Vector not in span$left(Sright)$ implies $Scup{vec{X}}$ be linearly independent
$begingroup$
Theorem:
Let S be a linearly independent set in a vector space. If $vec{X} in V$ and $vec{X} in span left(Sright)$ then $S cup {vec{X}}$ is linearly independent.
Proof:
It suffices to show, by contra positivity, that $vec{X}$ $in$ $spanleft(Sright)$.
Let $S = {vec{v}_{i}}_{i=1}^{n}$
$vec{0} = c_{X}vec{X} + sum_{i=1}^{n}c_{i}vec{v}_{i}$ with coefficients $c_{i} = c_{X} = 0, forall i in mathbb{Z}_{1}^{n}$.
Rearranging:
$vec{X} = frac{1}{c_{X}} sum_{i=1}^{n}c_{i}vec{v}_{i}$
Here, I am unable to move forward. The reciprocal of $c_{X}$ gives an undefined value.
Any help to move this proof forward is appreciated.
Thanks in advance.
linear-algebra proof-verification
asked Nov 24 '18 at 10:52
MathematicingMathematicing
2,44321854
$endgroup$
add a comment |
$begingroup$
Theorem:
Let S be a linearly independent set in a vector space. If $vec{X} in V$ and $vec{X} in span left(Sright)$ then $S cup {vec{X}}$ is linearly independent.
Proof:
It suffices to show, by contra positivity, that $vec{X}$ $in$ $spanleft(Sright)$.
Let $S = {vec{v}_{i}}_{i=1}^{n}$
$vec{0} = c_{X}vec{X} + sum_{i=1}^{n}c_{i}vec{v}_{i}$ with coefficients $c_{i} = c_{X} = 0, forall i in mathbb{Z}_{1}^{n}$.
Rearranging:
$vec{X} = frac{1}{c_{X}} sum_{i=1}^{n}c_{i}vec{v}_{i}$
Here, I am unable to move forward. The reciprocal of $c_{X}$ gives an undefined value.
Any help to move this proof forward is appreciated.
Thanks in advance.
linear-algebra proof-verification
asked Nov 24 '18 at 10:52
MathematicingMathematicing
2,44321854
$endgroup$
$begingroup$
You stated the theorem incorrectly: It is supposed to have $vec{X} notin Span(S)$, instead of $vec{X} in Span(S)$. The theorem as-stated is false (for example take $vec{X}=vec{0} in Span(S)$).
$endgroup$
– Michael
Nov 24 '18 at 10:57
1
$begingroup$
Your sentence $vec{0} = c_Xvec{X} + sum_{i=1}^n c_i vec{v}_i$ "with coefficients $c_i=c_X=0$ for all $i$" does not make sense (particularly the part in quotes) since if you define those coefficients to be zero you get $vec{0}=vec{0}$ and the equation is useless to work with. That is not the definition of linear independence...you need to show that if the equation holds, then all coefficients must be zero (you are not supposed to assume they are zero, you must prove they are zero).
$endgroup$
– Michael
Nov 24 '18 at 11:04
add a comment |
$begingroup$
Theorem:
Let S be a linearly independent set in a vector space. If $vec{X} in V$ and $vec{X} in span left(Sright)$ then $S cup {vec{X}}$ is linearly independent.
Proof:
It suffices to show, by contra positivity, that $vec{X}$ $in$ $spanleft(Sright)$.
Let $S = {vec{v}_{i}}_{i=1}^{n}$
$vec{0} = c_{X}vec{X} + sum_{i=1}^{n}c_{i}vec{v}_{i}$ with coefficients $c_{i} = c_{X} = 0, forall i in mathbb{Z}_{1}^{n}$.
Rearranging:
$vec{X} = frac{1}{c_{X}} sum_{i=1}^{n}c_{i}vec{v}_{i}$
Here, I am unable to move forward. The reciprocal of $c_{X}$ gives an undefined value.
Any help to move this proof forward is appreciated.
Thanks in advance.
linear-algebra proof-verification
asked Nov 24 '18 at 10:52
MathematicingMathematicing
2,44321854
$endgroup$
Theorem:
Let S be a linearly independent set in a vector space. If $vec{X} in V$ and $vec{X} in span left(Sright)$ then $S cup {vec{X}}$ is linearly independent.
