Vector Subspaces 

yes correct.. the question is confusing originally

Thank you! I was so confused

Well, in 2 and 4 thats obviously not possible, but in 3 it was implied as the multiples with c1 and c2 had to guarantee that the zero vector is in the space. that its closed was obvious from the equation as it is already set up as a linear combination. in case 1 you ofc could check the axioms for subspaces, but it was trivial and this way showed a new perspective with the association of the null space.

try to take b1 b2 b3 as xyz coordinate system and if you observe in 3 the equation is just showing the xz plane and as its a plane passing through origin its a subspace and in 4 again observe its the equation of a plane parallel to xz plane and passing through y=1 and as its not a plane passing through origin so its not a subspace I hope it helps correct if I am wrong

He doesn't really explain the base logic in this video and this is a problem I had for some time myself. The math you are looking at makes complete sense, but when you look at the vector (1,1,1) compared to (2,2,4), you will notice that the 2 and the 4 have been scaled differently and that these vectors are linearly independent. This means that they are not scalar multiples of each other but they were derived using a single "scaled" vector. If you scale a vector and get a new vector that is not a scalar multiple of the original, then you have left the realm of "linear" algebra. You should expect this kind of situation every time you see a non-linear function used to describe something in linear algebra. Bottom line justification: vector*scalar = scaled or else its violated linear rules. I hope this helps :)

Also, I just noticed that it may have been confusing to some when he put the vector (2,2,2) down. He did this without doing any math just to show an example of what a correctly scaled vector should look like. The fact that you cannot get (2,2,2) from (1,1,1) using the function in question shows that the function does not represent a subspace.

@Tenzin Ogyen in num 2 he literally is trying to show that the equation number 2 doesnt obey the rule (2) which its said that b1*b2-b3 = 0 thats mean b1.b2 = b3 and he input b1 and b2 with 1 = and equation become 1*1 = 1 after that he multiply by 2 and he get ( 2 2 2) which doesnt obey the rule b1*b2 = b3 because 2*2 is not the same as 2

ya 100%

Sorry, where did he explain that?

dont know if you guys gives a shit but if you are stoned like me atm then you can watch pretty much all the latest movies and series on InstaFlixxer. Have been binge watching with my girlfriend for the last couple of weeks xD

@Xavier Axl yea, I've been watching on instaflixxer for since november myself :D

@Xavier Axl yea, I have been watching on InstaFlixxer for years myself =)

This is how I think about it. x + y - z = 0 z = x + y This equation represents a plane that includes [0 0 0]. So, it's a subspace.

because after translation from linear equation into matrix form it is already in the form Ax=0(A=[1 1-1],x[b1 b2 b3] & b=0) and we know that Ax=0 is a condition for a special type of subspace can you guess its name ? well, of course, you can because in lecture 6 we've seen the NULLSPACE and from book NULLSPACE definition is "The nullspace of a matrix A consist of all vectors x such that Ax=0 ..."

@@chotirawee I also thought the same thing..but the reason he told I could not get it...if that's a null space of 1 1 -1 why should it be subspace of R^3 vector space

​ SOUMAM BANERJEE The plane that we are talking about is (without doubts) a vector space because you can take any linear combination of any vectors inside that plane and the resultant vectors still line on the plane. Of cause it is also correct to think about it as a null space of [1 1 -1] which, yes, give us the same good old plane. For the reason that this plane is inside R^3, well, I am not 100% sure... but I do believe that we can always think about R^3 as the whole 3-dimensional space for vectors with 3 components. So, this plane is inside R^3 because the vectors inside this plane consist of 3 components.

since it satisfies (1,1,1) it has to satisfy every multiple of (1,1,1). for eg he took (2,2,2)...you may find many examples where it satisfies but to be a valid subspace it has to satisfy every multiple of them too.

The subspace needs to be closed, which means every vector of it have to meet the condition. In order words one proof is not sufficient to finish it, but one counter-example is enough to falsificate it.

the only way to prove it is to transform the origin form of the column(b1 b2 b3) into a form that satisfy C1X1+C2X2=0, which is followed by every vector of the space.

Sir, what have you done? What have you used? Can you please share them?

@@truschaoperation1539 Truscha Operation lol this is a year ago now. I honestly don't even remember what the original problem /reason for me posting the comment from a year ago ; but since you asked. Professor Leonard, Krista King, Patrick JMT for all things math. Doc Schuster, Organic Chem Tutor, and Flipping Physics for all things physics Chad's Reviews videos from ASU for Chemistry. My how much happens in a year.

Any 2 vectors in standard position that are not co-linear can be considered to represent a plane in 3d vector space. This is because they share the starting point (0,0,0) and their end points can form a line (which would create a triangle). It actually takes a bit of matrix math to find the true equation for the plane being represented (the scalar triple product to be exact) and you would need that equation in order to plot the plane. If you decide to try it, remember that one of the points needed for the scalar triple product would be (0,0,0). In this video, he is able to say that the 2 vectors can represent a plane since he has noticed that they are linearly independent. I think he only mentions it here to help visualize what would happen if the first vector happens to represent a point that does not fall in that plane. The fact that the point in question is linearly dependent means that the vector it represents must be co-planar to the others.

plot both (1 0 -1) and (1 0 1) on academo.org/demos/3d-vector-plotter/ You should see they point in two different directions. If you imagine scaling either of them by c1 and c2 then adding them (joining them tip to tail or using parellogram method) you will see that by taking all combinations (all different values for c1 and c2) you will be able to fill out the whole x-z plane. (by picking different values for c1 and c2 you should be able to show yourself you can get any vector in the x-z plane; a 3d vector where the second component is zero).

He just plugged in numbers using intuition, which would not always be possible

1) was trivial and this showed an interesting perspective. 2) the factor was arbitrary, but doesn't matter. just call it C and you get C^2 which brings the same conclusion. 3) was actually quite systematic, as the zero vector has to be in the span and thus a linear combination of the c1 and c2 vectors needed to be able produce the inverse of that first vector in the equation to fulfill the first condition for sub-spaces 4) same as 3 basically.

I think the point from the video is correct. However, with the explanation from the video, it looks like your point only true when the 1 you said is linearly independent to other two vectors. In addition, 0.5V1 + 0.5 V2 =V3. Therefore, C1v1+0.5v1+C2v1+0.5v2+=(Yes, the video skips few step to make people confused). The next step is the answer for answer 3: if you factor Ci+0.5 as the one in the video, you can construct an equation without the 1 you said. I am not sure I am correct or wrong. This is only from what I think from the video.

@@jonathansum9084 Now I see no problem at all ! c1=-1/2 and c2=-1/2 produces the origin . I don't what was wrong with me at the time!

@@mehD its okay. you gotta lose to learn

smh pause the video!!

why

I'd say he is capable of teaching but the camera man is not qualified.

I found it easy to follow