# Adjugate vs transpose of a matrix

Adjugate(classical adjoint) of a matrix https://en.m.wikipedia.org/wiki/Adjugate_matrix is the inverse rotation without scaling. When a matrix is multipled by its inverse, off diagonal entries are zero and diagonal entries have the value scaled by matrix.

But according to the Matt's answer in Geometric interpretation of $\det(A^T) = \det(A)$ , transpose too is stated as inverse rotation without scaling. So adjugate is equal to transpose? Which is clearly wrong. Where am I going wrong?

• if $A$ is a matrix with real entries, its adjoint and its transpose are the same thing. Jan 14 at 17:48
• @Jay What exactly do you mean by "adjoint of a matrix is the inverse rotation without the scaling"? Also, in the second sentence it looks like that "adjoint" that you are thinking of is the classical adjoint (AKA adjugate) rather than the notion of "adjoint" discussed in the link. Jan 14 at 21:07
• @Jay At no point in the answer does Matt say that the transpose is "inverse rotation without scaling". If you point to where you think Matt says that in the answer, we might have a better idea of what you actually have in mind. Jan 14 at 21:08
• @Ben In the title of the third part
– Jay
Jan 15 at 3:29
• Adjugate, transpose, adjoint, inverse: all different but interrelated concepts. Jan 15 at 3:50

Let $$A^*$$ be the adjoint of $$A$$. The defining property of the adjoint is that it satisfies $$\langle Ax,y\rangle = \langle x,A^*y\rangle$$ for some inner product. For a real vector space this will be the transpose however for a complex vector space it will be the conjugate transpose.
To see this let’s restrict ourselves to real vector spaces with $$x$$ and $$y$$ column vectors so that we can represent the Euclidean inner product with matrix multiplication as $$\langle x,y \rangle = x^Ty$$. This means we have $$\langle Ax,y\rangle = (Ax)^Ty = x^TA^Ty=\langle x , A^Ty\rangle$$