# Can $ABA-BAB=I$?

Let $A,B\in M_{n\times n} (\mathbb{C})$. Is it possible that $ABA-BAB=I$?

I came across this interesting problem as I was studying for an exam. I guess in the case when $A$ and $B$ commute we have $A(A-B)B=I$ and I am not sure if it can happen.

Any ideas?

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Yes. Let $$A_n = \mathrm{diag}\left(\frac{1}{2}(1 + \sqrt{5}), \dots, \frac{1}{2}(1 + \sqrt{5})\right)$$ and $$B_n = \mathrm{diag}(1, \dots, 1).$$ This example will work for any $n$.

Edit: Now, for a general solution in the noncommutative case, combine my commutative general solution with Robert Israel's $2 \times 2$ noncommutative solution, taking the block diagonal matrices $$\tilde{A}_n = \begin{pmatrix} A_{n-2} & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ and $$\tilde{B}_n = \begin{pmatrix} B_{n-2} & 0 & 0 \\ 0 & -\frac{1}{2} & 1 \\ 0 & -\frac{7}{2} & 3 \end{pmatrix}.$$

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To explain, this is the commuting case, which mostly boils down to the behavior of eigenvalues. Replacing the matrices with eigenvalues, you seek $a^2 b - a b^2 = 1$. Pick any value for $b$ (other than zero) and you can solve for $a$. – Hurkyl May 21 '12 at 21:54

Indeed, you could have $A = I$ and then you need $B - B^2 - I = 0$, which is true e.g. for $B = \pmatrix{0 & 1\cr -1 & 1\cr}$.

EDIT: For a non-commuting example, try $A = \pmatrix{2 & 0\cr 0 & 1\cr}$, $B = \pmatrix{-1/2 & 1\cr -7/2 & 3\cr}$

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