0
$\begingroup$

Problem Formulation

under what conditions can we solve $\mathrm{trace}(\mathbf{AB})=0$ ? or more specifically, when will $\mathrm{trace}(\mathbf{AB})=0$ implies that $\mathrm{trace}(\mathbf{B})=0$.

Related Art

This occurs to me when $\mathbf{A}$ was a Positive semidefinite matrix, and $\mathbf{B}$ was a symmetric matrix with a parameter $\lambda$. I want to solve a partial derivatives equations which constrains that $\mathrm{trace}[\mathbf{AB}(\lambda)] = 0$, i.e., \begin{equation} \frac{\partial \mathcal{L}(i)}{\partial \gamma_i} = \mathrm{Tr}\left[ (\mathbf{I}_{d_i} + \gamma_i\mathbf{B}_i\mathbf{s}_i)^{-1}\mathbf{B}_i\mathbf{s}_i\right] - \mathbf{q}_i^T(\gamma_i^{-1}\mathbf{B}_i^{-1} + \mathbf{s}_i)^{-1}\frac{\mathbf{B}_i^{-1}}{\gamma_i^2}(\gamma_i^{-1}\mathbf{B}_i^{-1} + \mathbf{s}_i)^{-1}\mathbf{q}_i, \end{equation} where $\mathbf{s}_i$ and $\mathbf{B}_i$ are symmetric PSD matrix, $\mathbf{q}_i$ is a column vector, $\gamma_i$ is a scalar, $\mathbf{I}_d$ is an identity matrix with size $d$. I want to solve \begin{equation} \frac{\partial \mathcal{L}(i)}{\partial \gamma_i} = 0. \end{equation} with $\gamma_i$ derived in a closed-form.

Using cyclic product properties of the trace norm, I could only go this far, \begin{equation} \frac{\partial \mathcal{L}(i)}{\partial \gamma_i} = \mathrm{Tr}\left[ (\mathbf{B}_i^{-1} + \gamma_i\mathbf{s}_i)^{-1}\mathbf{B}_i^{-1}(\mathbf{B}_i^{-1} + \gamma_i\mathbf{s}_i)^{-1}\left( \mathbf{s}_i + \gamma_i\mathbf{s}_i\mathbf{B}_i\mathbf{s}_i - \mathbf{q}_i\mathbf{q}_i^T \right) \right] \end{equation} I sensed that the closed form solution of $\gamma_i$ might be related with \begin{equation} \mathrm{trace}\left( \mathbf{s}_i + \gamma_i\mathbf{s}_i\mathbf{B}_i\mathbf{s}_i - \mathbf{q}_i\mathbf{q}_i^T \right) = 0. \end{equation}

Counter Example

plus, I can also give a counter example. Let $$ \mathbf{A} = \begin{pmatrix} 2 & 1 \\\ 1 & 2 \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} 1+\lambda & -2 \\\ -2 & \lambda - 3 \end{pmatrix}, $$ where $\mathrm{trace}(\mathbf{AB})=0$ implies $\lambda=2$, $\mathrm{trace}(\mathbf{B})=0$ implies $\lambda=1$.

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ This is not the right site for this question. See the FAQ for alternatives. $\endgroup$
    – Chris Godsil
    Commented Jun 4, 2013 at 8:56
  • $\begingroup$ Dear Chris, I have modified my problem. I think it related to the problem of linalg and the optimization of logdet function, this problem happened mostly in $\ell_1$ minimization solvers. L $\endgroup$
    – liubenyuan
    Commented Jun 4, 2013 at 9:09
  • 1
    $\begingroup$ $\text{Tr}(AB)$ is a symmetric non-degenerate inner product on the space of all matrices, with signature $(\frac{(n+1)n}2,\frac{(n-1)n}2)$. Over $mathbb R$, the symmetric bilinear form $\text{Tr}(AB^\top)$ is positive definite. $\endgroup$
    – Peter Michor
    Commented Jun 4, 2013 at 11:55
  • $\begingroup$ Dear Peter, you mean that $\mathrm{Tr}(\mathbf{AB^T})$ is PD and therefore could not equal $0$ ? $\endgroup$
    – liubenyuan
    Commented Jun 4, 2013 at 12:52
  • $\begingroup$ Note that $A=0$, $B$ any matrix with nonzero trace, is a pretty good counterexample too. $\endgroup$
    – Julien
    Commented Jun 4, 2013 at 14:34

1 Answer 1

Reset to default
1
$\begingroup$

Here is a result, which I think is interesting in this context (for reference see books on Lie algebras):

Theorem: Let $V$ be a finite-dimensional vector space over a field $K$ of characteristic zero. Let $E\subseteq F$ be two linear subspaces of $End(V)$ and $M=\lbrace A \in End(V) \mid [A,F] \subseteq E\rbrace $. Assume that $tr (AB)=0$ for all $A\in M$. Then $B$ is nilpotent with $tr(B)=tr(B^2)=\cdots = tr(B^n)=0$ for all $n$.

Edit: $[A,B]=AB-BA$ in $End(V)$.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Sorry, dear Dietrich, I could not quite follow your answer. and what does End(V) mean ? Which book should I refer to ? However thanks for your answer, it seem like I could obtain $\mathrm{Tr}(B)=0$ with some additional constrains imposed on $\mathbf{A}$. $\endgroup$
    – liubenyuan
    Commented Jun 4, 2013 at 12:43
  • $\begingroup$ $End(V)$ is the space of all linear maps $A\colon V\rightarrow V$ (square matrices, if you want). $\endgroup$
    – Dietrich Burde
    Commented Jun 4, 2013 at 13:04
  • $\begingroup$ @Dietrich: Do you really mean $AF-FA$, not $\left[A,F\right]$ (which are different things, e. g., because $AF-FA$ is the same as $AF+FA$) ? $\endgroup$
    – darij grinberg
    Commented Jun 4, 2013 at 17:42
  • $\begingroup$ (If it is $\left[A,F\right]$ that Dietrich wanted to say, the result appears in C. Brookes' "Lectures on Lie Algebras", scribed by Eva Belmont people.ds.cam.ac.uk/eb525/lie-notes.pdf . It is Lemma 9.4 (3.25). The former number apparently can change while the latter is invariant. $x_s$ and $x_n$ are defined in Lemma 9.1 (3.22).) $\endgroup$
    – darij grinberg
    Commented Jun 4, 2013 at 17:47
  • $\begingroup$ @Darij: yes, sorry. $\endgroup$
    – Dietrich Burde
    Commented Jun 4, 2013 at 18:24

Not the answer you're looking for? Browse other questions tagged .