$A$ 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!

  • $\begingroup$ 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. $\endgroup$ – Aaron Apr 16 '14 at 14:51
  • $\begingroup$ Without hint. Thanks $\endgroup$ – 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.

  • 5
    $\begingroup$ 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. $\endgroup$ – user1551 Apr 16 '14 at 23:44
  • 2
    $\begingroup$ @ziangchen No, your inequality $(*)$ is not correct. consider for instance $T=\left[\matrix{0&0\cr-2&0}\right]$, $U=\left[\matrix{2&2\cr0&-1}\right]$. $\endgroup$ – Omran Kouba Apr 17 '14 at 17:23
  • 2
    $\begingroup$ @ziangchen, In fact, you have in general, for any real $n$-by$n$ matrices $\det(X^TX+Y^TY)\geq\det(X^TY\pm Y^TX)$. $\endgroup$ – Omran Kouba Apr 18 '14 at 5:48
  • 2
    $\begingroup$ 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. $\endgroup$ – Omran Kouba Apr 18 '14 at 6:32
  • 2
    $\begingroup$ @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. $\endgroup$ – 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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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