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Let $B_1=\{b_1,b_2,b_3\}$ is a basis of real vector space $V$. Show that a set of vectors $B_2=\{c_1,c_2,c_3\}$,such that $c_1=b_1+2b_2+3b_3,c_2=b_1+3b_2+2b_3,c_3=2b_1+3b_2+b_3$ is also a basis of the same vector space $V$. Find coordinates of vector $v=10b_1+15b_2+14b_3$ over basis $B_2$.

What is the procedure for finding coordinates of vector $v=10b_1+15b_2+14b_3$ over basis $B_2$?

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Hint: Just take $\ a,b,c\in{R}$ and solve a system of equations from $\ v=ac_1+bc_2+cc_3$

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  • $\begingroup$ Could you expand your answer? $\endgroup$ – user300046 Apr 24 '17 at 9:20
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    $\begingroup$ $\ 10b_1+15b_2+14b_3=(a+b+2c)b_1+(2a+2b+3c)b_2+(3a+2b+c)b_3 => a+b+2c=10, 2a+2b+3c=15, 3a+2b+c=14 $ $\endgroup$ – CTSnake Apr 24 '17 at 9:32
  • $\begingroup$ For showing that a set of vectors $B_2$ is also a basis of the same vector space $V$, we must show that real number triplet $(x_1,x_2,x_3)$ is equal to $(0,0,0)$, right? From $x_1b_1+x_2b_2+x_3b_3=x_1c_1+x_2c_2+x_3c_3$ we get the system $x_2+2x_3=0,2x_1+2x_2+3x_3=0,3x_1+2x_2=0$ that gives $x_1=x_2=x_3=0$. If $x_1\neq 0$ or $x_2\neq 0$ or $x_3\neq 0$, the statement wouldn't be true, right? $\endgroup$ – user300046 Apr 24 '17 at 12:09
  • $\begingroup$ You can't write this $\ x_1b_1+x_2b_2+x_3b_3=x_1c_1+x_2c_2+x_3c_3$ because in general coordinates of vectors aren't same for all basis. $\endgroup$ – CTSnake Apr 24 '17 at 12:13
  • $\begingroup$ I have described the first half of the problem. Is it correct? $\endgroup$ – user300046 Apr 24 '17 at 12:25

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