6
$\begingroup$

Suppose that $M$ is a $3$ by $2$ matrix in which every $2$ by $2$ submatrix is invertible. Is it true that $M$ always has a $2$ by $2$ submatrix $M_{1}$ such that $\left\Vert M_{1}\right\Vert ^{2}\left\Vert M_{1}^{-1}\right\Vert ^{2}\lambda_{2}\left(MM^{T}\right)\ge\left\Vert M\right\Vert ^{2}$ where $\left\Vert \cdot\right\Vert $ is the norm of a matrix (refer to here for the definition) and $\lambda_{2}\left(\cdot\right)$ is the second largest eigenvalue of a $3$ by $3$ symmetric matrix? Thanks

  • 0
    Sounds like this could follow from interlacing inequalities for singular values. You might want to simplify the inequality by expressing it in terms of the singular values of $M_1$ and $M$.2012-12-17
  • 0
    http://mathoverflow.net/questions/116665/would-this-hold-for-any-2-by-3-matrix2012-12-25
  • 0
    @peanae: A link is given for the definition of the norm, but the definition linked to is dependent on which norm is given to $\mathbb C^n$, which is not specified above. Are we to presume that $\mathbb C^n$ has the Euclidean norm?2013-07-24

1 Answers 1

4

The answer is affirmative. Your inequality is equivalent to $\sigma_1(M_1)/\sigma_2(M_1)\ge\sigma_1(M)/\sigma_2(M)$, where $\sigma_i(X)$ denotes the $i$-th largest singular value of a matrix $X$. We can actually prove a more general proposition:

Proposition. Suppose $n\ge1$ and every $n\times n$ submatrix of some $M\in\mathbb{C}^{(n+1)\times n}$ is invertible. Then there exists an $n\times n$ submatrix $M_1$ of $M$ such that $\sigma_1(M_1)/\sigma_n(M_1)\ge\sigma_1(M)/\sigma_n(M)$.

Proof. Denote the $n+1$ rows of $M$ by $b_1^\ast, \ldots, b_{n+1}^\ast$. Let $B_i=\sum_{k\not=i} b_kb_k^\ast$. Then each $B_i$ corresponds to a product of the form $M_1^\ast M_1$ where $M_1$ is the submatrix obtained by deleting the $i$-row of $M$. We have $\sum_i B_i=nM^\ast M$ and $$ \frac{\sigma_1^2(M)}{\sigma_n^2(M)} =\frac{\max_{\|u\|=1} u^\ast M^\ast Mu}{\min_{\|v\|=1} v^\ast M^\ast Mv} =\frac{\max_{\|u\|=1} \sum_i u^\ast B_iu}{\min_{\|v\|=1} \sum_i v^\ast B_iv}. $$ Suppose $u_0$ and $v_0$ are respectively the maximizer of the numerator and minimizer of the denominator of last expression. By assumption on submatrices of the form $M_1$, each $B_i$ is positive definite. Therefore \begin{align} \frac{\sigma_1^2(M)}{\sigma_n^2(M)} =\frac{\sum_i u_0^\ast B_iu_0}{\sum_i v_0^\ast B_iv_0} &\le\max_{i=1,\ldots,n+1}\frac{u_0^\ast B_iu_0}{v_0^\ast B_iv_0}\\ &\le\max_{i=1,\ldots,n+1}\frac{\max_{\|u\|=1} u^\ast B_iu}{\min_{\|v\|=1} v^\ast B_iv}\\ &=\max\left\{\frac{\sigma_1^2(M_1)}{\sigma_n^2(M_1)}: \ M_1 \textrm{ is an } n\times n \textrm{ submatrix of } M\right\}. \end{align} Hence the result.

  • 1
    user1551, please see http://tea.mathoverflow.net/discussion/1187/extending-from-a-plane-in-r3-again-and-again-and-again/ and email me if you want more information.2012-12-19
  • 0
    @WillJagy We'll see if you're qualified as a prophet ;-D2012-12-19
  • 0
    There won't be any sign here. Your answer got two upvotes, so the question cannot be deleted.2012-12-19
  • 0
    For the (believed) provenance of the question, see Suvrit's answer at http://mathoverflow.net/questions/878722012-12-19
  • 0
    @WillJagy Thanks for the information, but I'm not sure if how to formulate the OP's original question withiin the framework that Suvirt mentioned, because we are dealing with ratios rather then products of singular values. The general case, $\sigma_1(M_1)/\sigma_n(M_1)\ge\sigma_1(M)/\sigma_n(M)$, which straddles two non-consecutive singular values, is even more difficult. On one hand, we need to skip the intermediate ones, but on the other hand, majorisation is involved. Any idea?2012-12-20
  • 0
    user1551, I am over my head with this material. My one contribution is having enough reputation points to see deleted posts on bot MSE and MO, so I was able to gather up screenshot jpegs of all the self-deleted evidence. If this is a new area of investigation, I imagine that is a good thing. Suvrit just said he had no time to look at it but it is close enough to the others to suggest the same person asking.2012-12-20