I'm having a hard time understanding the problem that asks to write the adjoint application $f_A$ defined by $A:\begin{bmatrix}a & b \\ c & d\end{bmatrix}$ with respect to the euclidean product. Any tips?

  • $\begingroup$ Are we talking about finite-dimension vector spaces? $\endgroup$ – Sam Skywalker Jun 12 at 15:15
  • $\begingroup$ Yes, precisely two-dimensions vector space. $\endgroup$ – Kevin Jun 12 at 15:16
  • $\begingroup$ With dot product I mean that the vector space is defined by the euclidean product. $\endgroup$ – Kevin Jun 12 at 15:27

Bear in mind that a matrix stands for a linear map between two $\mathbb F$-vector spaces over the field $\mathbb F$. Since it is a square matrix and our dimension is two, we can assume that both spaces are the same, so $A:V\to V$.

The dual space of $V$, which we denote by $V^*$, is the space of linear maps from $V$ to the field $\mathbb F$. This is again a vector space and its dimension is that of $V$.

We can regard the dot product as a rule that associates to each vector $v\in V$ a linear map the following way: $$<\cdot,\cdot>: V\to V^*;\quad v\mapsto <v,\cdot>. $$

Recall that $<v,\cdot>:V\to \mathbb F$ is a linear map that eats vectors and gives scalars.

The transpose matrix of $A$, $A^ t$ is the finite-dimensional version of the adjoint. In this case, $A^t$ is the dual map of $A$. If you let $A^t$ act on a basis of the dual space $V^*$ (see the discussion on the dot product two paragraphs above), you get a basis of $V^{**}$, which is canonically isomorphic to $V$, our original vector space.

Do not hesitate to reply if you need further help.

  • $\begingroup$ Thank you very much! :) $\endgroup$ – Kevin Jun 12 at 20:52

The adjoint of $A$ is the transpose of $C$, that is, the $n×n$ matrix whose $(i,j)$ entry is the $(j,i)$ cofactor of $A$

${\displaystyle \operatorname {adj} (\mathbf {A} )=\mathbf {C} ^{\mathsf {T}}=\left((-1)^{i+j}\mathbf {M} _{ji}\right)_{1\leq i,j\leq n}.}$ Where $M_{ij}$ is minor of A.

${\displaystyle \mathbf {A} ={\begin{pmatrix}{a}&{b}\\{c}&{d}\end{pmatrix}}}$ Then ${\displaystyle \operatorname {adj} (\mathbf {A} )={\begin{pmatrix}{d}&{-b}\\{-c}&{a}\end{pmatrix}}}$.

  • $\begingroup$ Is this what you are looking for? $\endgroup$ – Vineet Jun 12 at 16:04
  • 1
    $\begingroup$ I am pretty sure he wants the other kind of adjoint, since he is saying "with respect to the euclidean product". $\endgroup$ – darij grinberg Jun 12 at 16:28
  • $\begingroup$ Yes, that one with euclidean product. I'm new to linear algebra, thanks! $\endgroup$ – Kevin Jun 12 at 16:42

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.