Mixing
vector representation by spanning sets.
$begingroup$
Prove that the representation of vector in terms of linearly dependent spanning is not unique.
I proceed like this
Let X be any vector which can be written as
$X=a_1.b_1+a_2.b_2+....+a_k.b_k$ where ${b_1,b_2,...,b_k}$ is linearly dependent vectors and ${a_1,a_2,...,a_k}$ are scalars not all zero .Now any other vector C can be written as a linear combination of ${b_1,b_2,..,b_k}$ so replacing any of the $b_i$ for which $a_i$ not equal to zero in representation of $X$ by $C$.we get a new representation of $X$.
linear-algebra
edited Jan 26 at 5:30
YuiTo Cheng
2,0542637
asked Jan 25 at 17:47
Daniel Daniel
114
$endgroup$
add a comment |
$begingroup$
Prove that the representation of vector in terms of linearly dependent spanning is not unique.
I proceed like this
Let X be any vector which can be written as
$X=a_1.b_1+a_2.b_2+....+a_k.b_k$ where ${b_1,b_2,...,b_k}$ is linearly dependent vectors and ${a_1,a_2,...,a_k}$ are scalars not all zero .Now any other vector C can be written as a linear combination of ${b_1,b_2,..,b_k}$ so replacing any of the $b_i$ for which $a_i$ not equal to zero in representation of $X$ by $C$.we get a new representation of $X$.
linear-algebra
edited Jan 26 at 5:30
YuiTo Cheng
2,0542637
asked Jan 25 at 17:47
Daniel Daniel
114
$endgroup$
$begingroup$
Welcome Math S.E. The people who answer questions here are passionate about mathematics. We love to help others learn it. But we hate it when people try to use us to avoid having to learn it themselves. For this reason, bare questions like this attract down and close votes. Show us you are trying to learn by editing the question to show what you've attempted and where you are stuck. You will also get much better help when we know where your problem lies.
$endgroup$
– Paul Sinclair
Jan 26 at 3:09
add a comment |
$begingroup$
Prove that the representation of vector in terms of linearly dependent spanning is not unique.
I proceed like this
Let X be any vector which can be written as
$X=a_1.b_1+a_2.b_2+....+a_k.b_k$ where ${b_1,b_2,...,b_k}$ is linearly dependent vectors and ${a_1,a_2,...,a_k}$ are scalars not all zero .Now any other vector C can be written as a linear combination of ${b_1,b_2,..,b_k}$ so replacing any of the $b_i$ for which $a_i$ not equal to zero in representation of $X$ by $C$.we get a new representation of $X$.
linear-algebra
edited Jan 26 at 5:30
YuiTo Cheng
2,0542637
asked Jan 25 at 17:47
Daniel Daniel
114
$endgroup$
Prove that the representation of vector in terms of linearly dependent spanning is not unique.
I proceed like this
Let X be any vector which can be written as
$X=a_1.b_1+a_2.b_2+....+a_k.b_k$ where ${b_1,b_2,...,b_k}$ is linearly dependent vectors and ${a_1,a_2,...,a_k}$ are scalars not all zero .Now any other vector C can be written as a linear combination of ${b_1,b_2,..,b_k}$ so replacing any of the $b_i$ for which $a_i$ not equal to zero in representation of $X$ by $C$.we get a new representation of $X$.
linear-algebra
linear-algebra
edited Jan 26 at 5:30
YuiTo Cheng
2,0542637
asked Jan 25 at 17:47
Daniel Daniel
114
edited Jan 26 at 5:30
YuiTo Cheng
2,0542637
asked Jan 25 at 17:47
Daniel Daniel
114
edited Jan 26 at 5:30
YuiTo Cheng
2,0542637
edited Jan 26 at 5:30
YuiTo Cheng
2,0542637
edited Jan 26 at 5:30
YuiTo Cheng
2,0542637
2,0542637
asked Jan 25 at 17:47
Daniel Daniel
114
asked Jan 25 at 17:47
Daniel Daniel
114
asked Jan 25 at 17:47
Daniel Daniel
114
114
$begingroup$
Welcome Math S.E. The people who answer questions here are passionate about mathematics. We love to help others learn it. But we hate it when people try to use us to avoid having to learn it themselves. For this reason, bare questions like this attract down and close votes. Show us you are trying to learn by editing the question to show what you've attempted and where you are stuck. You will also get much better help when we know where your problem lies.
$endgroup$
– Paul Sinclair
Jan 26 at 3:09
add a comment |
$begingroup$
Welcome Math S.E. The people who answer questions here are passionate about mathematics. We love to help others learn it. But we hate it when people try to use us to avoid having to learn it themselves. For this reason, bare questions like this attract down and close votes. Show us you are trying to learn by editing the question to show what you've attempted and where you are stuck. You will also get much better help when we know where your problem lies.
