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I know these vectors $\alpha_1=(1,1,0,0),\alpha_2=(0,0,1,1),\alpha_3=(1,0,0,4),\alpha_4=(0,0,0,2)$ form a basis for $R^4$ but to find the coordinates of each of the standard basis vectors in the ordered basis $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}$ I have done this much. In terms of the standard basis vectors $\alpha_1=c_1e_1+c_2e_2+c_3e_3+c_4e_4$ i.e. $$c_1\begin{pmatrix}1\\ 0 \\ 0 \\ 0 \end{pmatrix}+c_2\begin{pmatrix}0\\ 1 \\ 0 \\ 0 \end{pmatrix}+c_3\begin{pmatrix}0\\ 0 \\ 1 \\ 0 \end{pmatrix}+c_4\begin{pmatrix}0\\ 0 \\ 0 \\ 1 \end{pmatrix}=\begin{pmatrix}1\\ 1 \\ 0 \\ 0 \end{pmatrix}$$ gives $$\begin{pmatrix}c_1\\ c_2\\ c_3 \\ c_4 \end{pmatrix}= \begin{pmatrix}1\\ 1 \\ 0 \\ 0 \end{pmatrix}$$ similarly $\alpha_2=a_1e_1+a_2e_2+a_3e_3+a_4e_4$

gives $$\begin{pmatrix}a_1\\ a_2\\ a_3 \\ a_4 \end{pmatrix}= \begin{pmatrix}0\\ 0 \\ 1 \\ 1 \end{pmatrix}$$ thus for $\alpha_3=b_1e_1+b_2e_2+b_3e_3+b_4e_4$ and $\alpha_4=d_1e_1+d_2e_2+d_3e_3+d_4e_4$

I got $$\begin{pmatrix}b_1\\ b_2\\ b_3 \\ b_4 \end{pmatrix}= \begin{pmatrix}1\\ 0 \\ 0 \\ 4 \end{pmatrix} \qquad \begin{pmatrix}d_1\\ d_2\\ d_3 \\ d_4 \end{pmatrix}= \begin{pmatrix}0\\ 0 \\ 0 \\ 2 \end{pmatrix}$$

But I don't know whether I am right or not. Is my procedure correct? please help me

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So you would take elements in the standard basis and write them as a linear combination of your $\alpha_{i}$ vectors. So $(1, 0, 0, 0) = c_{1}\alpha_{1} + c_{2} \alpha_{2} + c_{3}\alpha_{3} + c_{4} \alpha_{4}$. The vector $(c_{1}, c_{2}, c_{3}, c_{4})$ represents $(1, 0, 0, 0)$ in terms of your basis on your $\alpha$ vectors.

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    $\begingroup$ No. The coordinates of $(1, 0, 0, 0)$ are $(c_{1}, c_{2}, c_{3}, c_{4})$. $\endgroup$ – ml0105 Mar 17 '14 at 5:46
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$$X=SX_B$$ S=$[\alpha_1,\alpha_2,\alpha_3,\alpha_4]$
X=$[e_1,e_2,e_3,e_4]$, e denote standard basis.
from above relation $X_B$ which can be found are required coordinates

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  • $\begingroup$ If you take S[a1,a2,a3,a4] and augment X[e1,e2,e3,e4] to it, then using Gauss Elimination you will easily get required. $\endgroup$ – ketan Mar 17 '14 at 6:05
  • $\begingroup$ is (0,0,1,-2) the answer? I have to be conform that my work is wheter right, so don't take me wrong way $\endgroup$ – mahavir Mar 17 '14 at 6:08

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