If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:11:16

let's say I've got me a set of vectors so let me call my set B and let's say I have the vectors v1 v2 all the way to through VK now let's say this isn't just any set of vectors or some interesting things about these vectors the first thing is that all of these guys have length of one so we could say the length the length of vector V I is equal to 1 for I is I is equal to well we could say between 1 and K or I is equal to 1 2 all the way to K all of these guys have lengths equal 1 or another way to say it is that the square of their lengths are 1 the square of VI so length is equal to 1 or VI dot VI is equal to 1 for I is you know any of these guys any I can be 1 2 3 all the way to K for I is equal to 1 2 all the way to K so that's the first interesting thing about it let me write in regular words all the vectors all the vectors in B have length length 1 or another way to say it is that they've all been normalized that's another way to say that is that they have all been normalized or they're all unit vectors normalized vectors or vectors that you've made their lengths 1 you've turned them into unit vectors they have all been normalized they have all been normalized so that's the first interesting thing about my set B and then the next interesting thing about my set B is that all of the vectors are orthogonal to each other so if you dotted with itself so if you dot a vector with itself you get length 1 but if you take a vector and you dot it with any other vector if you take VI and you were to dot it with V J so if you typical you know v2 and dotted v1 it's going to be equal to 0 for I does not equal J it's going to be zero all of these guys are orthogonal let me write that down all all of the vectors are orthogonal to each other orthogonal to each other each other and of course they're not diagonal to themselves because they all have length one so if you take the dot product with itself you get 1 if you take a dot product with the some other guy in your set you're going to get 0 or maybe I we can write it this way VI dot VJ for all the members of the set is going to be equal to it equals 0 for I does not equal J and then if these guys are the same vector I'm dotting with myself I'm going to have length 1 so it would equal length 1 for I is equal to J so I've got a special set all of these guys have length 1 and they are all orthogonal with each other they're normalized and they're all orthogonal and we have a special word for this this is called an ortho now orthonormal set so B is an ortho ortho for orthogonal orthonormal set orthonormal set normal for normalize everything is orthogonal they're all orthogonal relative to each other and everything has been normalized everything has length 1 now the first interesting thing about an orthonormal set is that it's also going to be a linearly independent set so B so if V is orthonormal B is also going to be linearly linearly independent independent and how can I show that to you well let's assume that it isn't linearly independent let's assume so we're clearly all of you know let me take let me take you know V I let me take VJ that are members of my set and let's assume that I does not equal that I does not well J now we already know that's an orthonormal set so VI dot VJ VI dot VJ is going to be equal to zero they're orthogonal these are two vectors in my set now let's assume that they are linearly dependent I want to prove that they're linearly independent and the way I'm going to prove that is by assuming they're linearly dependent and then arriving at a contradiction and then arriving at a contradiction so let's assume let's assume that VI and VJ are linearly sorry linearly dependent linearly dependent well then that means that I can represent one of these guys as a scalar multiple of the other and I can pick either way so let's just say for the sake of argument that I can represent VI let's say that VI is equal to some scalar C times V J times V J that's what linear dependency means that one of them can be represented as a scalar multiple of the other well if this is true then I can just substitute this back in for V I and what do I get I get C times V J which is just another way of writing V I because I assumed linearly linear dependence that dot VJ dot VJ has got to be equal to zero this guy was V I this is V J they're orthogonal to each other but this right here is just equal to C times V J dot VJ which is just equal to C times the length of V J V J squared and that has to equal zero right they're orthogonal so has to be equal to zero which implies which implies that the length of V J has to be equal to zero if we assume that this is some non-zero multiple and this has to be some nonzero flesh-eater for it there C does not equal zero why does this have to be a nonzero multiple because these were both nonzero vectors this is a non zero vector so this guy can't be zero this guy has length one so this is a nonzero vector there's no way that I could just put a zero here because if I put a zero then I would get zero vectors so secant be zero so see isn't zero then this guy right here has to be zero and so we get the length of VJ 0 which we know is false the length of V J is 1 this is an orthonormal set the length of all of the members of B are 1 so we reach a contradiction this is our contradiction VJ is not the zero vector it has length 1 contradiction so if you have a bunch of vectors that are orthogonal and they're nonzero they have to be linearly independent which is pretty interesting so if I have this set this orthonormal set right here it's also a set of linearly independent vectors so it can be a basis it could be a basis for a for a or a subspace so we can say let's say that B is the basis for some subspace V or we could say that V is equal to the span of we could the span let me write it this way the span of v1 v2 all the way to VK then we call B if it was just a set we call it an orthonormal set but it can be an orthonormal basis when it spans some subspace so we can write we can say that B is an ortho normal orthonormal basis for V now everything I've done is very abstract but let me do some quick examples for you just do you understand what an orthonormal basis looks like with real numbers so let's say I have two vectors let's say I have the vector or view one that is let's say we're dealing in r3 so it's 1/3 2/3 2/3 and 2/3 and let's say I have another vector V 2 that is equal to 2/3 1/3 and minus 2/3 and let's say let's say that say that B is the set of v1 and v2 so the first question is is what are the lengths of these guys so let's take the length the length of v1 squared is just v1 dot v1 which is just 1/3 squared which is just 1 over 9 plus 2/3 squared which is 4 over 9 plus 2/3 squared which is 4 over 9 which is equal to 1 so the length squared is 1 then that tells us that the length of our first vector is equal to 1 is the square of the length is when you take a square root so the length is 1 what about vector 2 well the length of vector 2 squared is equal to v2 dot v2 which is equal to let's see 2/3 squared is 4/9 plus 1/3 squared is 1/9 plus 2/3 squared is 4/9 so that is 9 9th which is equal to 1 which tells us that the length of V 2 the length of vector v2 is equal to 1 so we know that these guys are definitely normalized this is ami could call this a normalized set but is it an orthonormal set are these guys orthogonal to each other and to test that out we just take their dot product so v1 dot v2 is equal to 1/3 times 2/3 which is 2/9 plus 2/3 times 1/3 which is 2/9 plus 2/3 times minus 2/3 that's minus 4/9 minus 4/9 2 plus 2 minus 4 is 0 so it equals 0 so these guys indeed are orthogonal so B is an orthonormal set ortho normal set orthonormal set and if I have some subspace let's say that you know let's say that V is equal to the span of v1 and v2 then we can say that the basis for B or we could say that B is an ortho normal basis basis for V