The gram-schmidt process

[Anony523]

Wolframalpha provides this solution:
http://www.wolframalpha.com/input/?...i,+3i,+1-i,+2i),+(-1+7i,+6+10i,+11-4i,+3+4i)}


However when i compute the second term v2 = x2 - (x2.v1)/(v1.v1) * v1
The result is different from that of above. What's wrong?
http://www.wolframalpha.com/input/?.../+(1,i,2-i,-1).(1,i,2-i,-1))+*(1,+i,+2-i,+-1)

 

[Ishuda]

Wolframalpha provides this solution:
http://www.wolframalpha.com/input/?...i,+3i,+1-i,+2i),+(-1+7i,+6+10i,+11-4i,+3+4i)}


However when i compute the second term v2 = x2 - (x2.v1)/(v1.v1) * v1
The result is different from that of above. What's wrong?
http://www.wolframalpha.com/input/?.../+(1,i,2-i,-1).(1,i,2-i,-1))+*(1,+i,+2-i,+-1)

Please show your work so we can possibly see where things went wrong. Also, you may want to read
http://www.freemathhelp.com/forum/announcement.php?f=32

 

[stapel]

Wolframalpha provides this solution:
http://www.wolframalpha.com/input/?...i,+3i,+1-i,+2i),+(-1+7i,+6+10i,+11-4i,+3+4i)}

Should we assume that the original exercise and instructions were as follows?

. . .\(\displaystyle \mbox{Orthogonalize the following:}\)

. . . . .\(\displaystyle x_1\,=\, \langle \,1,\, i,\, 2\, -\, i,\, -1 \,\rangle\)

. . . . .\(\displaystyle x_2\,=\, \langle \,2\, +\, 3i,\, 3i,\, 1\, -\, i,\, 2i \,\rangle\)

. . . . .\(\displaystyle x_3\,=\, \langle \,-1\, +\, 7i,\, 6\, +\, 10i,\, 11\, -\, 4i,\, 3\, +\, 4i \,\rangle\)

Also, should we assume that you are using the period (the "." character) to indicate a dot (or "inner") product, and you are using the asterisk (the "*" character) to indicate scalar multiplication?

However when i compute the second term v2 = x2 - (x2.v1)/(v1.v1) * v1
The result is different from that of above. What's wrong?
http://www.wolframalpha.com/input/?.../+(1,i,2-i,-1).(1,i,2-i,-1))+*(1,+i,+2-i,+-1)

Are you keeping in mind that there is more than one orthogonalization for any given set of vectors? For instance, if you enter the same vectors, but in a different order, and ask Wolfram's Alpha solver to orthogonalize, you'll get a different set of vectors. This isn't "wrong"; it merely reflects the fact that the solver is using some method, perhaps not Gram-Schmidt, in order to complete the process. :wink:

 

[Anony523]

Should we assume that the original exercise and instructions were as follows?

. . .\(\displaystyle \mbox{Orthogonalize the following:}\)

. . . . .\(\displaystyle x_1\,=\, \langle \,1,\, i,\, 2\, -\, i,\, -1 \,\rangle\)

. . . . .\(\displaystyle x_2\,=\, \langle \,2\, +\, 3i,\, 3i,\, 1\, -\, i,\, 2i \,\rangle\)

. . . . .\(\displaystyle x_3\,=\, \langle \,-1\, +\, 7i,\, 6\, +\, 10i,\, 11\, -\, 4i,\, 3\, +\, 4i \,\rangle\)

Also, should we assume that you are using the period (the "." character) to indicate a dot (or "inner") product, and you are using the asterisk (the "*" character) to indicate scalar multiplication?


Are you keeping in mind that there is more than one orthogonalization for any given set of vectors? For instance, if you enter the same vectors, but in a different order, and ask Wolfram's Alpha solver to orthogonalize, you'll get a different set of vectors. This isn't "wrong"; it merely reflects the fact that the solver is using some method, perhaps not Gram-Schmidt, in order to complete the process. :wink:

Original exercise is S = {(1, i, 2-i,-1), (2+3i, 3i, 1-i, 2i), (-1+7i, 6+10i, 11-4i, 3+4i)}, find an orthonormal basis for span(S).

Isn't the orthonormal (normalized) basis unique for a given matrix?

I used conjugate instead and fixed the problem.
Though when i compute the orth on matlab:

Why is it different from http://www.wolframalpha.com/input/?...i,+3i,+1-i,+2i),+(-1+7i,+6+10i,+11-4i,+3+4i)}?

 

[stapel]

Original exercise is S = {(1, i, 2-i,-1), (2+3i, 3i, 1-i, 2i), (-1+7i, 6+10i, 11-4i, 3+4i)}, find an orthonormal basis for span(S).

Isn't the orthonormal (normalized) basis unique for a given matrix?

Did you read the exercise, where it said to find "an" orthonormal basis, not "the" orthonormal basis? Did you read my earlier reply, where I said:

...if you enter the same vectors, but in a different order, and ask Wolfram's Alpha solver to orthogonalize, you'll get a different set of vectors.