Gram Schmidt Process

alright thanks for watching and you might say whoa what happened here well I want to try to iPod a little bit for more legibility but I'm sure you'll be waiting for this Oh too small to break but anyway so welcome to the third part of my linear algebra extravaganza and today I will talk about the gram-schmidt process and what is this crap Schmidt process well remember that orthogonal sets or orthogonal basis are very nice and so which also begs the question is there any way of turning any basis into an orthogonal basis and in fact we can do that and the cool thing is it it just takes ideas from the previous video about orthogonal projections so let's do a quick read example fine an orthonormal now I'll explain what that is basis for the following set for W which is the span of the following three vectors u1 u2 u3 where you want is 1 0 1 0 u 2 with or without you ok it's 0 1 1 1 you three is minus one zero zero one what does that mean so W it's the span of those three vectors so in some sense it's all the information that's contained by those three vectors and the idea is as follows so we have those three vectors they can point at any direction let's say u 1 u 2 u 3 and what we want to do that with that is we want to turn this set into an orthonormal set in other words what you want to do is what a perpendicular is this set in some sense so with this gram-schmidt process we'll be able to turn those three ugly vectors into something that's very nice something that W 1 W 2 W 3 with the property that each of those vectors are perpendicular and not only that each of the vectors will have a length one so as I said it's a nice way of perpendicular rising the vectors and a really turning any basis into if you want a standard basis because those are just all the properties we need for the standard basis ok and how are we going to achieve that first of all forget about the word orthonormal for a second let's first find an orthogonal set for this for this vector space I said first find an orthogonal basis then the also normal thing will be very quick I promise pieces so we want to find the v1 v2 v3 orthogonal but with the six patents are said w so that's very important because it's easy to produce orthogonal sets just do the standard unit vectors but here we want to happen orthogonal set with the same span as W and well there's nothing we can really do let's just start with something so step one well our first vector v1 let let's just let it be u 1 so let V 1 equals to u 1 equals to 1 0 1 0 of course I erased a previous part but once you have your vector u 1 here just cross it out on your list because you will never use it again actually because from now on we'll work with V 1 okay and what's the second step here as I said we start with V 1 we want equals to u 1 and as I said we want to find some vector that's perpendicular to V 1 well let's see what other vectors we have so we do have the vector u 2 right which is 0 1 1 1 and now remember from the previous video a very nice way of finding a vector orthogonal to the Tsun is with the concept of orthogonal projections so let me just quickly recall in case you have not seen the other video if you have a line L and a vector X there's a very easy way of squishing X onto that line it's called this orthogonal projection X hat and then from there you can produce a vector orthogonal to that line namely X minus X hat this vectors X minus X hat and fact X minus X hat is perpendicular to the line well here we have the exact same issue we have a line here spanned by u1 we have our vector u 2 so all we really need to do is to construct you 2 hat and our answer for V 2 will be you know X minus X hat which is u 2 minus u 2 hat and so now let's try to achieve this u 2 hat now remember the definition of orthogonal projection we want a vector that's in this line so it has to be a multiple of V 1 so something times V 1 and to find that something all you do is you hug the vector u 2 with 2 V 1 so it's u 2 dot v1 over v1 dot v1 times v1 and then if you do that you know this calculation so u 2 is 0 1 1 1 dotted with 1 0 1 / so this vector you repeated actually four times so one zero one zero dotted with one zero one zero this huge number is just a number and you multiplied by the vector 1 0 1 0 if you calculate that you get 1/2 times 1 0 1 0 which is 1/2 0 1/2 0 and now it's very important that you leave the vector like that do not rescale it because you make it end up getting a vector that's not on this line so right now that doesn't quite work but in a second we can rescale so what we did so far is good we found you to hat which is 1/2 0 1/2 0 and now just to construct v2 you just do u 2 minus u 2 hat so let's do that therefore so V 2 is u 2 minus u 2 hat and that's 0 1 1 1 minus 1/2 0 1/2 0 which becomes minus 1/2 1 1/2 1 and now you do have a vector on this slide and the nice thing about lines is that you can multiply something and you still