How to Find the Basis of a Subspace

In this video we try to find the basis of a subspace as well as prove the set is a subspace of R3! Part of showing vector addition is closed under S was cut off, all it says is 2*y2 + 3*y3. We then make the substitution that it equals y1 on the following line.

[@dansheldon6955]

Thank you man, i would never have thought this (finding a basis for a linear subspace) was that easy to do.

[@ammaryildirim7721]

grateful for the existence of this video, thanks

[@onivvoisbored8515]

you're goated

[@forheuristiclifeksh7836]

6:39 linearccombi gives vectors

[@The-Collector-g7q]

Good video,but can I ask something. Since the definition of a bases states that not only must the list be linearly independent but must also span the entire vector space. How do we prove that or is it not necessary for this proof?

[@DrewWerbowski]

Yes, you’re right that we need to prove both the vectors are linearly independent and that they span the entire vector space the vectors are a basis for. There are multiple ways to show this and it depends on the question. I believe I have a video on my channel highlighting an example

[@The-Collector-g7q]

​@@DrewWerbowskiOkay I'll check it out, thanks for this one I learnt a great technique from it.

[@nadeemsukhera8476]

Thank you so much Sir

[@dansheldon6955]

By the way, could you explain to me one thing please. If we want to define a linear subspace by saying it is closed under addition and multiplying by a constant, do we really need to say there also has to be an [0, 0, ...0] (for R^n) element in our subspace? If it is proven that any vector of our space multiplied by some constant is also in the space, can't we just say that as the constant is basically every real number it is also zero as well, so [0, 0, ...0] vector is just V * c, where c = 0?

[@juliangrace-martin1740]

Thank you! This was really helpful.

[@MajesticQT]

Thanks!

[@RawanAlfaraj]

THANK YOU!!!!!!

[@pointymoustache]

thanks brodda

[@electronmaxwell8372]

Thanks man

[@namansingh7721]

How can the dimension be less than 3 if they are basis

[@DrewWerbowski]

The dimension can be less than 3 if the vector space is less than 3 (I.e the dimension of R2 is 2)

[@namansingh7721]

@@DrewWerbowski Thanks sir you almost saved my year. thanks a ton

[@mithilesh6793]

@@DrewWerbowski but the vector space was R³, so shouldn't the dimension = 3???

[@phoemi]

@@mithilesh6793 the example in this video isn't the entire R3, but merely a subspace of R3. The dimension of a vectorspace is defined as the number of vectors in a basis, and because the basis for this subspace is [2,1,0] + [3,0,1], the dimension is 2. If it were the entire R3 you're right, then the dimension would be 3 (you can take the standard vectors [1,0,0], [0,1,0], [0,0,1] as an example of a basis). i'm also still learning, so please correct me if i'm wrong !

[@Johnny-vx1eh]

Does it matter which variables I choose as my free ones? I’m having trouble finding anything regarding this as I have chosen different free variables and have a different answer

[@samiulhaqtaj5807]

It doesn't matter for example as in this case if u choose x1 and x2 free variables then from given equation find x3

[@dondavis9075]

4:50

[@Felix-vu9tz]

thanks man haha I was wondering when he was gonna actually answer the question, good tutorial either way tho

[@samiulhaqtaj5807]

Good boy