Proof:
It suffices to show, by contra positivity, that $vec{X}$ $in$ $spanleft(Sright)$.
Let $S = {vec{v}_{i}}_{i=1}^{n}$
$vec{0} = c_{X}vec{X} + sum_{i=1}^{n}c_{i}vec{v}_{i}$ with coefficients $c_{i} = c_{X} = 0, forall i in mathbb{Z}_{1}^{n}$.
Rearranging:
$vec{X} = frac{1}{c_{X}} sum_{i=1}^{n}c_{i}vec{v}_{i}$
Here, I am unable to move forward. The reciprocal of $c_{X}$ gives an undefined value.
Any help to move this proof forward is appreciated.
Thanks in advance.
linear-algebra proof-verification
linear-algebra proof-verification
asked Nov 24 '18 at 10:52
MathematicingMathematicing
2,44321854
asked Nov 24 '18 at 10:52
MathematicingMathematicing
2,44321854
asked Nov 24 '18 at 10:52
MathematicingMathematicing
2,44321854
asked Nov 24 '18 at 10:52
MathematicingMathematicing
2,44321854
asked Nov 24 '18 at 10:52
MathematicingMathematicing
2,44321854
2,44321854
$begingroup$
You stated the theorem incorrectly: It is supposed to have $vec{X} notin Span(S)$, instead of $vec{X} in Span(S)$. The theorem as-stated is false (for example take $vec{X}=vec{0} in Span(S)$).
$endgroup$
– Michael
Nov 24 '18 at 10:57
1
$begingroup$
Your sentence $vec{0} = c_Xvec{X} + sum_{i=1}^n c_i vec{v}_i$ "with coefficients $c_i=c_X=0$ for all $i$" does not make sense (particularly the part in quotes) since if you define those coefficients to be zero you get $vec{0}=vec{0}$ and the equation is useless to work with. That is not the definition of linear independence...you need to show that if the equation holds, then all coefficients must be zero (you are not supposed to assume they are zero, you must prove they are zero).
$endgroup$
– Michael
Nov 24 '18 at 11:04
add a comment |
$begingroup$
You stated the theorem incorrectly: It is supposed to have $vec{X} notin Span(S)$, instead of $vec{X} in Span(S)$. The theorem as-stated is false (for example take $vec{X}=vec{0} in Span(S)$).
$endgroup$
– Michael
Nov 24 '18 at 10:57
1
$begingroup$
Your sentence $vec{0} = c_Xvec{X} + sum_{i=1}^n c_i vec{v}_i$ "with coefficients $c_i=c_X=0$ for all $i$" does not make sense (particularly the part in quotes) since if you define those coefficients to be zero you get $vec{0}=vec{0}$ and the equation is useless to work with. That is not the definition of linear independence...you need to show that if the equation holds, then all coefficients must be zero (you are not supposed to assume they are zero, you must prove they are zero).
$endgroup$
– Michael
Nov 24 '18 at 11:04
$begingroup$
You stated the theorem incorrectly: It is supposed to have $vec{X} notin Span(S)$, instead of $vec{X} in Span(S)$. The theorem as-stated is false (for example take $vec{X}=vec{0} in Span(S)$).
$endgroup$
– Michael
Nov 24 '18 at 10:57
$begingroup$
You stated the theorem incorrectly: It is supposed to have $vec{X} notin Span(S)$, instead of $vec{X} in Span(S)$. The theorem as-stated is false (for example take $vec{X}=vec{0} in Span(S)$).
$endgroup$
– Michael
Nov 24 '18 at 10:57
1
1
$begingroup$
Your sentence $vec{0} = c_Xvec{X} + sum_{i=1}^n c_i vec{v}_i$ "with coefficients $c_i=c_X=0$ for all $i$" does not make sense (particularly the part in quotes) since if you define those coefficients to be zero you get $vec{0}=vec{0}$ and the equation is useless to work with. That is not the definition of linear independence...you need to show that if the equation holds, then all coefficients must be zero (you are not supposed to assume they are zero, you must prove they are zero).
$endgroup$
– Michael
Nov 24 '18 at 11:04
$begingroup$
Your sentence $vec{0} = c_Xvec{X} + sum_{i=1}^n c_i vec{v}_i$ "with coefficients $c_i=c_X=0$ for all $i$" does not make sense (particularly the part in quotes) since if you define those coefficients to be zero you get $vec{0}=vec{0}$ and the equation is useless to work with. That is not the definition of linear independence...you need to show that if the equation holds, then all coefficients must be zero (you are not supposed to assume they are zero, you must prove they are zero).