$endgroup$
– Paul Sinclair
Jan 26 at 3:09
$begingroup$
Welcome Math S.E. The people who answer questions here are passionate about mathematics. We love to help others learn it. But we hate it when people try to use us to avoid having to learn it themselves. For this reason, bare questions like this attract down and close votes. Show us you are trying to learn by editing the question to show what you've attempted and where you are stuck. You will also get much better help when we know where your problem lies.
$endgroup$
– Paul Sinclair
Jan 26 at 3:09
$begingroup$
Welcome Math S.E. The people who answer questions here are passionate about mathematics. We love to help others learn it. But we hate it when people try to use us to avoid having to learn it themselves. For this reason, bare questions like this attract down and close votes. Show us you are trying to learn by editing the question to show what you've attempted and where you are stuck. You will also get much better help when we know where your problem lies.
$endgroup$
– Paul Sinclair
Jan 26 at 3:09
add a comment |
1 Answer
1
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votes
$begingroup$
The answer you've provided to your question is correct. We can write it a little more mathematically, if you want, like this:
let $b_1,b_2,...b_n$ be your linearly dependent basis vectors. Because they are linearly dependent we can pick one of them ($b_1$, renumbering the basis if needed) and write it as a linear combination of the others:
$$ b_1 = sum_{i=2}^n alpha_i b_i $$
where the $alpha_i$ are not all zero for $i=2,ldots ,n$.
Now let $x = sum_{i=1}^n a_icdot b_i$ (using your notation in the question; I would prefer to use a greek letter for the co-efficients to make it obvious a glance what letter is doing what job). Then:
$$ begin{eqnarray}
x & = & sum_{i=1}^n a_i b_i \
& = & a_1 b_1 + sum_{i=2}^n a_i b_i \
& = & a_1 sum_{i=2}^n alpha_i b_i + sum_{i=2}^n a_i b_i \
& = & sum_{i=2}^n left( a_1alpha_i +a_i right) b_i \
& = & 0b_1 + sum_{i=2}^n left( a_1alpha_i +a_i right) b_i
end{eqnarray} $$
and so $x$ has two (different) representations in terms of the basis $b_1, b_2,ldots ,b_n$.
Note that this also holds for Schauder bases, where although there are an infinite number of basis vectors $b_1, b_2, ldots$ any finite linearly independent subset must provide a representation of elements of the space (Banach spaces may have Schauder bases) and for Hamel bases.
answered Jan 26 at 11:32
postmortespostmortes
2,16531222
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The answer you've provided to your question is correct. We can write it a little more mathematically, if you want, like this:
let $b_1,b_2,...b_n$ be your linearly dependent basis vectors. Because they are linearly dependent we can pick one of them ($b_1$, renumbering the basis if needed) and write it as a linear combination of the others:
$$ b_1 = sum_{i=2}^n alpha_i b_i $$
where the $alpha_i$ are not all zero for $i=2,ldots ,n$.
Now let $x = sum_{i=1}^n a_icdot b_i$ (using your notation in the question; I would prefer to use a greek letter for the co-efficients to make it obvious a glance what letter is doing what job). Then:
$$ begin{eqnarray}
x & = & sum_{i=1}^n a_i b_i \
& = & a_1 b_1 + sum_{i=2}^n a_i b_i \
& = & a_1 sum_{i=2}^n alpha_i b_i + sum_{i=2}^n a_i b_i \
& = & sum_{i=2}^n left( a_1alpha_i +a_i right) b_i \
& = & 0b_1 + sum_{i=2}^n left( a_1alpha_i +a_i right) b_i
end{eqnarray} $$
and so $x$ has two (different) representations in terms of the basis $b_1, b_2,ldots ,b_n$.
Note that this also holds for Schauder bases, where although there are an infinite number of basis vectors $b_1, b_2, ldots$ any finite linearly independent subset must provide a representation of elements of the space (Banach spaces may have Schauder bases) and for Hamel bases.
answered Jan 26 at 11:32
postmortespostmortes
2,16531222
$endgroup$
add a comment |
$begingroup$
The answer you've provided to your question is correct. We can write it a little more mathematically, if you want, like this:
let $b_1,b_2,...b_n$ be your linearly dependent basis vectors. Because they are linearly dependent we can pick one of them ($b_1$, renumbering the basis if needed) and write it as a linear combination of the others:
$$ b_1 = sum_{i=2}^n alpha_i b_i $$
where the $alpha_i$ are not all zero for $i=2,ldots ,n$.
