 We have described when a pair of vectors are considered orthogonal in this video We're going to talk about when a set of vectors is orthogonal So suppose we have some set of vectors v1 all the way up to VR and this lives inside of FN Dimensional space there. We say that the set v1 up to VR is Orthogonal it's an orthogonal set if each pair of distinct vectors is orthogonal. That is if vi dot vj Equal zero whenever the indices i and j are different from each other. So that's what we mean my own orthogonal set So I think a quick example of this Let's take three vectors. We'll say v1 v2 and v3 I Claimed that the set v1 v2 v3 is an orthogonal set of vectors and we can see that by just the quick calculations there What happens when you take v1 dot v2 you're gonna get one plus two minus three, which is equal to zero So that's an orthogonal pair. If I take v1 dot v3 And that one you're gonna get negative five plus eight minus three. That's likewise zero So v1 v3 is also an orthogonal pair. So v1 is orthogonal to v2 and v3 But to be an orthogonal set every pair must be orthogonal So the last pair we have to check is two and three It doesn't matter who goes first or who goes second even with complex vectors If you get zero one time if you flip it around you get zero the other time as well So v2 dot v3 you get negative five plus four plus It's actually it's a double negative. So it's a plus one So we're gonna get zero right there And so this in fact shows that this is an orthogonal set of vectors if any pair was not Zero if any pair of if any pair was not zero here Then that means those two vectors are not orthogonal and hence the whole set is not orthogonal with each other now One important caveat I have to mention here is when we define when we define a set of vectors we actually Require that this be a non zero Set of vectors when we define orthogonal sets and why is that? Well, like I mentioned before the zero vector is in fact orthogonal to everything and it's quite exceptional in that regard and As such we kind of have to we have to remove the zero vector from consideration when we talk about Orthogonal sets because zero is gonna be your thought with everything and so we do have to treat it exceptionally We're gonna see why that is in just a second later on in this video but We require that the when we define orthogonal sets that all the vectors be non zero and there are some special cases of Orthogonal sets I want to make mention to you We saw an example of an orthogonal set of size 3 if you just have a single vector Right here a single vector just by itself is considered orthogonal This is considered an orthogonal set if and only if V is not equal to zero Because if it were equal to zero Orthogonal sets are not allowed to have the zero vector and so any vector that contains the zero vector Any set of vectors contains a zero vector cannot be orthogonal so it's ruled out automatically But even if it's non zero then you just have one vector and so what about the pairs? Well The definition here says that each pair of distinct vectors is orthogonal if you have just a single vector There are no pairs of vectors and therefore there are no pairs which are not orthogonal This is sort of like a little bit of logical acrobatics here, but a single vector by itself is vacuously orthogonal because there's not a second vector For which it's not orthogonal to and since there exists no counter examples. It's considered orthogonal So the single vector by itself is an orthogonal set. That's that's something I want to mention here Also, if you take the empty set the empty set is also considered an orthogonal set for the same reason That there's no vectors inside that set and so in particular There's no pairs of vectors that come from the empty set And so since there's no pairs there exist no pairs which are not orthogonal There's no dot products that give you non zero So a single vector is considered orthogonal the empty set is considered orthogonal But then if you have two or more you do have to check like we did in this example right here On why you have to you have to check to make sure the pairs are zero or non zero to be orthogonal Each of these dot products has to be zero So, okay What why is it so important that we've thrown zero sure zero is exceptional But I could throw zero in there and it could be it's not going to complicate the issue Is it well it turns out that the main reason we don't want zero to be considered an orthogonal vector That is we don't want a set contained zero to be orthogonal is the following if s is a set of vectors We'll say v1 up to vp inside of fn If this set is orthogonal and these are all non zero vectors then s is actually an independent set This is a pretty cool argument here that if we were trying to show that these vectors are linearly independent We have to take some scalars like c1 v1 plus c2 v2 All the way down to cp vp Right, this is equal to the zero vector If you want to show this set of vectors is linearly dependent We then check are there any linear combinations that produce the zero vector and I claim that there aren't going to be any