EXAMPLE: Deciding if sets of vectors are subspaces

[@twa_le]

After so many years since this video has been uploaded, it still helps students out. Much appreciated!~

[@xkhokokox]

Why is 8:10 not a subspace. Aren't the only conditions for a subspace that the sum and scalair product are in thesame subspace?

[@mathslearningUofA]

domien van steendam There are in fact three conditions for a subspace: two of them are closure under sum and scalar multiplication, and the third is either "the set is non-empty" or "the set contains the zero vector". Even if you use "the set is non-empty" as your third law, then it is still true that every subspace must contain the zero vector because if there is any vector v in the subspace and you scalar multiply by 0 you get the zero vector. So if a set does not contain the zero vector then it is can't possibly be a subspace.

[@xkhokokox]

Hey thank you for your response. The reason of my question is because my teached wanted to proof that the kernel and im(T) are subspaces. He did only proof thesum and scalair though. Is there a reason he doesnt talk about the zero vector? Is the zero vector automatically implied so there's no need for mentioning?

[@mathslearningUofA]

domien van steendam Technically, the proof is supposed to contain a part that shows there are vectors in the set, even if it is "implied". For the image, you can probably get away with it because every input certainly has an image so it's not empty. But for the kernel it's not obvious that there are any vectors in it at all, so you must describe why T(0)=0 or you haven't really shown it's a subspace. I recommend checking with your teacher, though.

[@xkhokokox]

Thank you! :)

Good examples and explanations. Thanks for sharing!

[@Muse-mz4hb]

For problem b, doesn't the scalar have to be in the set as well, so the scalar k cannot be negative? I'm confused as to why we only add by positive values for addition because the set is >= 0, but it's not the same for scalar multiplication...

[@mathslearningUofA]

Vector spaces are made of points (also called vectors) and the sets within them are made of points. The scalars aren't actually part of the space itself, but are just used as tools for combining vectors together. So the rules for which vectors are part of a set do not apply to the scalars. Indeed, the definition of subspace requires that it be closed under scalar multiplication for ALL the possible scalars.

[@Muse-mz4hb]

Okay, that makes sense. Thanks!

[@anacousticguitar123]

how would you prove question A?

[@mathslearningUofA]

You'd need to prove each of the rules of subspaces separately. Check out my video called "proving that a set defined by an equation is a subspace"

[@mathslearningUofA]

Sorry you feel that way Jesse. The aim of the examples is not to explain the meaning of the concepts, but to show how to think about problems in order to solve them. With this problem, the aim was to show how you could tell if something was a subspace just by looking at how its definition is written.

[@dinas5254]

Loved it ! Thanks a heap you have no idea how much this has helped

[@lizardperson2293]

Can you please explain something to me? About problem B, if you add (-10,-10,-10) to your vector (1,1,1), you will get a negative vector, which is not in the subspace. Could you explain why people don't do that?

[@mathslearningUofA]

There's nothing stopping you doing something like that. But of course, (-10,-10,-10) isn't a vector in the subspace, so it doesn't really help. If you break it by vector addition, you need both original vectors to be in the set.

[@nomad_on_tour]

If you have a set { (x,y,z) ϵ R^3 | x+2y-3z^2=0 }, is this a set of solutions to homogeneous linear equations, hence a subspace of R^3.

[@mathslearningUofA]

Yes, that is the reason why the first one is a subspace (though there should not be a squared on the z. If there's a squared on the z then it's actually NOT a subspace because the equation's not linear!)

[@nomad_on_tour]

what if you consider the 3 criterion: If you substitute 0 in the equation then you will get 0. Whereas scalar multiplication k, since 0 is in the set (x,y,z) so if you multiply 0 x (x,y,z) = 0. So its true for these 2 criterion. I dont know how to prove closure under addition.

[@mathslearningUofA]

marshall12345able You haven't shown that it's closed under scalar multiplication for any POSSIBLE scalar k. You've only shown it when k =0. It has to be true for every possible k. Check out our video on proving that a set is a subspace using the three rules of the definiton: EXAMPLE: Proving that a set defined by a linear equation is a subspace

[@karabotshehla3967]

Wow thank you sir,its clear now.You just earned yourself a thumb up and a new subscriber

[@dans4evah]

a) looks like it is a homogeneous equation... how do you know it is the set of solutions?

[@mathslearningUofA]

I know because of how it's written {(x,y,z) | x+2y=3z}. Aloud this is "the set of (x,y,z)'s such that x+2y is equal to 3z". A point is part of the set when it satisfies the equation and it's not part of the set when it doesn't satisfy the equation. So this set is all the (x,y,z)'s that make the equation true, which is precisely what it means to be the set of solutions.

[@danamuise4117]

these seems so tedius... is there a faster way? ]

[@DrQlimakz]

what if 2^2 = 0 though?

[@georghieronymus9935]

Excellent tutorial!

[@unarine339]

Awesome :-) Thanx

[@bolurammogor9261]

thank you @mathslearningcentre

[@ATrekafied]

thanks so much, was very enlightening

[@blahblah77124]

you sound like Nasir Hussien English cricket Commentator !!!

[@iworkout4523]

Bless yall

[@童乙]

Really helpful!

[@yousuffarhan963]

Thank you.

[@footage6402]

For b it also isn't because it has no additive inverse.

[@annalam8624]

it helps me a lot!!! thank you so much:D

[@Amanda-lw6tw]

this is rlly helpful thx

[@fredweasleylives00]

Thank you :)

[@gadriel0000]

hey thanks. :) it helped me in my lesson. i'm only in the 10th grade and we're already learning this. This lesson is for college isn't it?

[@kenethmutta4461]

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