Now let $x = sum_{i=1}^n a_icdot b_i$ (using your notation in the question; I would prefer to use a greek letter for the co-efficients to make it obvious a glance what letter is doing what job). Then:
$$ begin{eqnarray}
x & = & sum_{i=1}^n a_i b_i \
& = & a_1 b_1 + sum_{i=2}^n a_i b_i \
& = & a_1 sum_{i=2}^n alpha_i b_i + sum_{i=2}^n a_i b_i \
& = & sum_{i=2}^n left( a_1alpha_i +a_i right) b_i \
& = & 0b_1 + sum_{i=2}^n left( a_1alpha_i +a_i right) b_i
end{eqnarray} $$
and so $x$ has two (different) representations in terms of the basis $b_1, b_2,ldots ,b_n$.
Note that this also holds for Schauder bases, where although there are an infinite number of basis vectors $b_1, b_2, ldots$ any finite linearly independent subset must provide a representation of elements of the space (Banach spaces may have Schauder bases) and for Hamel bases.
answered Jan 26 at 11:32
postmortespostmortes
2,16531222
$endgroup$
add a comment |
$begingroup$
The answer you've provided to your question is correct. We can write it a little more mathematically, if you want, like this:
let $b_1,b_2,...b_n$ be your linearly dependent basis vectors. Because they are linearly dependent we can pick one of them ($b_1$, renumbering the basis if needed) and write it as a linear combination of the others:
$$ b_1 = sum_{i=2}^n alpha_i b_i $$
where the $alpha_i$ are not all zero for $i=2,ldots ,n$.
Now let $x = sum_{i=1}^n a_icdot b_i$ (using your notation in the question; I would prefer to use a greek letter for the co-efficients to make it obvious a glance what letter is doing what job). Then:
$$ begin{eqnarray}
x & = & sum_{i=1}^n a_i b_i \
& = & a_1 b_1 + sum_{i=2}^n a_i b_i \
& = & a_1 sum_{i=2}^n alpha_i b_i + sum_{i=2}^n a_i b_i \
& = & sum_{i=2}^n left( a_1alpha_i +a_i right) b_i \
& = & 0b_1 + sum_{i=2}^n left( a_1alpha_i +a_i right) b_i
end{eqnarray} $$
and so $x$ has two (different) representations in terms of the basis $b_1, b_2,ldots ,b_n$.
Note that this also holds for Schauder bases, where although there are an infinite number of basis vectors $b_1, b_2, ldots$ any finite linearly independent subset must provide a representation of elements of the space (Banach spaces may have Schauder bases) and for Hamel bases.
answered Jan 26 at 11:32
postmortespostmortes
2,16531222
$endgroup$
The answer you've provided to your question is correct. We can write it a little more mathematically, if you want, like this:
let $b_1,b_2,...b_n$ be your linearly dependent basis vectors. Because they are linearly dependent we can pick one of them ($b_1$, renumbering the basis if needed) and write it as a linear combination of the others:
$$ b_1 = sum_{i=2}^n alpha_i b_i $$
where the $alpha_i$ are not all zero for $i=2,ldots ,n$.
Now let $x = sum_{i=1}^n a_icdot b_i$ (using your notation in the question; I would prefer to use a greek letter for the co-efficients to make it obvious a glance what letter is doing what job). Then:
$$ begin{eqnarray}
x & = & sum_{i=1}^n a_i b_i \
& = & a_1 b_1 + sum_{i=2}^n a_i b_i \
& = & a_1 sum_{i=2}^n alpha_i b_i + sum_{i=2}^n a_i b_i \
& = & sum_{i=2}^n left( a_1alpha_i +a_i right) b_i \
& = & 0b_1 + sum_{i=2}^n left( a_1alpha_i +a_i right) b_i
end{eqnarray} $$
and so $x$ has two (different) representations in terms of the basis $b_1, b_2,ldots ,b_n$.
Note that this also holds for Schauder bases, where although there are an infinite number of basis vectors $b_1, b_2, ldots$ any finite linearly independent subset must provide a representation of elements of the space (Banach spaces may have Schauder bases) and for Hamel bases.
answered Jan 26 at 11:32
postmortespostmortes
2,16531222
answered Jan 26 at 11:32
postmortespostmortes
2,16531222
answered Jan 26 at 11:32
postmortespostmortes
2,16531222
answered Jan 26 at 11:32
postmortespostmortes
2,16531222
2,16531222
add a comment |
add a comment |
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$begingroup$
Welcome Math S.E. The people who answer questions here are passionate about mathematics. We love to help others learn it. But we hate it when people try to use us to avoid having to learn it themselves. For this reason, bare questions like this attract down and close votes. Show us you are trying to learn by editing the question to show what you've attempted and where you are stuck. You will also get much better help when we know where your problem lies.
$endgroup$
– Paul Sinclair
Jan 26 at 3:09