Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I would like to know if anything can be said about the trace of a product of two matrices, where one matrix is a diagonal matrix, i.e., $$\text{trace}(DA)=...$$ Are there some bounds in terms of $\text{trace}(D)$ and $\text{trace}(A)$ ?

share|improve this question
1  
There is no bound in terms of $\text{tr}(D)$ and $\text{tr}(A)$. Consider, for example, $D = A = \left[ \begin{array}{cc} r & 0 \\ 0 & - r \end{array} \right]$ and let $r \to \infty$. –  Qiaochu Yuan Jul 1 at 21:46

2 Answers 2

up vote 6 down vote accepted

Note that, when $D$ is diagonal:

$$(DA)_{ii} = D_{ii} A_{ii}$$

So $tr(DA) = \sum_{i=1}^n D_{ii} A_{ii}$. About the best bound you can do for this is the Cauchy-Schwarz inequality, i.e.

$$|tr(DA)| \leq \left ( \sum_{i=1}^n D_{ii}^2 \right )^{1/2} \left ( \sum_{i=1}^n A_{ii}^2 \right )^{1/2}$$

If you want a result in terms of traces, you can use the fact that $\| x \|_2 \leq \| x \|_1$ to get

$$|tr(DA)| \leq tr(|D|) tr(|A|)$$

where $(|D|)_{ij} = |D_{ij}|$ and similar for $|A|$. The first bound is attained when the diagonals of $D$ and $A$ are proportional to each other. The second is attained when these diagonals only have one nonzero entry. So for example you could have $D=A=\begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix}$, then $|tr(DA)|=1=1 \cdot 1 = tr(|D|) tr(|A|)$.

Note that when $D$ and $A$ both have nonnegative diagonal entries, we get the nice result

$$tr(DA) \leq tr(D) tr(A)$$

which is probably more like what you were looking for.

share|improve this answer
    
This is exactly what I was looking for. Thanks. –  pisoir Jul 2 at 7:28

Well.. $$\mathrm{tr}(DA) = \sum_{l = 1}^n \sum_{k = 1}^n d_{lk}a_{kl} $$ But, if $l \neq k$, $d_{lk} = 0$, then $$\mathrm{tr}(DA) = \sum_{l = 0}^n d_{ll}a_{ll} $$

It is easy to compute.

E.g.: $$D = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -1 \end{pmatrix} \qquad A = \begin{pmatrix} 4 & 2 & 10 \\ -1 & 2 & -5 \\ 3 & -1 & -1 \end{pmatrix} \qquad \mathrm{tr}(DA) = 3 \cdot 4 + 2 \cdot 2+ (-1) \cdot (-1) = 17 $$ You can think of it as the dot product between the vectors formed by the diagonal entries of the matrices. Or even think of $D$'s entries as weights for the sum.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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