3

Suppose $A$ and $B$ are square matrices. I would like to find the trace or upper and lower bounds on the trace of $ABA^{T}$, where the lower diagonal and diagonal elements of $B=(b_{ij})$ are all zeros, for instance,

$$B = \begin{bmatrix} 0 & b_{1} & b_{2} & b_{3} \\ 0 & 0 & b_{1} & b_{2} \\ 0 & 0 & 0 & b_{1} \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$

and $A=(a_{ij})$ is a general matrix (in the sense that it is not diagonal and all elements are different from zero) assumed to be full rank. Thanks a lot!

Tobias
  • 606
Charles Chou
  • 391
  • I assume you mean the trace of $ABA^T$, not the trace of $\text{tr}(ABA^T)$. – Robert Israel Nov 18 '16 at 19:42
  • I don't understand: You have asked this question about an hour ago, you have deleted it and somebody has deleted the precise answer I gave. What has happened ? Could you explain me ? – Jean Marie Nov 18 '16 at 19:57
  • I apologize for any inconvenience caused. I revised the question a little bit., which is in more detailed form. When I post it here, it didn't allow me to post the same question, so I deleted the previous question. I deeply apologize for that. – Charles Chou Nov 18 '16 at 20:01
  • What sort of upper and lower bound do you mean? Is $B$ fixed? – copper.hat Nov 18 '16 at 20:10
  • I am looking for two constants c and d (may be very large) such that c<$\mbox{tr}\left( ABA^{T }\right)$<d. Thanks – Charles Chou Nov 18 '16 at 20:15
  • 1
    @CharlesChou: Unless you restrict $A,B$ in some meaningful way then the only constants for which it can be true are $c=-\infty, d=\infty$. – copper.hat Nov 18 '16 at 20:24
  • The trace of a matrix is invariant w.r.t. similarity transformations and therefore equal to the sum of the eigenvalues. On the other hand this sum is not larger than the norm of the matrix times the number of rows/columns. This gives you the estimation $-n|ABA^T| \leq \operatorname{tr}(ABA^T)\leq n|ABA^T|$ if $A,B\in\mathbb{R}^{n\times n}$. But this estimation is certainly too coarse for you. – Tobias Nov 18 '16 at 20:46
  • Do not delete and re-ask a question after it was answered. If the modifications you want to make are minor, you can edit the original version. If the modifications are substantial, ask an additional question, but don't delete the answered original. – Daniel Fischer Nov 18 '16 at 21:06

1 Answers1

0

Here is the answer I already gave:

let $$A=\pmatrix{k&1&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\}$$

It is a full rank matrix if $k\neq 0$.

$$\tag{1}ABA^T=\pmatrix{kb_1&kb_1&b_1+kb_2&b_2+kb_3\\0&0&b_1&b_2\\0&0&0&b_1\\0&0&0&0\\}$$

with trace$(ABA^T)=kb_1.$

If $b_1\neq 0$, this trace can take any nonzero value, in particular can be arbitrarily large.

Now, the non zero condition on all entries. In a nutshell, it suffices to add a convenient $\varepsilon\neq0$ to all entries of matrix $A$: this will modify the trace in an arbitrary small amount. In this way, the result stays the same: the trace can be arbitrarily large.

Remark: The computation in $(1)$ can be done very easily by block computations.

See also interesting answers to almost the same question (Simple way to find the trace or the bound of the trace of $ ABA^T$?)

Community
  • 1
Jean Marie
  • 81,803
  • Thanks for the input. Unfortunately, you are assuming a very particular form of A. The matrix A I have doesn't have this simple form, it is just a very general matrix in the sense that it is not diagonal and all elements are different from zero. Thanks – Charles Chou Nov 18 '16 at 20:13
  • @CharlesChou I think the answer is meant to say that if there's no upper or lower bound for this matrix, there's no chance you can find a nice formula for the lower or upper bound in the general case. But I'm not sure. – anderstood Nov 18 '16 at 20:15
  • All right, but in the previous version that you have deleted you hadn't this constraint... – Jean Marie Nov 18 '16 at 20:16
  • I have an answer to your non zero condition: see what I have added to my solution. – Jean Marie Nov 18 '16 at 20:23