Example of Gram-Schmidt Orthogonalization

Linear Algebra: Construct an orthonormal basis of R^3 by applying the Gram-Schmidt orthogonalization process to (1, 1, 1), (1, 0, 1), and (1, 1, 0). In addition, we show how the Gram-Schmidt equations allow one to factor an invertible matrix into an orthogonal matrix times an upper triangular matrix.

[Dominic]

you are the strongest math doctor ive ever seen

[yotube]

First midterm tomorrow. I've just been memorizing the orthogonalization formula this whole time. I actually understand where it came from now. Thanks a lot. Keep making videos! There are so few **good** higher-level-math videos on the web.

[KyleHeredia]

Thank you so much for help. I've been watching your videos here in AZ to help me with my linear algebra class. I appreciate you taking the time to make these videos to help others like me!

[yourmaster]

That was a much clearer explanation than my lecturer gave, thanks for a good clear video.

[micmarlen]

it seems he's going to beat me with that stick XD i just take "u" and subtract all the parallel apart XD 1:03

[MathDoctorBob]

That's exactly what you want, but you also need to extend the columns of R by 0s to get the dimensions to line up. All you need is that the columns of new Q form an orthonormal basis; that ensures that Q is orthogonal. Since the extra vectors zero out in the product, so you have a lot of freedom. - Bob

[lathspell]

When I saw that black stick I thought you were going to teach Kravmaga or something. Glad it turned out to be the Gram Schmidt orthogonalization :)

[INfAMm]

you are very muscly for a mathematician.  Thumbs up

[MathDoctorBob]

@MetallicAus You're welcome, and thanks again for the comment. -Bob

[MoreCores]

You are very thorough. Keep up the good work!

[chrisvolk]

Again, you killed it! Thanks a ton for all you do.

[MathDoctorBob]

Not in the span won't be enough. You need to have an orthonormal set at each step - so the vector you add must be orthogonal to the previous ones and have length 1.

[FlorisStoicaMarcu]

Great video, i really enjoyed your explanation!

[MathDoctorBob]

@ramhamla You're welcome. Thanks for the comment.

[karr]

Perfect explanation! Thank you so much

[MathDoctorBob]

That would make for a cool playlist. It's a hard thing to do since personality is a huge factor.

[drelos]

Thank you very much... I will try it out and if I find any problem I will ask in more details if I may... Have a nice day...

[MathDoctorBob]

We have no info on u parallel beforehand. Note that u parallel is the part of u pointing in the same direction of v. I can describe any vector by giving its length and its direction (unit vector), and then take the product. The direction of v is v/|v|, so we only need the length, which is given using the cosine formula. Thus u parallel is u.v/|v| v/|v| as written.

This works for any u and v, so no need to orient along any of the axes. Let me know if that helps.

[MathDoctorBob]

@zmh786 You're welcome! Thanks for the comment. - Bob

[mamumich]

haha great presentation, but you are so intimidating!