Unclear step when proving linear independence












0












$begingroup$


I don't understand how we choose the x-values to solve the equation system proving linear independence. For example I have this question:



enter image description here



Proving the linear dependence is trivial, but then this follows:



enter image description here



So I underand the first part, but how do we choose the three x values, namely x = 0, x = -1 and x = 1?






linear-algebra vector-spaces independence







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$endgroup$








  • 2




    $begingroup$
    We can chose any three different values for $x$.
    $endgroup$
    – Emilio Novati
    Dec 7 '18 at 16:09










  • $begingroup$
    So why do we sometimes choose 2 or other values? (at least in my exercises)
    $endgroup$
    – ToTom
    Dec 7 '18 at 16:11
















0












$begingroup$


I don't understand how we choose the x-values to solve the equation system proving linear independence. For example I have this question:



enter image description here



Proving the linear dependence is trivial, but then this follows:



enter image description here



So I underand the first part, but how do we choose the three x values, namely x = 0, x = -1 and x = 1?






linear-algebra vector-spaces independence







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    We can chose any three different values for $x$.
    $endgroup$
    – Emilio Novati
    Dec 7 '18 at 16:09










  • $begingroup$
    So why do we sometimes choose 2 or other values? (at least in my exercises)
    $endgroup$
    – ToTom
    Dec 7 '18 at 16:11














0












0








0





$begingroup$


I don't understand how we choose the x-values to solve the equation system proving linear independence. For example I have this question:



enter image description here



Proving the linear dependence is trivial, but then this follows:



enter image description here



So I underand the first part, but how do we choose the three x values, namely x = 0, x = -1 and x = 1?






linear-algebra vector-spaces independence







share|cite|improve this question









$endgroup$




I don't understand how we choose the x-values to solve the equation system proving linear independence. For example I have this question:



enter image description here



Proving the linear dependence is trivial, but then this follows:



enter image description here



So I underand the first part, but how do we choose the three x values, namely x = 0, x = -1 and x = 1?






linear-algebra vector-spaces independence




linear-algebra vector-spaces independence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 '18 at 16:01









ToTomToTom

526




526








  • 2




    $begingroup$
    We can chose any three different values for $x$.
    $endgroup$
    – Emilio Novati
    Dec 7 '18 at 16:09










  • $begingroup$
    So why do we sometimes choose 2 or other values? (at least in my exercises)
    $endgroup$
    – ToTom
    Dec 7 '18 at 16:11














  • 2




    $begingroup$
    We can chose any three different values for $x$.
    $endgroup$
    – Emilio Novati
    Dec 7 '18 at 16:09










  • $begingroup$
    So why do we sometimes choose 2 or other values? (at least in my exercises)
    $endgroup$
    – ToTom
    Dec 7 '18 at 16:11








2




2




$begingroup$
We can chose any three different values for $x$.
$endgroup$
– Emilio Novati
Dec 7 '18 at 16:09




$begingroup$
We can chose any three different values for $x$.
$endgroup$
– Emilio Novati
Dec 7 '18 at 16:09












$begingroup$
So why do we sometimes choose 2 or other values? (at least in my exercises)
$endgroup$
– ToTom
Dec 7 '18 at 16:11




$begingroup$
So why do we sometimes choose 2 or other values? (at least in my exercises)
$endgroup$
– ToTom
Dec 7 '18 at 16:11










1 Answer
1






active

oldest

votes


















1












$begingroup$

You can choose arbitrary values for $x$. It's purely for simplification purposes. In this case choosing $x=0$ in $alpha_1+alpha_2cdot 0+alpha_2 cdot (1+0+0^2) = 0$ simplifies to $alpha_1+alpha_2=0$. The values are usually chosen as a solution to one of the polynomials. In this case $x$ and $x^2+x+1$. The second polynomial doesn't have real solutions so you can't pick a nice number there instead they use $x=1$ and $x=-1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay, so even for a different exercise, here we could also have chosen -3, 5 and 8? imgur.com/a/l15TkgH
    $endgroup$
    – ToTom
    Dec 7 '18 at 16:12








  • 1




    $begingroup$
    As long as it's within the domain of these functions then yes.
    $endgroup$
    – DreaDk
    Dec 7 '18 at 16:14










  • $begingroup$
    Thank you a lot
    $endgroup$
    – ToTom
    Dec 7 '18 at 16:16











