4
$\begingroup$

Let $u_1,...,u_k$ be unit vectors in $\mathbb R^n$, with $k \geq n$, such that there exist scalars $c_1,...,c_k$ so that, for all $x \in \mathbb R^n$, $$\sum\limits_i c_i \langle x , u_i \rangle^2 = |x|^2,$$ or equivalently $$\sum\limits_i c_i \langle x , u_i \rangle u_i = x$$ for all $x \in \mathbb R^n$. This is the same as saying that the vectors $\sqrt{c_i}u_i$ form a Parseval frame. In these notes by Ball, it is asserted that this implies that if $T$ is a linear map of determinant $1$, then for at least one $i$, $|T u_i| \geq 1$. Is there an easy way to see this?

$\endgroup$
7
  • 2
    $\begingroup$ The idea is the following: The determinant is the product of the eigenvalues. Since the determinant is $1$, at least one of the eigenvalues is greater than $1$. Let $u$ be the corresponding eigenvector. Write $u$ as a sum of the $u_i$'s and see what happens if they all shrink. $\endgroup$ – Michael Burr Feb 26 '17 at 17:35
  • 1
    $\begingroup$ I was never able to figure this out. In particular, I do not understand the above comment. It seems to me that, if this strategy were to work, we ought to be able to prove this if we relax the requirement that $det(T)=1$ to $T$ merely having an eigenvalue $\lambda$ with $|\lambda| \geq 1$. But this is clearly false, as seen by taking, e.g., $n=2$, $c_1 = c_2 = 1$, $u_1 = {\sqrt{2} \over 2} (1,1)$, $u_2 = {\sqrt{2} \over 2} (1,-1)$, and $T : (x,y) \mapsto (x,0)$. $\endgroup$ – Justthisguy Sep 30 '17 at 1:01
  • $\begingroup$ Where in the notes is this assertion? $\endgroup$ – littleO Sep 30 '17 at 1:32
  • $\begingroup$ The first full paragraph on page 16. $\endgroup$ – Justthisguy Sep 30 '17 at 2:52
  • 1
    $\begingroup$ Yes I agree that the determinant is $0$. What I don't understand about your comment is that it seems to imply that all we need is an eigenvalue $\lambda$ with $|\lambda| \geq 1$, but perhaps I am misunderstanding. $\endgroup$ – Justthisguy Sep 30 '17 at 16:00
3
+50
$\begingroup$

The statement $\sum_ic_i\langle x,u_i\rangle u_i=x$ for all $x$ can be expressed in matrix form as $$ \sum_ic_i u_iu_i^t=I_n $$ with $u^t$ representing the transpose of $u$ and $I_n$ is the identity matrix. As $uu^t$ has trace $\lVert u\rVert^2$, we can take the trace of the above expression, $$ \sum_ic_i\lVert u_i\rVert^2=n.\qquad{\rm(1)} $$ For an $n\times n$ matrix $T$, $$ \sum_ic_i(Tu_i)(Tu_i)^t=TT^t $$ and, taking the trace again, $$ \sum_ic_i\lVert Tu_i\rVert^2={\rm tr}(TT^t). $$ The symmetric matrix $TT^t$ has nonnegative eigenvalues summing to its trace and whose product is ${\rm det}(T)^2$. By the AM-GM inequality, $$ \sum_ic_i\lVert Tu_i\rVert^2\ge n\,\lvert{\rm det}(T)\rvert^{2/n}.\qquad{\rm(2)} $$ It is assumed that the $c_i$ are nonnegative so, if $T$ has determinant $1$, then it cannot strictly reduce the size of each $u_i$, since (2) would then contradict (1).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.