on this line at least if it starts at the origin so now you are allowed to rescale so maybe let's multiply this by 2 and you get minus 1/2 1/2 wonderful so you started with you one right and then you now you calculated a vector that's perpendicular to u 1 so it's great so you have this situation here V 1 and then we have V 2 that's perpendicular to it so we are on the right track and by the way two things now that you use you two you can literally cross it on the off of your list and I suggest that otherwise you might be tempted to use it again and and now we just want to continue so we are V 1 V 2 and now let's try to find a vector that's perpendicular to here's the last third step so again the pictures as follows we have V 1 we have V 2 they are perpendicular the idea is now well we wanna you know given the last P first piece of information that we have let's find a vector perpendicular to those two so and really the only thing that we can still use this is u3 and now we want to do the same spiel again with the orthogonal projections if we wanted me to well this span a plane and all we have to do is a follow it first calculate the orthogonal projection of U 3 on that plane and then the answer is played simple it the answer then v3 is U 3 minus u 3 hat because precisely V 3 will be perpendicular to the plane and therefore in particular perpendicular to V 1 and V 2 so now let us do that so you three have in this case I didn't remember what it did with orthogonal projections you three have has to be a vector in that plane therefore a linear combination of v1 and v2 and all you do is take you three and you hug it with v1 so you three dotted with me one over V one dotted with me one and you three dotted with V 2 over V two dotted with V 2 let students nasty calculation so we have minus 1 0 1 0 1 dotted with 1 0 1 0 over 1 0 1 0 dotted with 1 0 1 0 times 1 0 1 0 us minus 1 0 0 1 . it with - 1 - 1 2 / - 1 2 1 2 dotted with - 1 - 1 - x - 1 - 1 - by the way do you see how nice it is that are we scaled it otherwise would have nasty fractions and then after the dust settles we have the following equals 2 so minus 1/2 times 1 0 1 0 plus 3 tens of minus 1/2 1/2 it's very intense right because we have 10 and in the end you get minus 4/5 3/5 minus 1/5 and 3/5 again at this point do not rescale you vector because again it would change your position of you better but once he found you 3 hat remember the answer is you three - you three have so we get V 3 is U 3 - you three hunt you can calculate this to be minus 1/5 minus 3/5 1/5 and 2/5 if you like rescale it to - 1 - 3 1 2 and therefore we did find our orthogonal basis maybe it's the follow illusion this set V 1 V 2 V 3 which is 1 0 1 0 - 1 - 1 2 - 1 - 3 1 2 is enough talking of a TSA's perpendicular basis for EE and in fact if you like you can always check if it's you know indeed orthogonal so ok and we're almost done well how do you find an orthonormal basis and in case you don't know orthonormal means each vector has length 1 turns out it's the easiest thing you know easiest step of them all so step 4 normalize and here's a cute fact and let me use my red fact if you take a vector and you divide by its length automatically this has length 1 okay unless it's a 0 which means that for each of those three vectors we just have to divide by the length then W want be the one divided by its length which is one zero one zero and the length of this is square root of 1 squared plus zero squared plus 1 squared plus 0 squared which is 1 over root 2 which is 1 over root 2 0 1 over root 2 0 and then W 2 which is after do my taxes V 2 over length of V 2 which is 1 over root n minus 1 2 1 2 which is minus 1 over root n 2 over root N 1 over root n + 2 over root n and lastly W 3 which is V 3 over length the week 3 and it's 1 over root 15 of this vector minus 1 minus 3 1 2 which is minus 1 over root 15 minus 3 over root 15 for ball roots okay 1 over root 15 + 2 or root 15 and of course do not rescale it because the point was we chose the scaling exactly for it to have length 1 and then once you have that you get W 1 W 2 W 3 is an orthonormal basis oh and basis for W it's all orthonormal which means that those three vectors are perpendicular and each of them has length 1 and remember in the first video I convinced you hopefully why so great in orthogonal or orthonormal sets which you know which makes this great because we're able to construct you know explicit orthonormal basis alright so if you like this linear algebraic stir McGann's ax and there's more to come and if you like math in general please make sure to subscribe to my channel thank you very much

Channel: Dr Peyam

Views: 8,924

Rating: 4.9303136 out of 5

Keywords: gram-schmidt, gram, schmidt, process, linear algebra, orthogonal, orthonormal, basis, orthogonal projection, dot product, normalization, unit vector, vector space

Id: hYtlRLWM1xo

Length: 18min 50sec (1130 seconds)

Published: Wed Feb 14 2018