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Let $ A, B \in M_n(\mathbb{C}) $ such that there exists $ \alpha \in \mathbb{C}^* $ and an odd natural number $ k $ such that $ A^k = \alpha I_n.$ If $ C = 2(I_n - B),$ prove that: $ A^2 B + B A^2 = A C A $ if and only if $ C = I_n. $

My attempt: Assuming that $ C = I_n $ we are done. Now we assume $ A^2 B + B A^2 = A C A $ and we prove $ C = I_n $.

$A^k = \alpha I_n \quad \text{with} \quad \alpha \neq 0,$ $A$ is invertible. Specifically:

$ A^k = \alpha I_n \implies A^{k-1} = \alpha A^{-1} \implies A^{-1} = \frac{1}{\alpha} A^{k-1}. $ Thus, $A$ is invertible.

From the definition of $C$: $ C = 2(I_n - B) \implies I_n - B = \frac{C}{2} \implies B = I_n - \frac{C}{2}. $

Substitute $B = I_n - \frac{C}{2}$ into the equation: $ A^2 B + B A^2 = A C A. $ Thus: $ A^2 \left(I_n - \frac{C}{2}\right) + \left(I_n - \frac{C}{2}\right) A^2 = A C A. $

Expanding the left-hand side (LHS): \begin{align*} A^2 \left(I_n - \frac{C}{2}\right) + \left(I_n - \frac{C}{2}\right) A^2 &= A^2 I_n - \frac{A^2 C}{2} + I_n A^2 - \frac{C A^2}{2} \\ &= 2 A^2 - \frac{A^2 C + C A^2}{2}. \end{align*}

Setting LHS equal to the right-hand side (RHS): $ 2 A^2 - \frac{A^2 C + C A^2}{2} = A C A. $

Thus, $ 4 A^2 = A^2 C + C A^2 + 2 A C A. $

I got stuck here. How can I continue it?

We can also prove that $A$ is diagonalizable since $\lambda^k = \alpha$ has all distinct roots.

user1551
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math.enthusiast9
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1 Answers1

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You are trying to prove $A^2 B + B A^2 = A \big(2I-2B\big) A\implies B=\frac{1}{2}I$. Since each side is homogenous in $A$ we can take WLOG that $\alpha =1$, i.e. let $\beta^k = \alpha$ and divide each side by $\beta^2$ to get $(\beta^{-1}A)^2 B + B (\beta^{-1}A)^2 =$ $ (\beta^{-1}A) \big(2I-2B\big) (\beta^{-1}A)= (A')^2 B + B (A')^2 = (A') \big(2I-2B\big) (A')$

$T:M_n\big(\mathbb C\big)\longrightarrow M_n\big(\mathbb C\big)$ given by $T\big(X\big)=A'X+XA'$
then $(A')^2 B + B (A')^2 = A' \big(2I-2B\big) A'\iff T^2\big(B\big) = 2(A')^2$ but
$T^2\big(\frac{1}{2}I\big)=T\big(A' \big)=2(A')^2\implies B=\frac{1}{2}I$ since $T$ is injective
in fact with $\omega$ being a primitive $k$th root of unity $T$ has eigenvalues in $\in \big\{\omega^j + \omega^r\big\}$ for $1\leq j, r\leq k$ and this set does not include zero (this is where we use $k$ is odd). Ref e.g Eigenvalues of linear operator $F(A) = AB + BA$

user8675309
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  • Can you please explain better starting form the part T is injective? Could you also include in the proof how T is injective? – math.enthusiast9 Jan 31 '25 at 05:26
  • I had $\alpha =1$ WLOG but did not explicitly say so-- I added that. Check if $\lambda$ is a $k$th root of unity then $(-\lambda)^k= (-1)^k \cdot \lambda^k = -1$ since $k$ is odd; since you know $T$ has eigenvalues in $\big{\omega^j + \omega^r\big}$ ths means $T$ does not have an eigenvalue zero hence is injective. I trust that you recognize $A'$ is annihilated by $x^k -1$ hence is diagonalizable with all eigenvalues being $k$th roots of unity, – user8675309 Jan 31 '25 at 18:23