$endgroup$
– Michael
Nov 24 '18 at 11:04
add a comment |
1 Answer
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$begingroup$
You want $vec{x} notin operatorname{span}(S)$, as the assumtion of course (as you do have in the title).
In that case we can show that $S cup {vec{x}}$ is lin. independent:
Suppose no, then there are vectors $vec{s_1},dots vec{s_n} in S$ and coefficients $c_1,ldots c_n, c$, not all $0$, such that
$$sum_{i=1}^n c_i vec{s_i} + cvec{x} = 0$$
If $c =0$ we have an immediate contradiction with the fact that $S$ is linearly independent, because not all the $c_i$ are $0$.
So we must have $c neq 0$ and so $frac{1}{c}$ is a well-defined scalar.
But then $$vec{x} = sum_{i=1}^n frac{1}{c}(-c_i) s_i$$ and we'd have that $vec{x} in operatorname{span}(S)$ contrary to the (corrected) assumption.
answered Nov 24 '18 at 11:06
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
$begingroup$
I made a mistake in the theorem. The vector X is not supposed to be in the span of the set S.
$endgroup$
– Mathematicing
Nov 24 '18 at 14:11
$begingroup$
@Mathematicing I worked from the assumption that $vec{x}$ indeed is not in the span of $S$.
$endgroup$
– Henno Brandsma
Nov 24 '18 at 14:13
add a comment |
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$begingroup$
You want $vec{x} notin operatorname{span}(S)$, as the assumtion of course (as you do have in the title).
In that case we can show that $S cup {vec{x}}$ is lin. independent:
Suppose no, then there are vectors $vec{s_1},dots vec{s_n} in S$ and coefficients $c_1,ldots c_n, c$, not all $0$, such that
$$sum_{i=1}^n c_i vec{s_i} + cvec{x} = 0$$
If $c =0$ we have an immediate contradiction with the fact that $S$ is linearly independent, because not all the $c_i$ are $0$.
So we must have $c neq 0$ and so $frac{1}{c}$ is a well-defined scalar.
But then $$vec{x} = sum_{i=1}^n frac{1}{c}(-c_i) s_i$$ and we'd have that $vec{x} in operatorname{span}(S)$ contrary to the (corrected) assumption.
answered Nov 24 '18 at 11:06
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
$begingroup$
I made a mistake in the theorem. The vector X is not supposed to be in the span of the set S.
$endgroup$
– Mathematicing
Nov 24 '18 at 14:11
$begingroup$
@Mathematicing I worked from the assumption that $vec{x}$ indeed is not in the span of $S$.
$endgroup$
– Henno Brandsma
Nov 24 '18 at 14:13
add a comment |
$begingroup$
You want $vec{x} notin operatorname{span}(S)$, as the assumtion of course (as you do have in the title).
In that case we can show that $S cup {vec{x}}$ is lin. independent:
Suppose no, then there are vectors $vec{s_1},dots vec{s_n} in S$ and coefficients $c_1,ldots c_n, c$, not all $0$, such that
$$sum_{i=1}^n c_i vec{s_i} + cvec{x} = 0$$
If $c =0$ we have an immediate contradiction with the fact that $S$ is linearly independent, because not all the $c_i$ are $0$.
So we must have $c neq 0$ and so $frac{1}{c}$ is a well-defined scalar.
But then $$vec{x} = sum_{i=1}^n frac{1}{c}(-c_i) s_i$$ and we'd have that $vec{x} in operatorname{span}(S)$ contrary to the (corrected) assumption.
answered Nov 24 '18 at 11:06
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
$begingroup$
I made a mistake in the theorem. The vector X is not supposed to be in the span of the set S.
$endgroup$
– Mathematicing
Nov 24 '18 at 14:11
$begingroup$
@Mathematicing I worked from the assumption that $vec{x}$ indeed is not in the span of $S$.
$endgroup$
– Henno Brandsma
Nov 24 '18 at 14:13
add a comment |
$begingroup$
You want $vec{x} notin operatorname{span}(S)$, as the assumtion of course (as you do have in the title).
