# bases in $\mathbb{R}^{2}$ - linear algebra

Find the bases $B_{1}$ and $B_{2}$ in $\mathbb{R}^2$ such that $u=(1,2)_{B_{1}}^{t}=(1,1)_{B_{2}}^{t}$ and $v=(-1,1)_{B_{1}}^{t}=(3,1)^{t}_{B_{2}}$.

Thanks :)

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I got your question wrong first time. Fixed my answer now. –  Dennis Gulko Nov 30 '12 at 10:43

If $B_1=\left(\binom{a}{b},\binom{c}{d}\right)$, $B_2=\left(\binom{x}{y},\binom{z}{w}\right)$, $u=\binom{\alpha}{\beta}$ and $v=\binom{\gamma}{\delta}$ then the given conditions can be formulated as: $$\begin{pmatrix}a&c\\b&d\end{pmatrix}\binom{\alpha}{\beta}=\binom{1}{2},\hspace{5pt} \begin{pmatrix}x&z\\y&w\end{pmatrix}\binom{\alpha}{\beta}=\binom{1}{1}\\ \begin{pmatrix}a&c\\b&d\end{pmatrix}\binom{\gamma}{\delta}=\binom{-1}{1},\hspace{5pt} \begin{pmatrix}x&z\\y&w\end{pmatrix}\binom{\gamma}{\delta}=\binom{3}{1}$$ Denoting $[I]^{B_1}_e=\begin{pmatrix}a&c\\b&d\end{pmatrix}$, $[I]^{B_2}_e=\begin{pmatrix}x&z\\y&w\end{pmatrix}$ we get: $$\left([I]^{B_1}_e\right)^{-1}\binom{1}{2}=\binom{\alpha}{\beta}=\left([I]^{B_2}_e\right)^{-1}\binom{1}{1}\\ \left([I]^{B_1}_e\right)^{-1}\binom{-1}{1}=\binom{\gamma}{\delta}=\left([I]^{B_2}_e\right)^{-1}\binom{3}{1}$$ Hence we can find $P=[I]^{B_2}_e\left([I]^{B_1}_e\right)^{-1}$ from $P\binom{1}{2}=\binom{1}{1}$ and $P\binom{-1}{1}=\binom{3}{1}$: $P=\frac13\begin{pmatrix}-5&4\\-1&2\end{pmatrix}$.
We see that the answer is not unique: choose any basis $B_1$ of $\mathbb{R}^2$, then you will get $[I]^{B_2}_e=P[I]^{B_1}_e$.

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+1 I really like this answer. It is arranged step by step. –  B. S. Nov 30 '12 at 14:23