Let $A \in \mathbb{R}^{n\times m}$, and $B\in \mathbb{R}^{m\times m}$. Let $A'$ denote the transpose of $A$.

From this we know that : $A'\in \mathbb{R}^{m\times n}$ and $AB\in \mathbb{R}^{n\times m}$ and hence $ABA' \in \mathbb{R}^{n\times n}$.

The Trace of a matrix is defined as the sum of its diagonal entries. $\textbf{My Question :}$

Does anyone know bounds (using any usual matrix norm) for

  • $\text{Trace}(ABA')$

  • Or more generally $\text{Trace}(AB)$

  • 1
    $\begingroup$ We can bound $|Tr(ABA')| \leq n \sigma_{Amax}^2|\lambda_{Bmax}|$ where $\sigma_{Amax}$ and $\lambda_{Bmax}$ are the largest (by absolute value) singular or eigen values of $A$ and $B$, respectively. $\endgroup$ Mar 30, 2020 at 17:47
  • $\begingroup$ @NinadMunshi That is an amazing bound, and very useful, have you got a link to this/ or make it into an answer ? Thank you very much by the way. $TR(ABA')$ is the one i am most interested in. $\endgroup$ Mar 30, 2020 at 19:15

3 Answers 3


Whenever you see a matrix trace, you should think inner product, because

$$ \operatorname{Tr}(A^T B) = \langle A, B\rangle_F = \langle A,B\rangle_{\mathbb R^m \otimes \mathbb R^n}$$

that is, the trace of the product of two matrices is equal to their frobenius inner product, which in turn is the induced inner product on the tensor product of Hilbert spaces.

Since it is an inner product, the Cauchy-Schwartz inequality applies:

$$ |\langle A, B \rangle_F |^2 \le \|A\|_F^2\|B\|_F^2$$

with equality if and only if $A$ and $B$ are linearly dependent matrices, i.e. scaler multiples of each other. In your case, we have

$$ |\operatorname{Tr}(ABA^T)| = |\operatorname{Tr}(A^TA B)| = |\langle A^TA , B\rangle_F| \le \|A^TA\|_F\|B\|_F$$

The last term can be further bounded by

$$\begin{aligned} \|A^TA\|_F\|B\|_F &\le \|A\|_F^2\|B\|_F = \Big(\sum\nolimits_i \sigma_i^2(A)\Big)\cdot\sqrt{\sum\nolimits_j \sigma_j^2(B)} \\ &\le rank(A)\cdot \sigma^2_{\max}(A)\cdot rank(B)\cdot\sigma_{\max}(B)\\ &\le m\cdot \min(m,n)\cdot \sigma^2_{\max}(A)\cdot \sigma_{\max}(B) \end{aligned}$$


$$tr(C'B)= (vec(C))'(vec(B)) \leq \bigg[(vec(C))'(vec(C)) \bigg]\bigg[(vec(B))'(vec(B)) \bigg]$$ by Cauchy Schwarz inequality. Now let $C=A'$ you get a upper bound of $tr(AB)$.

Similary for $tr(ABA')=tr(BA'A)$


The trace is merely the sum of the eigenvalues, so if we could find an estimate on the largest eigenvalue of $ABA'$, we could say that

$$|Tr(ABA')| \leq n\cdot|\lambda_{ABA'max}|$$

To that end we know that the eigenvalues of $ABA'$ are going to be some combination of the products of the singular values of $A$ and the eigenvalues of $B$. Take the biggest one of each to retrieve the bound:

$$|Tr(ABA')| \leq n \cdot \sigma_{Amax}^2\cdot|\lambda_{Bmax}|$$

$Tr(AB)$ is not defined because it is not a square matrix (In more abstract terms, $AB$ is not a trace-class operator).

  • 1
    $\begingroup$ Let $A$ be a matrix filled with ones, and $B$ be nilpotent, in particular say $B =\mathbf e_1 \mathbf e_2^T$ then $0\lt \Big \vert \text{trace}\big(ABA'\big)\Big\vert \leq n \cdot \sigma_{Amax}^2 \cdot \big \vert \lambda_{Bmax}\big\vert = 0$ which is a contradiction. The eigenvalues of a non-normal $B$ cannot play a role in an upper bound. $\endgroup$ Oct 8, 2020 at 18:33

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