# when does $\det(AB^T+BA^T)\le \det(AA^T+BB^T)$ hold?

When does the following matrix inequality hold?

$$\det(AB+B^TA^T)\le \det(AA^T+BB^T)$$

$A$ and $B$ are any real matrices.

My reply gives a counter example. The question is under what condition does that hold?

[Update]

The inequality actually I would like to prove is not the above one. It should be

$$\det(AB^T+BA^T)\le \det(AA^T+BB^T)$$

I also changed the title. Thanks julien.

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This is basically a matrix version of the inequality $(a-b)^2\ge 0$, which becomes $2ab\le a^2+b^2$. Try expanding $0\le \det\left( (A-B)(A-B)^T\right).$ – Giuseppe Negro May 28 '13 at 18:55
I am still trying. $0 \le \det((A-B)(A-B)^T) \le \det(AA^T+BB^T-BA^T-AB^T) \le \det(AA^T+BB^T) + \det(-BA^T-AB^T)$ Well, the second term of rhs has a negative sign. – Nature4D May 28 '13 at 19:10
@GiuseppeNegro: But are we allowed to perform like $0\le\det(U-V) \ \implies\ \det V\le \det U$? – Berci May 28 '13 at 19:15
@GiuseppeNegro Unlike the seven people who upvoted your comment, I don't understand it. First, the determinant is multilinear, not linear. Second, when you expand, you get $AB^T+BA^T$, not $AB+B^TA^T$. What did I miss? – 1015 May 28 '13 at 19:23
Sorry if I generated confusion. That comment is only a guess off the top of my head, not a hint. I thought that, with some extra assumptions on $A, B$ this can lead somewhere but I am not sure at all. – Giuseppe Negro May 28 '13 at 20:13

I assume that $A,B$ are real $n\times m$ matrices. But the same proof gives the same result as below for complex matrices with the adjoint instead of the transpose.

For every real $n\times m$ matrices $A,B$, we have $$|\det (AB^T+BA^T)|\leq \det(AA^T+BB^T).$$

Proof: if $x\in\ker (AA^T+BB^T)$, then $\|A^Tx\|^2+\|B^Tx\|^2=0$, whence $A^Tx=B^Tx=0$ and so $x\in \ker(AB^T+BA^T)$. In other words, if $\det(AA^T+BB^T)=0$, we do have $\det(AB^T+BA^T)=0$ and the inequality holds. So we will assume that $AA^T+BB^T$ is invertible from now on.

The matrix $C=(A-B)(A-B)^T$ is positive semidefinite. So $$0\leq (Cx,x)=((AA^T+BB^T)x,x)-((AB^T+BA^T)x,x)$$ whence $$((AB^T+BA^T)x,x)\leq ((AA^T+BB^T)x,x).$$ Changing $B$ for $-B$, we get altogether

$$|((AB^T+BA^T)x,x)|\leq ((AA^T+BB^T)x,x).$$

Denote $C:=AB^T+BA^T$ which is symmetric, and $D:=AA^T+BB^T$ which is positive definite and has a unique positive definite square root $D^{1/2}$. Now the inequality above entails, with $x=D^{-1/2}y$: $$|(D^{-1/2}CD^{-1/2}y,y)|\leq(y,y)=\|y\|^2\qquad\forall y\in\mathbb{R}^n.$$ Since $T=D^{-1/2}CD^{-1/2}$ is symmetric, it is diagonalizable in an orthonormal basis and the above shows that its spectrum is contained in $[-1,1]$. A fortiori $$|\det (D^{-1/2}CD^{-1/2})|\leq 1\qquad \Rightarrow \qquad |\det C|\leq \det D.$$ QED.

Note: the inequality you wanted initially requires that $A,B$ be square matrices of the same size. As you observed, the inequality is not true in general. The transpose was not at the right place. It follows in particular from the inequality above that the inequality you first wanted holds when $A,B$ are both symmetric, or both skew-symmetric. Of course, it is true also if $AB=0$. One can add some more solutions. All $(A,B)$ such that the symmetric matrix $AB+B^TA^T$ is singular or has an odd number of negative eigenvalues. But I doubt there is a nice general characterization. Although I might be wrong.

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I think I should get the equation wrong. The inequality you proved makes sense. From your proof, I learned a lot! Thanks! I need to double check the inequality. – Nature4D May 28 '13 at 22:19
@Nature4D I've added several other sufficient conditions. May I ask how you came up with the question? – 1015 May 29 '13 at 1:01
It comes from Cybernetics in which people represent status of a system with matrices. :) – Nature4D May 29 '13 at 1:20
Wow, cybernetics... If you have a link to share where this inequality is actually useful in cybernetrics, I'll be happy to see that. – 1015 May 29 '13 at 1:21
When I have a system whose state equation is $e=Q^{-1}(A-L_xC)e + Q^{-1}\phi,\quad Q=E+L_yC$, I try to use Lynapunov stability theory to prove the system is asymptotically stable. – Nature4D May 29 '13 at 3:06

I found a counter example by Matlab. I know that is cheap..... :-|

$A = 0.4582 \quad 0.1567 \quad 0.0519\\ 0.8204 \quad 0.9111 \quad 0.0512 \\ 0.9831 \quad 0.1020 \quad 0.6691$

$B = 0.8492 \quad 0.0900 \quad 0.0787\\ 0.8890 \quad 0.0126 \quad 0.7477\\ 0.7770 \quad 0.2320 \quad 0.0607$

f = @(A,B)(det(A*A'+B*B')-det(A*B+B'*A'))

f(A, B) gives -1.9356

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It seems if A and B are symmetric, the inequality would hold... – Nature4D May 28 '13 at 19:42

As $AA^T+BB^T\succeq0$, if $(AA^T+BB^T)x=0$, we must have $A^Tx=B^Tx=0$ and $(AB^T+BA^T)x=0$ too. So, it suffices to prove the inequality when $AA^T+BB^T\succ0$.

Write $AA^T+BB^T=\pmatrix{A&B}\pmatrix{A^T\\ B^T}$ and $AB^T+BA^T=\pmatrix{A&B}\pmatrix{0&I\\ I&0}\pmatrix{A^T\\ B^T}$. Let $\pmatrix{A&B}=U\pmatrix{\Sigma&0}V^T$ be a singular value decomposition, where $\Sigma$ is a diagonal matrix. Since $AA^T+BB^T$ is invertible, so is $\Sigma$. Consequently, the inequality in question can be rewritten as $$\det\left[\pmatrix{I&0}\ \underbrace{V^T\pmatrix{0&I\\ I&0} V}_S\ \pmatrix{I\\ 0}\right]\le1.$$ Hence it suffices to prove that for any symmetric square root $S$ of $I_{2n}$, its leading principal $n\times n$ minor is bounded above by $1$. This actually follows immediately from the interlacing inequality for eigenvalues of submatrices of Hermitian matrices, but we will offer another proof here.

Let $S=\pmatrix{X&Y^T\\ Y&Z}$, where $X$ and $Z$ are symmetric matrices of the same sizes. By comparing the first subblocks on both sides of $S^2=I_{2n}$, we get $X^TX+Y^TY=I$. Therefore $I\succeq X^TX\succeq 0$ and $\det(X)^2\le1$. QED.

The above proof has the merit that a similar argument (with $S$ being a skew-symmetric square root of $-I_{2n}$) can be used to show that

$\det(AB^T\color{red}{-}BA^T)\le \det(AA^T+BB^T)$.

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