Sum of vectors in any linear space
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Suppose that vectors $x_1,x_2,…,x_n$ have the following property: for each $i$ the sum of all vectors except $x_i$ is parallel to $x_i$. If at least two of the vectors $x_1,x_2,dots ,x_n$ are not parallel, what is the value of the sum $x_1+x_2+dots +x_n=$
I know that answer is $x_1+x_2+dots +x_n=0$ but I am not really understand why. So I can make these steps: $s = x_1+x_2+dots +x_n$ and $s-x_1 = x_2 + dots + x_n$. So $s-x_1 midmid x_1$. But what after?
I am think now (after made question) that:
$x_1 + x_2 midmid s - x_1 + s - x_2$
$dots$
$x_1 + dots + x_n midmid s_1 + dots + s_n - x_1 - x_2 - cdots - x_n$
$s midmid s_1 + cdots + s_n - (s) = s_1 + cdots + s_{n-1}$
where $s_i$ -- just sum of $x_1,cdots,x_n$. Am I right?
linear-algebra abstract-algebra analytic-geometry
asked Jan 16 at 9:06
Just do itJust do it
18918
$endgroup$
add a comment |
$begingroup$
Suppose that vectors $x_1,x_2,…,x_n$ have the following property: for each $i$ the sum of all vectors except $x_i$ is parallel to $x_i$. If at least two of the vectors $x_1,x_2,dots ,x_n$ are not parallel, what is the value of the sum $x_1+x_2+dots +x_n=$
I know that answer is $x_1+x_2+dots +x_n=0$ but I am not really understand why. So I can make these steps: $s = x_1+x_2+dots +x_n$ and $s-x_1 = x_2 + dots + x_n$. So $s-x_1 midmid x_1$. But what after?
I am think now (after made question) that:
$x_1 + x_2 midmid s - x_1 + s - x_2$
$dots$
$x_1 + dots + x_n midmid s_1 + dots + s_n - x_1 - x_2 - cdots - x_n$
$s midmid s_1 + cdots + s_n - (s) = s_1 + cdots + s_{n-1}$
where $s_i$ -- just sum of $x_1,cdots,x_n$. Am I right?
linear-algebra abstract-algebra analytic-geometry
asked Jan 16 at 9:06
Just do itJust do it
18918
$endgroup$
$begingroup$
"To be parallel to" is the same as saying "to be a scalar multiple of", but then the claim fails for $;n=2;$ because according to the last condition these two aren't parallel and the first condition is fulfilled in an empty way...
$endgroup$
– DonAntonio
Jan 16 at 9:30
add a comment |
$begingroup$
Suppose that vectors $x_1,x_2,…,x_n$ have the following property: for each $i$ the sum of all vectors except $x_i$ is parallel to $x_i$. If at least two of the vectors $x_1,x_2,dots ,x_n$ are not parallel, what is the value of the sum $x_1+x_2+dots +x_n=$
I know that answer is $x_1+x_2+dots +x_n=0$ but I am not really understand why. So I can make these steps: $s = x_1+x_2+dots +x_n$ and $s-x_1 = x_2 + dots + x_n$. So $s-x_1 midmid x_1$. But what after?
I am think now (after made question) that:
$x_1 + x_2 midmid s - x_1 + s - x_2$
$dots$
$x_1 + dots + x_n midmid s_1 + dots + s_n - x_1 - x_2 - cdots - x_n$
$s midmid s_1 + cdots + s_n - (s) = s_1 + cdots + s_{n-1}$
where $s_i$ -- just sum of $x_1,cdots,x_n$. Am I right?
linear-algebra abstract-algebra analytic-geometry
asked Jan 16 at 9:06
Just do itJust do it
18918
$endgroup$
Suppose that vectors $x_1,x_2,…,x_n$ have the following property: for each $i$ the sum of all vectors except $x_i$ is parallel to $x_i$. If at least two of the vectors $x_1,x_2,dots ,x_n$ are not parallel, what is the value of the sum $x_1+x_2+dots +x_n=$
I know that answer is $x_1+x_2+dots +x_n=0$ but I am not really understand why. So I can make these steps: $s = x_1+x_2+dots +x_n$ and $s-x_1 = x_2 + dots + x_n$. So $s-x_1 midmid x_1$. But what after?
I am think now (after made question) that:
$x_1 + x_2 midmid s - x_1 + s - x_2$
$dots$
$x_1 + dots + x_n midmid s_1 + dots + s_n - x_1 - x_2 - cdots - x_n$
$s midmid s_1 + cdots + s_n - (s) = s_1 + cdots + s_{n-1}$
where $s_i$ -- just sum of $x_1,cdots,x_n$. Am I right?
linear-algebra abstract-algebra analytic-geometry
linear-algebra abstract-algebra analytic-geometry
asked Jan 16 at 9:06
Just do itJust do it
18918
asked Jan 16 at 9:06
Just do itJust do it
18918
edited Jan 16 at 9:24
Just do it
asked Jan 16 at 9:06
Just do itJust do it
18918
asked Jan 16 at 9:06
Just do itJust do it
18918
asked Jan 16 at 9:06
Just do itJust do it
18918
18918
$begingroup$
"To be parallel to" is the same as saying "to be a scalar multiple of", but then the claim fails for $;n=2;$ because according to the last condition these two aren't parallel and the first condition is fulfilled in an empty way...
