Unclear step when proving linear independence
$begingroup$
I don't understand how we choose the x-values to solve the equation system proving linear independence. For example I have this question:
Proving the linear dependence is trivial, but then this follows:
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
asked Dec 7 '18 at 16:01
ToTomToTom
526
$endgroup$
add a comment |
$begingroup$
I don't understand how we choose the x-values to solve the equation system proving linear independence. For example I have this question:
Proving the linear dependence is trivial, but then this follows:
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
asked Dec 7 '18 at 16:01
ToTomToTom
526
$endgroup$
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
add a comment |
$begingroup$
I don't understand how we choose the x-values to solve the equation system proving linear independence. For example I have this question:
Proving the linear dependence is trivial, but then this follows:
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
asked Dec 7 '18 at 16:01
ToTomToTom
526
$endgroup$
I don't understand how we choose the x-values to solve the equation system proving linear independence. For example I have this question:
Proving the linear dependence is trivial, but then this follows:
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
asked Dec 7 '18 at 16:01
ToTomToTom
526
asked Dec 7 '18 at 16:01
ToTomToTom
526
asked Dec 7 '18 at 16:01
ToTomToTom
526
asked Dec 7 '18 at 16:01
ToTomToTom
526
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $x=-1$.
answered Dec 7 '18 at 16:10
DreaDkDreaDk
6461318
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
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 $x=-1$.
answered Dec 7 '18 at 16:10
DreaDkDreaDk
6461318
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $x=-1$.
answered Dec 7 '18 at 16:10
DreaDkDreaDk
6461318
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $x=-1$.
answered Dec 7 '18 at 16:10
DreaDkDreaDk
6461318
$endgroup$
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 $x=-1$.
answered Dec 7 '18 at 16:10
DreaDkDreaDk
6461318
answered Dec 7 '18 at 16:10
DreaDkDreaDk
6461318
answered Dec 7 '18 at 16:10
DreaDkDreaDk
6461318
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
add a comment |
$begingroup$
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
add a comment |
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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