The multiplicative conjugate of an invertible matrix is invertible

Question

If $A,B,C$ are $n \times n$ (real) matrices and $A$ and $B$ are invertible, with $AB=BC$, prove that $C$ is also invertible.

My attempted proof is $(B^{-1})(AB) = (B^{-1})(BC)$. Then $A(B^{-1})(B) = C$. So $A=C$ and $C$ is invertible.

I'm pretty sure the second line is not valid but am unsure how to fix the proof.

Answer 1

Hint:

You should have written have $B^{-1}AB = C$

So what is $(B^{-1}A^{-1}B)C $ or $C(B^{-1}A^{-1}B)$?

Answer 2

Hint: Show that $C$ has the trivial nullspace $\{0\}$. A proof by contradiction would work well.

Answer 3

$$A^{-1}BC= A^{-1}AB=B$$ so $$B^{-1}(A^{-1}BC)=\operatorname{Id}$$ so $$C^{-1} = B^{-1}A^{-1}B$$

Answer 4

$AB=BC\Rightarrow 0\neq \det A \det B=\det B \det C\rightarrow \det C\neq 0\Rightarrow C$ is invertible.

Answer 5

The matrices $A$ and $B$ are invertible, hence $\det A\ne0$ and $\det B\ne 0$. Then

$$\det B\cdot \det C=\det (BC)=\det (AB)=\det A\cdot \det B\ne 0,$$ hence $\det C\ne 0$ and thus $C$ is invertible.

Answer 6

Hint: $$AB=BC \implies C = B^{-1}AB$$ $$det(B^{-1}AB)= det(B^{-1})det(A)det(B)=det(B^{-1}B)det(A)=det(A) \neq 0 \ [\text{all the matrices are invertible}]$$

So, $det(C) \neq 0 \implies \ C$ is invertible.