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how to find eigenvectors and eigenvalues of R matrix

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Let V a vector space over F2 and $R(R\\in M_{2\\times 2}(F2))$ such that $R(v,w)= (w,v)$.\nProve that there exist a basis $B={(1,0),(0,1)}$ of V such that:

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There exists an $\\lambda \\in F$ such that $R(v,v)=\\lambda\\cdot v$ for all $v\\in B$\nThere exists an $\\beta \\in F$ such that $R(v,w)=(\\beta,v)$ for all $v\\in B$ and $w\\in B$

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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.

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A:

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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 $u_i$.\nIn 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 $r^{ -1}$.

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Postal 3 Crack Keygen Autocad

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