Let $R$ be a ring. Then I would like to know the units in $M_n(R)$.

Here is my idea. The determinant is a homomorphism from $M_n(R)$ to $R$. If $A$ is a unit, then there is a $B$ such that $AB = I$. Then taking determinants one finds that $\det(A)$ is a unit in $R$. So therefore the units in $M_n(R)$ are those matrices $A$ where $\det(A)$ is a unit in $R$.

Is this correct? Can this be done more explicitly?

  • 3
    $\begingroup$ You want $R$ to be commutative. $\endgroup$
    – Pedro Tamaroff
    Nov 10 '15 at 14:08
  • $\begingroup$ I don't know if det is still a homomorphism when $R$ is not commutative $\endgroup$ Nov 10 '15 at 14:08
  • 2
    $\begingroup$ You seem to be confusing the statement with its converse. You've argued that if $A$ is a unit, then $\det(A)$ is also a unit. However, you need to actually argue the converse! Start with the assumption that $\det(A)$ is a unit and show that $A$ is as well. $\endgroup$ Nov 30 '15 at 15:07

When $R$ is commutative, you have $$ \mathbf{A}\, \mathrm{adj}(\mathbf{A}) = \mathrm{adj}(\mathbf{A})\, \mathbf{A} = \det(\mathbf{A})\, \mathbf I_n \qquad $$ where $\mathrm{adj}(\mathbf{A})$ is the adjugate or classical adjoint of $\mathbf{A}$, that is, the transpose of its cofactor matrix.

This equation proves that

If $\det(\mathbf{A})$ is invertible in $R$, then $\mathbf{A}$ is invertible in $M_n(R)$.

because then $\mathbf{A}^{-1}=\det(\mathbf{A})^{-1}\mathrm{adj}(\mathbf{A})$.

Your argument based on $\det(\mathbf{AB})=\det(\mathbf{A})\det(\mathbf{B})$ proves the converse:

If $\mathbf{A}$ is invertible in $M_n(R)$, then $\det(\mathbf{A})$ is invertible in $R$.


Hint: Use the cofactor matrix if R is commutative

  • $\begingroup$ So if my conclusion correct if $R$ is commutative? $\endgroup$
    – John Doe
    Nov 10 '15 at 14:24

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.