $d,r\geq 1$.

I wonder if one can find a closed formula for the maximizer/maximum of the function $$f(x):=\langle x^{\otimes r}, A x^{\otimes r}\rangle$$ over $x\in\mathbb R^d$ under the constraint $\||x\||=1$ (unit sphere).

Of course when $r=1$ the maximum is reached at the eigenvector associated with the maximal eigenvalue of $A$, which is the maximal value of $f$. But what if $r\geq 2$?

When $r\geq 2$, this is still true when $A$ is the $r$-fold product of a $d\times d$ matrix $B$, namely $A=B^{\otimes r}$.

So my question is: What can we say when $r\geq 2$ and $A$ not the $r$-fold product of a $d\times d$ matrix? Somehow we are looking at eigenvectors of $A$ that are tensor products, hence the title.