Linear Algebra: Gram-Schmidt

so in this video we're going to tackle the gram-schmidt process this is the process by which we find an orthonormal orthonormal basis given a set of vectors so in a previous video we went over what that was I'll just recap very briefly an orthonormal basis is where we have multiple vectors they are all at right angles to each other that's where we get the word orthogonal they're all linearly independent obviously because they're at right angles and they're all unit vectors as well so given just a random set of vectors how are we going to go about making them all right angles and making them all unit vectors here's what it's going to look like geometrically so say we have two random vectors here's V 1 and here's V 2 okay we've already learned about projection and that looks like this if we will project 2 V 2 U onto V 1 make so it's like casting a shadow down this piece here becomes the projection onto V 1 of V 2 and this piece here that's V V 2 that's the projection that just becomes V 2 minus the projection onto V 1 V 2 but it turns out this piece here is exactly what we want because if we take this piece here we now have V 1 and again V 2 minus projection and that's what we're looking for in an orthonormal basis right now there these two are at a right angle and the only thing that's left to do we also want them to be unit vectors so we just have to shorten them by however much brings them down to unit vector of shape so that will give us u 1 and u 2 this is the process that we're trying to do if we have more than two vectors it's basically the same just now we have another vector in some other direction third dimension fourth dimension we need to do that we basically need to do the same thing we'll take that original vector subtract the projection onto V one and then we'll also have to subtract the projection onto V 2 as well and it just keeps adding up so in the case of three vectors such as these three that I have right here which we're going to do our example on to get our first vector we start with I'm going to label these v1 v2 and v3 and what we're trying to get is our orthonormal basis and what we're going to try to get I'm going to call them u1 u2 and u3 so u1 is just going to be v1 that's what we're starting with and if you remember in my drawing I didn't change that one at all but we just need to shorten it down to unit vector size so divided by its length okay so you two as we saw that's going to be v2 minus the projection onto V 1 and V 2 and because now we have unit vector for u1 which is in the same direction we're going projected onto you one that's unit vector and we remember the formula for that is u 1 dot V 2 times u 2 I mean you one let me say that again u 1 dot V 2 times u 1 ok but that's that's not exactly it I'm going to call this u 2 prime because we're not exactly done the real u 2 is when we take this result u 2 prime and we need to divide that by its length I could have put big parentheses around this and divide that by its length but hopefully that makes it clear we do this process to it and then we divide by the length and 4 u 3 hopefully you can see the pattern here now we take v3 we're going to subtract subtract the projection of v3 onto YouTube and the projection of v3 on to u1 okay and that we're not done yet that's going to be prime and the real u3 comes when we divide by its length to get it down to unit vector okay sound good we're going to try to do that with these three vectors now so we'll start with this step and v1 so the length of v1 here we have two two and one so the length that is going to be 2 squared plus 2 squared plus 1 squared that's 4 plus 4 plus 1 that's 9 and then the square root of that is 3 the length of this is 3 so my U 1 is going to be 1 over 3 times my V 1 and that's 2/3 2/3 1/3 cool all right now things are going to get a little bit more complicated so for YouTube Prime I'm going to take v2 I'm just going to fill this in with what these values are that's minus 2 1 2 and the dot product of U 1 right here we have V 2 okay um I better write this all out so that's the vector 2/3 2/3 1/3 dotted with minus 2 1 2 times you to view over as well times that one and that's going to equal minus 2 1 2 we have 2/3 minus 2 that's minus 4/3 plus 2 thirds plus 2/3 again and that actually equals 0 so our job is easy because that's just 0 times this vector so that's just 0 and our result is again just the same as V 2 minus 2 1 2 and don't forget our last step we need to normalize this one as well so it's a unit vector and the length of that is going to be 1 over the square root of 2 plus 1 plus 2 all squared so that's again that's just the square root of 9 which is 3 so that's our real you - that was you - prime and so are you - our final answer is minus 2/3 2/3 1/3 2/3 cool okay so we have u1 u2 now for you three all right we're going to start with a vector 3 which is 1800 we're subtracting the dot product of u2 dotted with v3 so that's going to be pretty easy because these are both zeros is just 18 times whatever this versatile u is minus 2/3 so 18 times minus 2/3 cross out by threes that's 6 times minus 2 or minus 12 okay and you one dotted with v3 so here's v3 here's you 118 times two thirds again that's just 12 and you won again is 2/3 2/3 1/3 all right here that's the minus sign that's a minus sign so this becomes plus okay so my vector becomes 1800 plus these threes all cancel so this becomes 3 skin so that's 4x minus 8 for 8 & 8 8 for the result of that is that looks like a times that's a plus okay 18 minus 8 minus 8 minus 8 is 2 4 minus 8 is minus 4 & 8 minus 4 is positive 4 and again my last step I need to divide by the length to make sure it's unit vector so 3 that's going to be 1 over the square root of 2 squared that's 4 plus 16 plus 16 that's 16 plus 16 is 32 plus 4 is 36 yay 36 is a perfect square and that just gives us 6 so one six times two minus four four and that's equal to one-third minus 2/3 and 2/3 all right so we've reached the end um that is u3 this guy right here and now we have our orthonormal basis I'm going to write it down here as u1 u2 and u3 just so you can see u 1 is 2/3 2/3 1/3 u 2 is minus 2 1 2 and that's a 1 - 2 - awesome so here we have our translated version of V 1 V 2 and V 3 into an orthonormal basis they're all at right angles to each other and they're all unit vectors as well so that's all we have to do for that this is the Graham current process I hope this example has been helpful for you check out other videos as well for more linear algebra stuff thanks for watching this video be sure to check out the rest of the videos in this series and any of the other math related videos on our channel if you're not subscribed to our channel click this link right here for more help with linear algebra check out world wide differential equations with linear algebra by Robert McGowan or elementary linear algebra by Bruce gooberstein both are available at an affordable price in digital formats on our website just click this link right here you

Channel: Center of Math

Views: 81,157

Rating: 4.8732395 out of 5

Keywords: math, center of math, mathematics, High-definition Pre-recorded Media And Compression, Linear Algebra (Field Of Study), Gram--Schmidt Process, step by step, help, free, tutor, vector, matrix, Mathematics (Field Of Study)

Id: Aslf3KGq2UE

Length: 13min 53sec (833 seconds)

Published: Fri Jun 27 2014