3

Let $A $ and $B$ be real symmetric matrices of order $n$, satisfying $A^2B=ABA.$ Proof that $AB=BA.$ I only find that $A^2B=BA^2=ABA.$ Then I don't know what to do next.Can you help me solve this puzzle?

Fred
  • 77,394
Gyh
  • 39
  • 1
    How did you show that $A^2B=BA^2$? Compare also with this post. – Dietrich Burde Jan 24 '23 at 08:32
  • 1
    Multiply the initial relation by $A^{-1}$ on the left. – Abezhiko Jan 24 '23 at 08:33
  • Are $A$ and $B$ invertible? – seaver Jan 24 '23 at 08:34
  • Hint: you can assume that $A$ is diagonal. – Aphelli Jan 24 '23 at 09:37
  • 5
    There are many ways to solve this problem. Method 1: prove that $B$ commutes with $A^3$ and argue that $A=p(A^3)$ for some polynomial $p$. Method 2: prove that $\ker(A)$ and $\ker(A)^\perp$ are invariant subspaces of both $A$ and $B$; then prove that $AB=BA$ on these two subspaces. Method 3: first, show that $A$ commutes with $AB-BA$; then apply Jacobson's lemma and note that $AB-BA$ is skew-symmetric. – user1551 Jan 24 '23 at 09:55
  • $A^2B=ABA \implies A(AB) = A(BA) \implies A^{-1}A(AB) = A^{-1}A(BA) \implies AB=BA$ surely? – tai Jan 24 '23 at 10:41
  • 1
    @tai Of course, the result is obvious when $A$ is invertible, but here we don't know this. – Marcos Jan 24 '23 at 10:44
  • 1
    Another option: write out $\big \Vert AB-BA\big\Vert_F^2=\text{trace}\big(BA^2B\big) +\text{trace}\big(AB^2A\big)-\text{trace}\big(BABA\big)-\text{trace}\big(ABAB\big)=0$ where the zero comes from applying the identity in OP to the 2 subtracted items (as well as cyclic property of trace). – user8675309 Jan 24 '23 at 17:35
  • @user1551 How did you proof this puzzle using Method 3 ? Could you explain it in more detail to make it an answer？ – Gyh Jan 25 '23 at 02:59
  • 1
    @Gyh Any nilpotent skew-symmetric matrix must be zero. – user1551 Jan 25 '23 at 11:21
  • 1
    One more option (loosely related to @user1551 's method 2). Define bilinear form $\langle \mathbf x,\mathbf y\rangle:= \mathbf x^T (ABA)\mathbf y$. Now select non-zero $\mathbf v,\mathbf w$, coming from 2 (arbitrary) distinct eigenspaces of $A$. The identity in the OP also tells us we can compute $\langle \mathbf v, \mathbf w\rangle$ two different ways, inferring $\langle \mathbf v, \mathbf w\rangle = 0 =\mathbf v^T B\mathbf w$. Conclude $B$ respects $A's$ eigenspaces $\implies AB=BA$. (The case $A=\lambda I$ trivially holds.) – user8675309 Jan 25 '23 at 18:52

2 Answers2

2

If $A$ is invertible or $A=0$ then it is obvious. Assume now that $A\ne 0$ and $\det A=0$.

$A$ is symmetric and hence diagonalizabe by an orthnormal matrix $U$. Set $A=U^{-1}LU$, where$L$ is diagonal. Expressing $L$ and $\hat B=UBU^{-1}$ (which is also symmetric) in block form, $$ L=\left( \begin{array}{cc}D & 0 \\ 0 & 0\end{array} \right),\qquad UBU^{-1}=\left( \begin{array}{cc}B_1 & B_2 \\ B_3 & B_4\end{array} \right)=\hat B, $$ where $D$ is diagonal and invertible, we have $$ U^{-1}L^2\hat B U=U^{-1}L^2UB=A^2B=ABA=U^{-1}LUBU^{-1}LU=U^{-1}L\hat BLU $$ or $$ L^2\hat B=L\hat BL $$ or $$ \left( \begin{array}{cc}D^2 & 0 \\ 0 & 0\end{array} \right)\left( \begin{array}{cc}B_1 & B_2 \\ B_3 & B_4\end{array} \right)=\left( \begin{array}{cc}D & 0 \\ 0 & 0\end{array} \right)\left( \begin{array}{cc}B_1 & B_2 \\ B_3 & B_4\end{array} \right)\left( \begin{array}{cc}D & 0 \\ 0 & 0\end{array} \right) $$ or $$ \left( \begin{array}{cc}D^2B_1 & D^2B_2 \\ 0 & 0\end{array} \right)=\left( \begin{array}{cc}DB_1D & 0 \\ 0 & 0\end{array} \right). $$ Hence $B_2=0$ and $DB_1=B_1D$, and also $B_3=B_1^T=0$, in which case $$ BA=U^{-1}\hat BUU^{-1}\left( \begin{array}{cc}D & 0 \\ 0 & 0\end{array} \right)U= U^{-1}\hat B\left( \begin{array}{cc}D & 0 \\ 0 & 0\end{array} \right)U=U^{-1}\left( \begin{array}{cc}B_1 & 0 \\ 0 & B_4\end{array} \right)\left( \begin{array}{cc}D & 0 \\ 0 & 0\end{array} \right)U=U^{-1}\left( \begin{array}{cc}B_1D & 0 \\ 0 & 0\end{array} \right)U=U^{-1}\left( \begin{array}{cc}DB_1 & 0 \\ 0 & 0\end{array} \right)U=U^{-1}\left( \begin{array}{cc}D & 0 \\ 0 & 0\end{array} \right)\left( \begin{array}{cc}B_1 & 0 \\ 0 & B_4\end{array} \right)U=U^{-1}\left( \begin{array}{cc}D & 0 \\ 0 & 0\end{array} \right)\hat BU=U^{-1}\left( \begin{array}{cc}D & 0 \\ 0 & 0\end{array} \right)UBU^{-1}U=AB $$

Yiorgos S. Smyrlis
  • 83,933
-1

Since $A^{2}B=ABA$ and both these matrices are symmetric, we obtain $A^{2}B=BA^{2}$.

Multiply by $A$ on both sides to obtain $BA^{3}=A^{3}B$ (Using the relation $ABA=A^{2}B$).

Carrying this process out repeatedly, we see that $A^{n}B=BA^{n} \forall n \geq 2$.

Using Cayley-Hamilton, we see that the characteristic polynomial kills $A$, i.e. $p(A)=0$. So, now we can write $A$ as a linear combination of $I$ and $A^{n}$ for $n \geq 2$. Using the fact we proved above, we're done.

dormordo
  • 59
  • From $ABA=A^2B$ I get $A^2BA=A^3B$ not $BA^3=A^3B,.$ – Kurt G. Jan 24 '23 at 11:39
  • $BA^2=A^2B$ implies $BA^3=A^2BA$, but $A(ABA)=A(A^2B)$. – dormordo Jan 24 '23 at 11:42
  • This does not always work, because the coefficient of $A$ in the characteristic and minimal polynomials of $A$ may be zero. Consider $A=\pmatrix{1\ &-1}$ for instance. – user1551 Jan 24 '23 at 13:46
  • I think there's a fix for that. Since A is symmetric, the minimal polynomial would have only distinct roots. This implies both the constant and the coefficient of A cannot be zero. Now, if the coefficient of A is zero, just multiply the min poly by A and you have a relation with higher powers of A. – dormordo Jan 24 '23 at 14:51
  • Yes, it's a valid fix. – user1551 Jan 24 '23 at 16:55