In that case we can show that $S cup {vec{x}}$ is lin. independent:
Suppose no, then there are vectors $vec{s_1},dots vec{s_n} in S$ and coefficients $c_1,ldots c_n, c$, not all $0$, such that
$$sum_{i=1}^n c_i vec{s_i} + cvec{x} = 0$$
If $c =0$ we have an immediate contradiction with the fact that $S$ is linearly independent, because not all the $c_i$ are $0$.
So we must have $c neq 0$ and so $frac{1}{c}$ is a well-defined scalar.
But then $$vec{x} = sum_{i=1}^n frac{1}{c}(-c_i) s_i$$ and we'd have that $vec{x} in operatorname{span}(S)$ contrary to the (corrected) assumption.
answered Nov 24 '18 at 11:06
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
You want $vec{x} notin operatorname{span}(S)$, as the assumtion of course (as you do have in the title).
In that case we can show that $S cup {vec{x}}$ is lin. independent:
Suppose no, then there are vectors $vec{s_1},dots vec{s_n} in S$ and coefficients $c_1,ldots c_n, c$, not all $0$, such that
$$sum_{i=1}^n c_i vec{s_i} + cvec{x} = 0$$
If $c =0$ we have an immediate contradiction with the fact that $S$ is linearly independent, because not all the $c_i$ are $0$.
So we must have $c neq 0$ and so $frac{1}{c}$ is a well-defined scalar.
But then $$vec{x} = sum_{i=1}^n frac{1}{c}(-c_i) s_i$$ and we'd have that $vec{x} in operatorname{span}(S)$ contrary to the (corrected) assumption.
answered Nov 24 '18 at 11:06
Henno BrandsmaHenno Brandsma
106k347114
answered Nov 24 '18 at 11:06
Henno BrandsmaHenno Brandsma
106k347114
answered Nov 24 '18 at 11:06
Henno BrandsmaHenno Brandsma
106k347114
answered Nov 24 '18 at 11:06
Henno BrandsmaHenno Brandsma
106k347114
106k347114
$begingroup$
I made a mistake in the theorem. The vector X is not supposed to be in the span of the set S.
$endgroup$
– Mathematicing
Nov 24 '18 at 14:11
$begingroup$
@Mathematicing I worked from the assumption that $vec{x}$ indeed is not in the span of $S$.
$endgroup$
– Henno Brandsma
Nov 24 '18 at 14:13
add a comment |
$begingroup$
I made a mistake in the theorem. The vector X is not supposed to be in the span of the set S.
$endgroup$
– Mathematicing
Nov 24 '18 at 14:11
$begingroup$
@Mathematicing I worked from the assumption that $vec{x}$ indeed is not in the span of $S$.
$endgroup$
– Henno Brandsma
Nov 24 '18 at 14:13
$begingroup$
I made a mistake in the theorem. The vector X is not supposed to be in the span of the set S.
$endgroup$
– Mathematicing
Nov 24 '18 at 14:11
$begingroup$
I made a mistake in the theorem. The vector X is not supposed to be in the span of the set S.
$endgroup$
– Mathematicing
Nov 24 '18 at 14:11
$begingroup$
@Mathematicing I worked from the assumption that $vec{x}$ indeed is not in the span of $S$.
$endgroup$
– Henno Brandsma
Nov 24 '18 at 14:13
$begingroup$
@Mathematicing I worked from the assumption that $vec{x}$ indeed is not in the span of $S$.
$endgroup$
– Henno Brandsma
Nov 24 '18 at 14:13
add a comment |
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$begingroup$
You stated the theorem incorrectly: It is supposed to have $vec{X} notin Span(S)$, instead of $vec{X} in Span(S)$. The theorem as-stated is false (for example take $vec{X}=vec{0} in Span(S)$).
$endgroup$
– Michael
Nov 24 '18 at 10:57
1
$begingroup$
Your sentence $vec{0} = c_Xvec{X} + sum_{i=1}^n c_i vec{v}_i$ "with coefficients $c_i=c_X=0$ for all $i$" does not make sense (particularly the part in quotes) since if you define those coefficients to be zero you get $vec{0}=vec{0}$ and the equation is useless to work with. That is not the definition of linear independence...you need to show that if the equation holds, then all coefficients must be zero (you are not supposed to assume they are zero, you must prove they are zero).
$endgroup$
– Michael
Nov 24 '18 at 11:04