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1 Answer
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1 Answer
1






active

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active

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1












$begingroup$

You can choose arbitrary values for $x$. It's purely for simplification purposes. In this case choosing $x=0$ in $alpha_1+alpha_2cdot 0+alpha_2 cdot (1+0+0^2) = 0$ simplifies to $alpha_1+alpha_2=0$. The values are usually chosen as a solution to one of the polynomials. In this case $x$ and $x^2+x+1$. The second polynomial doesn't have real solutions so you can't pick a nice number there instead they use $x=1$ and $x=-1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay, so even for a different exercise, here we could also have chosen -3, 5 and 8? imgur.com/a/l15TkgH
    $endgroup$
    – ToTom
    Dec 7 '18 at 16:12








  • 1




    $begingroup$
    As long as it's within the domain of these functions then yes.
    $endgroup$
    – DreaDk
    Dec 7 '18 at 16:14










  • $begingroup$
    Thank you a lot
    $endgroup$
    – ToTom
    Dec 7 '18 at 16:16
















1












$begingroup$

You can choose arbitrary values for $x$. It's purely for simplification purposes. In this case choosing $x=0$ in $alpha_1+alpha_2cdot 0+alpha_2 cdot (1+0+0^2) = 0$ simplifies to $alpha_1+alpha_2=0$. The values are usually chosen as a solution to one of the polynomials. In this case $x$ and $x^2+x+1$. The second polynomial doesn't have real solutions so you can't pick a nice number there instead they use $x=1$ and $x=-1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay, so even for a different exercise, here we could also have chosen -3, 5 and 8? imgur.com/a/l15TkgH
    $endgroup$
    – ToTom
    Dec 7 '18 at 16:12








  • 1




    $begingroup$
    As long as it's within the domain of these functions then yes.
    $endgroup$
    – DreaDk
    Dec 7 '18 at 16:14










  • $begingroup$
    Thank you a lot
    $endgroup$
    – ToTom
    Dec 7 '18 at 16:16














1












1








1





$begingroup$

You can choose arbitrary values for $x$. It's purely for simplification purposes. In this case choosing $x=0$ in $alpha_1+alpha_2cdot 0+alpha_2 cdot (1+0+0^2) = 0$ simplifies to $alpha_1+alpha_2=0$. The values are usually chosen as a solution to one of the polynomials. In this case $x$ and $x^2+x+1$. The second polynomial doesn't have real solutions so you can't pick a nice number there instead they use $x=1$ and $x=-1$.






share|cite|improve this answer









$endgroup$



You can choose arbitrary values for $x$. It's purely for simplification purposes. In this case choosing $x=0$ in $alpha_1+alpha_2cdot 0+alpha_2 cdot (1+0+0^2) = 0$ simplifies to $alpha_1+alpha_2=0$. The values are usually chosen as a solution to one of the polynomials. In this case $x$ and $x^2+x+1$. The second polynomial doesn't have real solutions so you can't pick a nice number there instead they use $x=1$ and $x=-1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 16:10









DreaDkDreaDk

6461318




6461318












  • $begingroup$
    Okay, so even for a different exercise, here we could also have chosen -3, 5 and 8? imgur.com/a/l15TkgH
    $endgroup$
    – ToTom
    Dec 7 '18 at 16:12








  • 1




    $begingroup$
    As long as it's within the domain of these functions then yes.
    $endgroup$
    – DreaDk
    Dec 7 '18 at 16:14










  • $begingroup$
    Thank you a lot
    $endgroup$
    – ToTom
    Dec 7 '18 at 16:16


















  • $begingroup$
    Okay, so even for a different exercise, here we could also have chosen -3, 5 and 8? imgur.com/a/l15TkgH
    $endgroup$
    – ToTom
    Dec 7 '18 at 16:12








  • 1




    $begingroup$
    As long as it's within the domain of these functions then yes.
    $endgroup$
    – DreaDk
    Dec 7 '18 at 16:14










  • $begingroup$
    Thank you a lot
    $endgroup$
    – ToTom
    Dec 7 '18 at 16:16
















$begingroup$
Okay, so even for a different exercise, here we could also have chosen -3, 5 and 8? imgur.com/a/l15TkgH
$endgroup$
– ToTom
Dec 7 '18 at 16:12






$begingroup$
Okay, so even for a different exercise, here we could also have chosen -3, 5 and 8? imgur.com/a/l15TkgH
$endgroup$
– ToTom
Dec 7 '18 at 16:12






1




1




$begingroup$
As long as it's within the domain of these functions then yes.
$endgroup$
– DreaDk
Dec 7 '18 at 16:14




$begingroup$
As long as it's within the domain of these functions then yes.
$endgroup$
– DreaDk
Dec 7 '18 at 16:14












$begingroup$
Thank you a lot
$endgroup$
– ToTom
Dec 7 '18 at 16:16




$begingroup$
Thank you a lot
$endgroup$
– ToTom
Dec 7 '18 at 16:16


















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