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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?

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ Thank you very much! :) $\endgroup$ – Kevin Jun 12 at 20:52
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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 ${\displaystyle \operatorname {adj} (\mathbf {A} )={\begin{pmatrix}{d}&{-b}\\{-c}&{a}\end{pmatrix}}}$.

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  • $\begingroup$ Is this what you are looking for? $\endgroup$ – Vineet Jun 12 at 16:04
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    $\begingroup$ 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

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