Determinant inequality $\det(A^2+B^2+(A-B)^2)\ge 3\det(AB-BA)$

$$A$$ and $$B$$ are two $$2\times2$$ reals matrices. then

$$\det \Big(A^2+B^2+(A-B)^2\Big)\ge 3\det\left(AB-BA\right).$$

Well, it is seems interesting, but it is really hard to get started.

Thank you very much!

• Where did you encounter this problem? If it came from a book , the location in the book is often a hint on how to solve it. – Aaron Apr 16 '14 at 14:51
• Without hint. Thanks – ziang chen Apr 17 '14 at 0:35

We will use the following Lemmas:

${\bf Lemma~1.}$ Let $R$, $S$ be two real $2\times 2$ matrices. Then $$\Re(\det(R+iS))=\det(R )-\det(S).$$

$Proof.$ Indeed, if $R_1,R_2$ are the column vectors of $R$, and $S_1,S_2$ are the column vectors of $S$, then using the bi-linearity of the determinant we have \eqalign{ \det(R+iS)&=\det(R_1+iS_1,R_2+iS_2)\cr &=\det(R)-\det(S)+i(\det(R_1,S_2)+\det(R_2,S_1)).} and Lemma 1 follows. $\qquad\square$

${\bf Lemma~2.}$ Let $T$, $U$ be two real $2\times 2$ matrices. Then $$\vert \det(T+iU) \vert^2=\det(T^2+U^2 )-\det(TU-UT).$$ In particular, $$\det(TU-UT)\leq \det(T^2+U^2 ).\tag{*}$$

$Proof.$ Indeed, \eqalign{ \vert \det(T+iU) \vert^2&=\det(T-iU)\det(T+iU)\cr&=\det((T-iU) (T+iU)) \cr&=\det(T^2+U^2+i(TU-UT)) } Now, we apply Lemma 1. with $R=T^2+U^2$ and $S=TU-UT$.$\qquad\square$

Let us come to the proposed question. We will apply $(*)$ with $$T=\sqrt{3}(A-B),\quad U= A+B$$ We check easily that \eqalign{ TU-UT&=2\sqrt{3}(AB-BA)\cr T^2+U^2&=2(A^2+B^2+(A-B)^2).} From this the proposed inequality follows.

• As a remark, while the inequality $\det(X^2+Y^2)\ge \det(XY-YX)$ does not hold for 3x3 or larger matrices, the analogous inequality $\det(X^TX+Y^TY)\ge \det(X^TY-Y^TX)$ does hold in general. – user1551 Apr 16 '14 at 23:44
• @ziangchen No, your inequality $(*)$ is not correct. consider for instance $T=\left[\matrix{0&0\cr-2&0}\right]$, $U=\left[\matrix{2&2\cr0&-1}\right]$. – Omran Kouba Apr 17 '14 at 17:23
• @ziangchen, In fact, you have in general, for any real $n$-by$n$ matrices $\det(X^TX+Y^TY)\geq\det(X^TY\pm Y^TX)$. – Omran Kouba Apr 18 '14 at 5:48
• Let $A=X^TX+Y^TY-(X^TY+Y^TX)$. $$u^TAu=\Vert Xu-Yu\Vert^2$$ so $A$ is symmetric semi-definite positive, and so $X^TX+Y^TY\geq X^TY+Y^TX$, the inequality for determinants follows. – Omran Kouba Apr 18 '14 at 6:32
• @user1551, you are right, but if $M\geq0$, $N$ symmetric, and $M\geq\pm N$, then surely $\det(M)\geq\det(N)$. My previous comment can be adapted so that these conditions are satisfied. Thank you. – Omran Kouba Apr 18 '14 at 11:37

As shown in Omran Kouba's answer, if we put $T=\sqrt{3}(A-B)$ and $U=A+B$, the inequality in question can be rewritten as $$\det(T^2+U^2)\ge\det(TU-UT).$$ Note that For any two $2\times2$ matrices $A$ and $B$, the following identities hold: $\renewcommand{\tr}{\operatorname{tr}}$ $\renewcommand{\adj}{\operatorname{adj}}$

\begin{align} \det(A+B) &\equiv \det(A) + \det(B) + \tr(A\adj(B))\tag{1}\\ \tr(A\adj(B)) &\equiv \tr(B\adj(A)).\tag{2} \end{align}

Let $t=\tr(T\adj(U))=\tr(U\adj(T))$. Put $(A,B)=(T,U),\ (T,-U)$ and $\left(TU-UT,\ U^2-T^2\right)$ in $(1)$, we obtain \begin{cases} \det(T-U) &= \det(T)+\det(U)-t,\\ \det(T+U) &= \det(T)+\det(U)+t,\\ \det((T-U)\det(T+U)) &= \det(TU-UT) + \det(T^2-U^2). \end{cases} Therefore $$-\det(T^2-U^2) = \det(TU-UT) + t^2 - (\det(T)+\det(U))^2.\tag{3}$$ Now, put $(A,B)=(T^2,U^2)$ and also $(T^2,-U^2)$ in $(1)$ and sum up, we get $$\det(T^2+U^2) + \det(T^2-U^2) = 2\det(T)^2+2\det(U)^2.$$ Thus \begin{align} &\det(T^2+U^2)\\ =& 2\det(T^2)+2\det(U^2) - \det(T^2-U^2)\\ =& 2\det(T^2)+2\det(U^2) + \det(TU-UT) + t^2 - (\det(T)+\det(U))^2\quad \text{ by } (3)\\ =& \det(TU-UT) + (\det(T)-\det(U))^2 + t^2\\ \ge& \det(TU-UT). \end{align}