I've been reviewing some linear algebra, and for and invertible matrix $A$, we have the following formula for the inverse: $$A^{-1} = \frac{1}{\text{det}(A)}\text{adj}(A).$$

However, in the context of bounded operators, we have that the adjoint is simply the transpose of $A$, since $$(Ax,y) = (Ax)^Ty = x^TA^Ty = (x,A^Ty).$$

These two definitions of the 'adjoint' appear to be different. Why are they both given the term adjoint?

Thanks for any insight.

  • 3
    $\begingroup$ Some call the first the adjugate. $\endgroup$ Jun 4, 2019 at 17:44
  • 1
    $\begingroup$ Because math is horrible like that. They are simply different uses of the same term in similar contexts. Most books, such as "Linear Algebra Done Right" even explicitly warn about this point. $\endgroup$
    – The Count
    Jun 4, 2019 at 17:44
  • 1
    $\begingroup$ adjoint is just a 'word', meaning (place) replacement. Due to the context of the theory (as these are different fields), they are used in both areas. However, they are not related. $\endgroup$
    – Maurice
    Jun 4, 2019 at 17:46
  • $\begingroup$ Ok, thanks a lot. $\endgroup$
    – D.B.
    Jun 4, 2019 at 17:51
  • $\begingroup$ It's not rare to encounter overloaded terms in mathematics. E.g. normal vector, normal subgroup, normal distribution, normal operator and normal form are all "normal" stuffs, but they are essentially unrelated concepts (except that the "normal" in normal operator has a similar meaning to the "normal" in normal vector). $\endgroup$
    – user1551
    Jun 4, 2019 at 18:02

2 Answers 2


As the comments have already pointed out, there is no relation. I always call $\operatorname{adj}(A)$ the adjugate of $A$ rather than the adjoint to try to avoid this confusion.

Note that there are plenty of other cases of overloaded terminology, even within linear algebra (for instance, "generalized eigenvector" comes to mind). In many cases you just need to figure out from context which definition is intended.

EDIT: Also a word of warning: the adjoint of $A$ is $A^T$ only for the Euclidean inner product.


They are separate definitions. The first "adjoint" as in $A\operatorname{adj}(A)=\det(A)I$ is a word of French origin. In linear algebra books written in English, some authors opt to call it "adjugate" instead, while the others just adopt the term "adjoint" directly.

Both "adjoint" and "adjugate" were already in use in the early 20th century (see, e.g. Muir's The theory of determinants in the historical order of development, vol. 1, 1906). My impression is that they were more or less equally popular in older books, but more recent texts (say, those published in or after the 1980's) tend to settle on "adjugate", probably to differentiate it from the other "adjoint", the adjoint of an operator/matrix with respect to an inner product.

When one wants to use the term "adjoint" in both definitions that appear in your question but also to disambiguate between them, one may call the first adjoint "classical adjoint" and the second one "Hermitian adjoint" (if the inner product in question is the usual one on $\mathbb C^n$).


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