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Choice of vectors from basis in Gram-Schmidt process
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Say I have a basis for $mathbb{c}^{2}$ composed of the vectors $(1,1), (4i,2i )$ with complex inner product. When I construct my orthogonal basis using the Gram-Schmidt process how do I make a choice of which of these vectors are $u_{1}$ and $u_{2}$ because this will obviously have an impact when we carry out the inner product and swapping would give different answers.
linear-algebra vector-spaces vectors gram-schmidt
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add a comment |
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Say I have a basis for $mathbb{c}^{2}$ composed of the vectors $(1,1), (4i,2i )$ with complex inner product. When I construct my orthogonal basis using the Gram-Schmidt process how do I make a choice of which of these vectors are $u_{1}$ and $u_{2}$ because this will obviously have an impact when we carry out the inner product and swapping would give different answers.
linear-algebra vector-spaces vectors gram-schmidt
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1
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Why does getting different answers matter?
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– amd
Jan 9 at 20:01
2
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You pick whichever you think will make the calculations easier. There is no "correct" choice, because there is no "correct" or unique orthonormal basis.
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– Arturo Magidin
Jan 9 at 20:02
1
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The Gram-Schmidt process operates on an ordered basis, so whatever order your ordered basis is in.
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– user3482749
Jan 9 at 20:06
add a comment |
$begingroup$
Say I have a basis for $mathbb{c}^{2}$ composed of the vectors $(1,1), (4i,2i )$ with complex inner product. When I construct my orthogonal basis using the Gram-Schmidt process how do I make a choice of which of these vectors are $u_{1}$ and $u_{2}$ because this will obviously have an impact when we carry out the inner product and swapping would give different answers.
linear-algebra vector-spaces vectors gram-schmidt
$endgroup$
Say I have a basis for $mathbb{c}^{2}$ composed of the vectors $(1,1), (4i,2i )$ with complex inner product. When I construct my orthogonal basis using the Gram-Schmidt process how do I make a choice of which of these vectors are $u_{1}$ and $u_{2}$ because this will obviously have an impact when we carry out the inner product and swapping would give different answers.
linear-algebra vector-spaces vectors gram-schmidt
linear-algebra vector-spaces vectors gram-schmidt
asked Jan 9 at 19:57
user571032
1
$begingroup$
Why does getting different answers matter?
$endgroup$
– amd
Jan 9 at 20:01
2
$begingroup$
You pick whichever you think will make the calculations easier. There is no "correct" choice, because there is no "correct" or unique orthonormal basis.
$endgroup$
– Arturo Magidin
Jan 9 at 20:02
1
$begingroup$
The Gram-Schmidt process operates on an ordered basis, so whatever order your ordered basis is in.
$endgroup$
– user3482749
Jan 9 at 20:06
add a comment |
1
$begingroup$
Why does getting different answers matter?
$endgroup$
– amd
Jan 9 at 20:01
2
$begingroup$
You pick whichever you think will make the calculations easier. There is no "correct" choice, because there is no "correct" or unique orthonormal basis.
$endgroup$
– Arturo Magidin
Jan 9 at 20:02
1
$begingroup$
The Gram-Schmidt process operates on an ordered basis, so whatever order your ordered basis is in.
$endgroup$
– user3482749
Jan 9 at 20:06
1
1
$begingroup$
Why does getting different answers matter?
$endgroup$
– amd
Jan 9 at 20:01
$begingroup$
Why does getting different answers matter?
$endgroup$
– amd
Jan 9 at 20:01
2
2
$begingroup$
You pick whichever you think will make the calculations easier. There is no "correct" choice, because there is no "correct" or unique orthonormal basis.
$endgroup$
– Arturo Magidin
Jan 9 at 20:02
$begingroup$
You pick whichever you think will make the calculations easier. There is no "correct" choice, because there is no "correct" or unique orthonormal basis.
$endgroup$
– Arturo Magidin
Jan 9 at 20:02
1
1
$begingroup$
The Gram-Schmidt process operates on an ordered basis, so whatever order your ordered basis is in.
$endgroup$
– user3482749
Jan 9 at 20:06
$begingroup$
The Gram-Schmidt process operates on an ordered basis, so whatever order your ordered basis is in.
$endgroup$
– user3482749
Jan 9 at 20:06
add a comment |
1 Answer
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You can take the vectors in any order you like. You will still get an orthonormal basis from the Gram-Schmidt process (though, in general, a different one).
answered Jan 9 at 20:33
Chris CusterChris Custer
12.6k3825
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$begingroup$
You can take the vectors in any order you like. You will still get an orthonormal basis from the Gram-Schmidt process (though, in general, a different one).
answered Jan 9 at 20:33
Chris CusterChris Custer
12.6k3825
$endgroup$
add a comment |
$begingroup$
You can take the vectors in any order you like. You will still get an orthonormal basis from the Gram-Schmidt process (though, in general, a different one).
answered Jan 9 at 20:33
Chris CusterChris Custer
12.6k3825
$endgroup$
add a comment |
$begingroup$
You can take the vectors in any order you like. You will still get an orthonormal basis from the Gram-Schmidt process (though, in general, a different one).
answered Jan 9 at 20:33
Chris CusterChris Custer
12.6k3825
$endgroup$
You can take the vectors in any order you like. You will still get an orthonormal basis from the Gram-Schmidt process (though, in general, a different one).
answered Jan 9 at 20:33
Chris CusterChris Custer
12.6k3825
answered Jan 9 at 20:33
Chris CusterChris Custer
12.6k3825
answered Jan 9 at 20:33
Chris CusterChris Custer
12.6k3825
answered Jan 9 at 20:33
Chris CusterChris Custer
12.6k3825
12.6k3825
add a comment |
add a comment |
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1
$begingroup$
Why does getting different answers matter?
$endgroup$
– amd
Jan 9 at 20:01
2
$begingroup$
You pick whichever you think will make the calculations easier. There is no "correct" choice, because there is no "correct" or unique orthonormal basis.
$endgroup$
– Arturo Magidin
Jan 9 at 20:02
1
$begingroup$
The Gram-Schmidt process operates on an ordered basis, so whatever order your ordered basis is in.
$endgroup$
– user3482749
Jan 9 at 20:06