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Q: how to find eigenvectors and eigenvalues of R matrix Let V a vector space over F2 and $R(R\in M_{2\times 2}(F2))$ such that $R(v,w)= (w,v)$. Prove that there exist a basis $B=\{(1,0),(0,1)\}$ of V such that: There exists an $\lambda \in F$ such that $R(v,v)=\lambda\cdot v$ for all $v\in B$ There exists an $\beta \in F$ such that $R(v,w)=(\beta,v)$ for all $v\in B$ and $w\in B$ The answer is very simple but I dont know how to solve this problem. I know $R(v,w)=(w,v)$ is a scalar multiple of the identity and it is a linear transformation. A: In general, if $T:\mathbb V\to\mathbb V$ is a linear map and $(u_1,\ldots,u_n)$ is a basis for $\mathbb V$, then $T(u_1),\ldots,T(u_n)$ will be a basis for $\mathbb V$ and hence, $T$ maps each $u_i$ to a multiple of $u_i$. In your case, $R(u_1) = ru_2$ for some $r\in F$, and hence $T(u_1)=ru_2$. Similarly, $T(u_2)=ru_1$. Furthermore, $R(u_1,u_2) = Ru_2$ so $T(u_1,u_2) = Ru_1$. Since $(u_1,u_2)$ and $(u_2,u_1)$ are two bases of $\mathbb V$, you can use the above two equations to uniquely determine $r$ and $r^{ -1}$. [Resistant arterial hypertension treated by nifedipine (NIF) associated to clonidine (CLO)]. The authors present a new group of patients affected by arterial hypertension resistant to previous treatments (3-20 years). Nifed c6a93da74d

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