And so what you do is you're going to pick your favorite vector inside this inside this list We'll take the vi right here and we're going to dot it with everything in this calculation right here vi Dot zero on the right vi dot everything else on the left Well, my properties of the dot product or the Hermitian product or whatever your product is we can distribute this in all the pieces So we end up with vi dot c1 v1 and then you're going to get the next one vi Dot c2 v2 Right, and then you continue down in this list right at some point in the list though. You're going to get vi dot ci vi so it's going to be dotted it with itself and then you continue on till the very end vi Dot cp vp Now the right hand side is going to be easy if you take any vector dot zero you're going to get zero like so Each of these scalars can come out Here we can take each we can factor each one of these scalars out again by properties of the inner product there So you're going to get c1 times vi dot v1 Right and you continue down this list like I said somewhere on this list There's going to be the ith position ci times vi dot vi Like so and then you get all the way down to the end. Whoops You get a cp times vi dot vp Now by assumption this is an orthogonal set So as long as you pick different vectors the dot product should be zero So this inner product is zero The second one the third one they're all zero the pth one is zero. They're all zero with the exception of this one right here vi dot vi We have no assumption about what happens when you dot the vector with itself So the left hand side becomes ci times vi dot vi Like so and this is going to equal zero So we have a product of two numbers. There's ci which is a scalar and then vi dot vi which is a scalar uh, and so We have these the product two numbers that equals zero so the zero product property applies We either get ci equals zero or We get that vi dot vi Equals zero, but it turns out the second condition is impossible because we have an inner product We have the positive definite condition The only way that the the inner product of the vector with itself equals zero is if that vector was zero Aha, this is why we actually care so much about the vectors non means to be non zero We need it to prove that orthogonal sets are actually linearly independent and so since The dot product is not zero. That means the scalar was zero And so then if we come back to our original our original Equation right here, then it turns out that one by one each of these scalars must have been zero And so the only solution to this homogeneous system was the trivial solution and that's exactly what one expects To when you have a linear independent set The only way to get this combination be zero is to be the zero itself So orthogonal sets are in fact linearly independent But we have to make sure that we only have non zero numbers If we if we throw the zero vector it doesn't it doesn't obstruct the orthogonal condition per se But it will affect the linear independence, which we want orthogonal sets to be linearly independent For the following reason if orthogonal sets are linearly independent automatically if our orthogonal set is then a spanning set Then we can construct what's called an orthogonal basis A basis remember is a linearly independent spanning set. We'll let if our basis is orthogonal for example the standard basis The standard basis for f n Is orthogonal And you can see that because your usual vectors e i right This is the vector with a one in the ith position. It's got zeros everywhere else, right The standard base is an orthogonal is an orthogonal basis And we are going to like orthogonal bases better than just regular bases There's there's some benefits we get from being orthogonal One of them, which is that orthogonal sets are automatically linearly independent Another thing I want to mention here is we said that a set of vectors is orthonormal If one all the vector if the if the set is orthogonal and every vector inside Of the set is a unit vector. Remember a unit vector means that the length of the vector is equal to one So an orthogonal set of unit vectors is called orthonormal And the reason we call orthonormal here is that orthonormal is this amalgamation between the word orthogonal and the word normal Remember that if you take any vector any nonzero vector v and you divide it by its uh, excuse me if you take any Any nonzero vector v and you divide it by its norm Then that will give you a unit vector This is a unit vector here and so this is the process of normalization So if you if you take away the norm the length of the vector you get this normalized vector a unit vector So orthonormal will be those Sets of vectors which are orthogonal and consist only of unit vectors And so therefore we can also talk about the idea of an orthonormal basis Which are going to be some of our favorite bases in existence, right And what is shocker the standard basis for f n is the canonical example of an orthonormal basis We'll see in the future the benefit of orthonormal bases Which we can construct uh, if we can come up with orthogonal sets that span