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Let $A$ be a $m \times n$ matrix with real entries, and let $B$ be a $n \times n$ real symmetric matrix with absolute eigenvalues $\le 1$.

Are there (ideally sharp) bounds for the inequality $$\det(AA^T) \ge \det(ABA^T)$$ in terms of $\det(AA^T)$ and functions of $B$ such as e.g. its rank?

In my particular problem $B$ is idempotent, if that makes for an interesting special case.

Lalit Tolani
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ntessore
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  • Is $B$ symmetric? – eyeballfrog Jun 01 '21 at 14:15
  • @eyeballfrog Yes it should have said that from the beginning. Apologies, I edited the question. – ntessore Jun 01 '21 at 14:17
  • What exactly do you mean by "bounds for the inequality"? Are you asking for bounds on $\det(AA^T) - \det(ABA^T)$? – Ben Grossmann Jun 01 '21 at 14:34
  • @BenGrossmann Yes, exactly, ideally in the form of a function $f$ such that $\det(AA^T) - \det(ABA^T) \le f(\det AA^T, B)$ and there are some $A$, $B$ such that equality holds. – ntessore Jun 01 '21 at 14:36
  • A possibly helpful step: if $B$ is symmetric and idempotent, then there exists a (rectangular) matrix $U$ with orthonormal columns such that $B = UU^T$. If $U$ has size $n \times r$ with $r \geq m$, then the Cauchy-Binet formula gives us $$ \det(ABA^T) = \det((AU)(AU)^T) = \sum_{S \in \binom{[r]}{m}} \det((AU)_{[m],S})^2. $$ If $r<m$, then the determinant must be zero. – Ben Grossmann Jun 01 '21 at 14:38
  • Note that $r$ is the rank of $B$. – Ben Grossmann Jun 01 '21 at 15:00
  • As @BenGrossmann points out in his answer below, there is no lower bound. I am still interested in an upper bound on $\det(ABA^T)$ in terms of $\det(AA^T)$ and $B$. – ntessore Jun 01 '21 at 16:02
  • @ntessore For $B$ symmetric and idempotent, it must hold that $\det(ABA^T) \leq \det(AA^T)$. Is this helpful? – Ben Grossmann Jun 01 '21 at 18:53

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One approach is as follows.

Note that $\det(AA^T)$ is the squared product of the singular values of $A$. On the other hand, as I note in my comment, if $B$ is symmetric and nilpotent, $\det(ABA^T)$ can be expressed as $\det((AU)(AU)^T)$ for some matrix $U$ with $r$ orthonormal columns, where $r$ is equal to the rank of $B$. In other words, $\det(A^TBA)$ is the squared product of the singular values of $AU$.

The Cauchy interlacing theorem for singular values implies that if $\sigma_1,\dots,\sigma_m$ denote the (non-zero) singular values of $A$ and $\tau_1,\dots,\tau_m$ denote the singular values of $AU$, then we must have $$ \sigma_k \geq \tau_{k} \geq \sigma_{(n-r) + k} $$ for all $k = 1,\dots,m$. With that, we can consider the fact that $$ \det(AA^T) = \sigma_1^2 \cdots \sigma_m^2, \quad \det(ABA^T) = \tau_1^2\cdots \tau_m^2 $$ To derive a lower bound for $\det(ABA^T)$ in terms of the singular values of $A$ and of $AU$.

Concretely: we have $\tau_k \geq \sigma_{(n-r) + k}$ for all $k = 1,\dots,m+r-n$. With that, we have $$ \det(ABA^T) \geq \sigma_{n-r} \cdot \sigma_{n-r+1}\cdots\sigma_m \cdot \tau_{m+r-n+1}\cdots\tau_m. $$

Proof that $\det(AA^T) \geq \det(ABA^T)$: note that both $ABA^T$ and $A(I - B)A^T$ are both (symmetric and) positive semidefinite. If $\lambda_i$ denotes the $i$th eigenvalue in ascending order, then any two symmetric matrices $P,Q$ will satisfy $$ \lambda_i(P + Q) \geq \lambda_i(P) + \lambda_1(Q). $$ Thus, we have $$ \lambda_i(AA^T) = \lambda_i(ABA^T + A(I - B)A^T) \geq \lambda_i(ABA^T) + \lambda_1(A(I - B)A^T) \\ \geq \lambda_i(ABA^T) + 0. $$ Thus, each eigenvalue of $AA^T$ is greater than or equal to the corresponding eigenvalue of $ABA^T$, and all of these eigenvalues are non-negative. It follows that $\det(AA^T)$ (the product of the eigenvalues of $AA^T$) is greater than $\det(ABA^T)$ (the product of the eigenvalues of $ABA^T$).

Ben Grossmann
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  • Thank you, note however that this would require knowledge of $A$ to form $AU$. – ntessore Jun 01 '21 at 15:49
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    That is true; at least, we would need to know the singular values of $A$ and some singular values of $AU$. I think that $\det(AA^T)$ and complete knowledge of $B$ will not be sufficient to determine a lower bound; would you want a proof of this? – Ben Grossmann Jun 01 '21 at 15:52
  • Oh... I think I can see that you are right. This changes my outlook a little. I guess you can salvage the argument to say something about an upper bound, although still not independently of $A$? – ntessore Jun 01 '21 at 16:00
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    @ntessore See my latest edit. – Ben Grossmann Jun 01 '21 at 19:42
  • This is a very nice way of showing the upper bound – ntessore Jun 01 '21 at 19:58
  • I think the upper bound is also sharp for $B$ idempotent: If $A = USV^T$ is the SVD of $A$ and $B = Q\Lambda Q^T$ is the ED of $B$, the matrix $W = V^TQ$ is orthogonal, and we can write block matrices $$S=\begin{pmatrix}S_r&0\end{pmatrix};,\quad W=\begin{pmatrix}W_r&\ldots\\ldots&\ldots\end{pmatrix};,\quad \Lambda=\begin{pmatrix}I_r&0\0&0\end{pmatrix}$$ where $r$ is the rank of $B$. We then have $\det(ABA^T) = \det(S_rW_rW_r^TS_r^T)$ which is equal to $\det(S_rS_r^T) = \det(AA^T)$ for $W_r = I_r$. – ntessore Jun 02 '21 at 09:17