$endgroup$
– DonAntonio
Jan 16 at 9:30
add a comment |
$begingroup$
"To be parallel to" is the same as saying "to be a scalar multiple of", but then the claim fails for $;n=2;$ because according to the last condition these two aren't parallel and the first condition is fulfilled in an empty way...
$endgroup$
– DonAntonio
Jan 16 at 9:30
$begingroup$
"To be parallel to" is the same as saying "to be a scalar multiple of", but then the claim fails for $;n=2;$ because according to the last condition these two aren't parallel and the first condition is fulfilled in an empty way...
$endgroup$
– DonAntonio
Jan 16 at 9:30
$begingroup$
"To be parallel to" is the same as saying "to be a scalar multiple of", but then the claim fails for $;n=2;$ because according to the last condition these two aren't parallel and the first condition is fulfilled in an empty way...
$endgroup$
– DonAntonio
Jan 16 at 9:30
add a comment |
1 Answer
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The fact that $s-x_kVert x_k$ gives that $exists lambda_kinmathbb{R}$ such that $s-x_k=lambda_k x_kiff s=(1+lambda_k)x_k$.
Assume that $x_1notVert x_2$. From the previous statements it follows that $s=(1+lambda_1)x_1=(1+lambda_2)x_2Rightarrow s=0.$
answered Jan 16 at 9:31
IuliaIulia
1,205414
$endgroup$
$begingroup$
I got it, thank you)
$endgroup$
– Just do it
Jan 16 at 9:40
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The fact that $s-x_kVert x_k$ gives that $exists lambda_kinmathbb{R}$ such that $s-x_k=lambda_k x_kiff s=(1+lambda_k)x_k$.
Assume that $x_1notVert x_2$. From the previous statements it follows that $s=(1+lambda_1)x_1=(1+lambda_2)x_2Rightarrow s=0.$
answered Jan 16 at 9:31
IuliaIulia
1,205414
$endgroup$
$begingroup$
I got it, thank you)
$endgroup$
– Just do it
Jan 16 at 9:40
add a comment |
$begingroup$
The fact that $s-x_kVert x_k$ gives that $exists lambda_kinmathbb{R}$ such that $s-x_k=lambda_k x_kiff s=(1+lambda_k)x_k$.
Assume that $x_1notVert x_2$. From the previous statements it follows that $s=(1+lambda_1)x_1=(1+lambda_2)x_2Rightarrow s=0.$
answered Jan 16 at 9:31
IuliaIulia
1,205414
$endgroup$
$begingroup$
I got it, thank you)
$endgroup$
– Just do it
Jan 16 at 9:40
add a comment |
$begingroup$
The fact that $s-x_kVert x_k$ gives that $exists lambda_kinmathbb{R}$ such that $s-x_k=lambda_k x_kiff s=(1+lambda_k)x_k$.
Assume that $x_1notVert x_2$. From the previous statements it follows that $s=(1+lambda_1)x_1=(1+lambda_2)x_2Rightarrow s=0.$
answered Jan 16 at 9:31
IuliaIulia
1,205414
$endgroup$
The fact that $s-x_kVert x_k$ gives that $exists lambda_kinmathbb{R}$ such that $s-x_k=lambda_k x_kiff s=(1+lambda_k)x_k$.
Assume that $x_1notVert x_2$. From the previous statements it follows that $s=(1+lambda_1)x_1=(1+lambda_2)x_2Rightarrow s=0.$
answered Jan 16 at 9:31
IuliaIulia
1,205414
answered Jan 16 at 9:31
IuliaIulia
1,205414
answered Jan 16 at 9:31
IuliaIulia
1,205414
answered Jan 16 at 9:31
IuliaIulia
1,205414
1,205414
$begingroup$
I got it, thank you)
$endgroup$
– Just do it
Jan 16 at 9:40
add a comment |
$begingroup$
I got it, thank you)
$endgroup$
– Just do it
Jan 16 at 9:40
$begingroup$
I got it, thank you)
$endgroup$
– Just do it
Jan 16 at 9:40
$begingroup$
I got it, thank you)
$endgroup$
– Just do it
Jan 16 at 9:40
add a comment |
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$begingroup$
"To be parallel to" is the same as saying "to be a scalar multiple of", but then the claim fails for $;n=2;$ because according to the last condition these two aren't parallel and the first condition is fulfilled in an empty way...
$endgroup$
– DonAntonio
Jan 16 at 9:30