Subspaces of Three Dimensional Space

[@doodelay]

Man she's easy to listen to. Very, very clear and gentle instruction

[@orientaldagger6920]

Yes. Very gentle instruction but I bet if you get the wrong answer she'll kick your ass.

[@dheerajsharma224]

@@orientaldagger6920 🤣🤣

[@shanayaasaumyasuneja9427]

This was so very well explained, thank you!

[@sabarim4904]

It's b . The New York Post New York 6//";:

[@yogiv1876]

Beautiful explained. 👏🏽👏🏽👏🏽

[@charlesabernathy1962]

I have only viewed it twice, as a Broadway theater goer said when asked whether or not he understood a play.

[@zhijianli8975]

thanks for your informative tutorial, it's real useful for me.

[@quirkyquester]

Thank you very much, this is amazingly well explained.

[@BarkanUgurlu]

Please correct me if I am wrong. But the last answer may not be v2. The plane that is spanned by v1 and v2 is -12x + 6y - 2z = 0. The intersection with the x-y plane is y = 2x (Oh I get it. OK)

[@ايسرالعنزي-ك7و]

ILove chinese All respect miss🙏

[@bhushandatre1508]

in the subspace not including x1 and x2 , you took x1+x2 but it includes 0 also and we can just choose perfect scalars so that x1 and x2 sums up to be 0 so basically x1 + x2 is containing x1 and x2 isnt it ?

[@Tman1000-be7op]

a subspace not including x1 and x2 means x1 and x2 do not show up in the subspace. it is not the case that any linear combination of x1 and x2 must not show up in the subspace.

[@khaledqaraman]

Great explanation... Thanks!

[@musaabdullahi3261]

Yeah.. Also grateful Linan... Thanks

[@quantumpizza576]

Even if all of space collapses onto itself, those lines are gonna be straight forever lol

[@ektabansal5332]

Very nicely explained!

[@ruby_wired]

The question didn't specify the dimensions of subspace S, I chose the zero vector, not sure whether to feel dumb or smart. Great explanation though, going through questions is an essential companion to lectures and these videos are criminally under-viewed.

[@crocopie]

Dude, these recitation videos are a life saver!

[@cats_of_neptune]

dont x1 and x2 go through the zero vector? implying that they also belong to that subspace?

[@zoomzune]

@@cats_of_neptune i got confused by this at first too - but i think x1 and x2 just represent the coordinates (not lines through) - (0,1,3) and (2,4,0). thus, they aren't actually vectors, just points on the xyz plane. so, they don't belong to the subspace of the line through x1 + x2

[@onlyinafricahappiness1615]

Seems to me x1 + x2 must go through origin when one takes o combinations of both x1 and x2. Which means then the subspace created by x1 + x2 meets the two vectors at the origin and therefore there inside the new subspace at least at one point

[@AbhishekMishra-jd4bn]

i would be great if you show this using a 3D software.

[@yiyu9519]

love this course

[@crocopie]

I have a problem. Why is x1+x2 part of V3?

[@quirkyquester]

V3 = all the linear combination of {x1, x2}. in other words, v3 = {x1 space = infinite line(length) * infinitely horizontally expanding (width) , x2 space is calculated the same way as x1}. If you still don't understand, you should at least know that V3 comes from the linear combination of {x1, x2}. so all combination such as addition and multiplications of {x1, x2} gets us V3.

[@bassmaiasa1312]

V3 is a plane in 3D space. I fnd it helps to compare any plane in 3D to the simple X-Y plane. You could figure out a pair of perpendicular unit vector axes of V3 that would operate just like the X-Y axis, it's just a little tricky. So every 3D point in V3 would correspond to 2D scalars of the unit vector axes. It's impossible to escape the X-Y plane by adding any two vectors in the X-Y plane because any two vectors in the X-Y plane will always have 0 as the z-value -- you can't get off the plane. By the same reason, it is impossible to escape the V3 plane by adding any two points in V3. All you've done is take the X-Y plane and rotate it in 3D space, otherwise nothing has really changed. So you can't escape V3 by X1 + X2. What's tricky for me is to understand that the same logic will apply to a subspace in any Rn. I know it does but I don't know why.

[@SpeaksYourWord]

The subspace S should be all the lines on v3 plane excluding the two lines v1 and v2. How can it be a single line?

[@bronzdragon]

S is a subspace of V3, which is a plane, right? If it were a plane minus those two lines, I could take two vectors on that line, and add them together to get onto that line. For example, [0,1,2] + [0,0,1] would give me x1, which is on v1. This is not allowed. We would have to take a subspace in a smaller dimension that v3 is (because if it were a 2d subspace, it would have to fill the full plane). The shape of that subspace is a line. Ergo, S must be a line.

[@bassmaiasa1312]

Hissss.. I was with Ms. Chen up to the last question. I got distracted because I know how to find the equation for V3, but since she hasn't mentioned the cross-product, I figured she was just asking for a generic description of the intersection -- a line in R3 that passes through the origin. (Lazy BrainFail: "There's nothing special about the X-Y plane, it's just the intersection of two planes.") I didn't look deeper into the specific properties of the X-Y plane, so I failed! I was thinking generally and failed to look for clues in the specific case. Fail!

[@murphysun4087]

So great thank you so much!

[@Abhi-qi6wm]

Could I take the entire V3 plane with no V1 and V2 for the 3rd part of Q2?

[@prasanthrajumanikonda]

Yes....the 3rd question seems wrong...as v3 is plane that has no v1 and v2...then intersection of v3 and xy plane must be null set

[@maxim_ml]

that wouldn't be a subspace

[@SpeaksYourWord]

@@maxim_ml Why not?

[@lilyrussell7900]

还好玩的，居然记了笔记

[@swethachilveri4123]

thank you

[@andrec.2935]

Excelente!

[@fvenf2kbl1xdwxz30]

Says can't draw straight. Draws lines more straight than my orientation.

[@onurdemir353]

thanks a lot :))

[@sohebsk2196]

❤