# Finding a basis such that a vector expressed with respect to that basis is equal to a given vector

Find a basis $\alpha = \{ \vec\alpha_1, \vec\alpha_2, \vec\alpha_3 \}$ of $P_2(R)$, such that $[2 + 5x + 4x^2]_\alpha =\left(\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}\right)$

I have approached the problem the following way: \begin{align*} \vec v &= 2 + 5x + 4x^2\\ s &= \{1, x, x^2\}\\ [\vec v]_s &= \left(\begin{matrix} 2 \\ 5 \\ 4 \end{matrix}\right)\\ [\vec v]_\alpha &= \left(\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}\right)\\ _sC_\alpha &=\left( \begin{matrix} \uparrow & \uparrow & \uparrow \\ \alpha_1 & \alpha_2 & \alpha_3 \\ \downarrow & \downarrow & \downarrow \end{matrix} \right) \end{align*}

Therefore: \begin{align*} {_sC_\alpha}^{-1} [\vec v]_s &= [\vec v]_\alpha\\ [\vec v]_s &= _sC_\alpha [\vec v]_\alpha \end{align*} So $$\left(\begin{matrix} 2 \\ 5 \\ 4 \end{matrix}\right) = \left(\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right) \left(\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}\right)$$ For which one possible solution is $$\left(\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right) = \left(\begin{matrix} 2 & 0 & 0 \\ 0 & 5/2 & 0 \\ 0 & 0 & 4/3 \end{matrix}\right)$$

My question basically boils down to this: Is this a valid answer? It appears to me like under the Field R there exist infinitely many such bases.

PS: Is it just me or does the latex formatting for \bordermatrix not work on this site?

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Yes, this is valid. What you are doing boils down to solving the system $$(a_1+b_1x + c_1x^2) + 2(a_2+b_2x+c_2x^2) + 3(a_3+b_3x+c_3x^2) = 1+5x+4x^2$$ subject to the condition that the solution yield a basis (which in your case is obvious, as they are clearly linearly independent); equating coefficients, this becomes exactly your system: \begin{align*} a_1 + 2a_2 + 3a_3 &= 1\\ b_1 + 2b_2 + 3b_3 &= 5\\ c_1 + 2c_2 + 3c_3 &= 4. \end{align*} Since this is a system of three equations with $9$ unknowns, there will generally be many solutions (though not all are necessarily linearly independent).
Yes, that works fine. But notice that you could have solved it more simply by starting with a standard basis $\rm\ \{a,\ b\:x,\ c\:x^2\}\:.\:$ Then $\rm\ a + 2\:b\: x + 3\:c\: x^2\ =\ 2 + 5\: x + 4\: x^2\$ yields your solution.