How to write the adjoint application defined by a matrix?

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?

• Are we talking about finite-dimension vector spaces? – Sam Skywalker Jun 12 at 15:15
• Yes, precisely two-dimensions vector space. – Kevin Jun 12 at 15:16
• With dot product I mean that the vector space is defined by the euclidean product. – 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: $$<\cdot,\cdot>: V\to V^*;\quad v\mapsto .$$

Recall that $$:V\to \mathbb F$$ 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.

• Thank you very much! :) – 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 $${\displaystyle \operatorname {adj} (\mathbf {A} )={\begin{pmatrix}{d}&{-b}\\{-c}&{a}\end{pmatrix}}}$$.

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