welcome back to linear algebra today we will talk about these spanning set their own so what does the spanning set theorem say well it says if s equal to vectors V 1 through VPN V and our H is equal to the span of our vectors V 1 through VPN if a vector V K in our set is a linear combination of the remaining vectors then we say that s minus that vector V K still spans H so the span of v1 through VPN is VK is equal to the span of v1 through VPN and our second condition says that look if H is not just the zero vector then there is some subset of H that is a basis so we will have a basis in there so let's prove this s minus VK still spans H so again s is equal to V 1 to V P let's take an arbitrary vector X and this arbitrary vector X is going to be a linear combination of C 1 V 1 all the way up to C PvP so this vector X is in our span it's arbitrary though so we're going to pick a VK that is a linear combination of the other ones for notation terms we're going to pick that VK and we're going to make VK be the last vector in our set so we're going to call it V P so V P is going to be a linear combination of all the other vectors so this will be a1 times v1 all the way up to a P minus 1 times V P minus 1 so V P is a linear combination of the rest well we can do now is we can substitute this in for our representation of the vector X so now we're going to have X is equal to c1 v1 all the way up to CP but when this V P goes in here we're going to replace this with a1 v1 all the way up to a P minus 1 V P minus 1 so now our vector X is a representation of just the vectors v1 through VPN us one but X was arbitrary so that means that we can do this for any vector X so this means that the span is still the same so because I took an arbitrary vector and I can rewrite that vector as a combinations of v1 through VPN 1 that means that the span is still the same so that's the proof of Part 1 the second part says that if H is not the zero vector then there's a subset that's a basis okay so there's two conditions one we can have that s is linearly independent so if s is linearly independent then we're done because it's a basis ok the second condition which is more important is what if s is linearly dependent so what happens if it's independent or if it if it is dependent okay if it is dependent then we can rewrite one of the vectors as a linear combination of the others so we're just gonna take s and we're going to remove that vector VK and now there's two conditions one if it's now linearly independent we're done so that's good too once we remove the vector if it's still linearly dependent then what we do is we just repeat the process and we keep going until it becomes linearly independent now we know that this zero vector cannot be a basis so what will happen is we'll keep reducing and we'll never quite hit zero but we will eventually find a basis for H so if the spanning set reduces to one vector it is not going to be the zero vector because it is going to be linearly independent because we're just reducing the nearly dependent said until we get something that's linearly independent so essentially we're removing all of these redundant vectors okay so that's the spanning set there so we have V 1 V 2 and V 3 these are three vectors and we have H is the span of these and we know that if we take for v1 add 5 V 2 and subtract 3 V 3 we're going to hit the zero vector I'm asking you to find three distinct bases for H so we're gonna have a basis one we're gonna have a basis to and we're going to have a basis three so how do we do this well we know that we can rewrite one of these vectors as a linear combination of the other so what this means is we can pick any one of these vectors and we can remove it so one of our bases could be the set containing V 1 and V 2 because we just remove vectors number 3 V 2 so a second basis you can remove vector 2 instead so we're left with the one on V 3 for our third basis we could remove V 1 so we're gonna be left with V 2 and V 3 so these are three distinct phases for H or basis for H it doesn't even matter what these are we just have to make sure that these two are also not linearly dependent I'm going to save the calculations because it would take a bit of time but these are not linearly dependent these are in fact linearly independent sets that span the space that we're looking for so this is good these three are good ok next question we have three vectors v1 v2 and v3 we have 1 0 1 0 1 1 and 0 1 0 and H is going to equal the set of vectors that are the form ABB so essentially the first entry is a number doesn't matter what it is but the second and third entries are always going to be equal to each other so we could rewrite these vectors vector ABB as a linear combination of V 1 V 2 and V 3 so we can take a times zero zero plus B minus a times zero one one plus a times zero one zero so we can prove these are equal this is just a zero zero plus zero B minus a B minus a and then we add together zero a zero okay so we add the three together this should be equal to a B be okay that looks good it does not look good because one of these entries was written wrong okay so that should be a 1 there which means this should be an A now it is good okay problem two fixed okay so is V 1 B 2 and B 3 a basis for H well what this means is that our v1 v2 and v3 so these are linearly independent and we can't write these as a linear combination of each other so that's good but there's one key point these vectors they have to be in H so these have to be an element of H what's the problem well this v3 0 1 0 is not in H because this is not of the form a V be the second and third entries are different therefore this cannot be a basis for H because v3 isn't even in H so of course you can't have a basis for something if you take a vector from the basis or we take a vector from the space we we have to have it from the space v3 is not in the space so where we pull eb-3 from well we can pull it out of our butts but that's not good mathematically so don't do that it also might hurt a bit so not a basis all right that's the spanning set theorem lecture if you have any questions please leave them in the comments below and I'll ask them the best that I can