Template talk:Matrix classes

An "adjugate matrix" is not a class of matrix

The adjugate of a matrix is an operation done on an arbitrary square matrices: For the case of invertible matrices, it happens to rqual $\operatorname {adj} (A)=\det(A)A^{-1}$. For the non-invertible cases, see the article. It is not a class of matrix, unless you mean (over algebraically-closed fields) the set of square matrices which are either invertible, zero, or have rank equal to 1 - which are precisely the matrices which occur as adjugates of some matrix. Actually, over general fields, it's also required that the determinant be an $n-1$st power, which is guaranteed in the algebraically-closed case, but can fail for certain values of $n$ over certain other fields like the reals or finite fields.

To see why those are the only possible values of $\operatorname {adj} (A)$, consider the different possible nullities of $A$:

- If the nullity of $A$ is 0, then $A$ is invertible. We get that if $\operatorname {adj} (B)=A$ then $B=\lambda A^{-1}$. It follows from some tedious calculation that $\lambda ^{n-1}=\det(A)$.

- Now consider the case where $A$ has nullity 1. In that case, $A\operatorname {adj} (A)=\det(A)I=0$, so the columns of $\operatorname {adj} (A)$ are all in $\ker(A)$. It follows that the rank of $\operatorname {adj} (A)$ is at most 1. Since $\operatorname {adj} (A)$ has rank at most 1, we get that $\operatorname {adj} (A)=uv^{T}$ for two column vectors $u$ and $v$. We finish this case by noting that it's easy to see that all rank-1 matrices $uv^{T}$ arise this way.

- If the nullity of $A$ is greater than 1, then a change-of-basis shows that all its minors are 0, so we get $\operatorname {adj} (A)=0$. Svennik (talk) 13:06, 14 April 2026 (UTC)