Linear Algebra 6f: Linear Dependence Example 2 - Geometric Vectors

[MaxPicAxe]

Honestly, I love how you explain all this before even touching matrices.

[xoppa]

Nice video. Can't wait to watch your future videos, on vector calculus and complex analysis, as well as linear algebra part 2. I am watching the full playlist on linear algebra to fill some significant holes in my mathematical background and because i love the subject - linear algebra ties together the geometric and algebraic world, both very important subjects in their own right. You are a wonderful teacher. Thanks for making higher mathematics accessible to a wider audience.

[angelmendez-rivera]

Actually, the two expressions are in fact equal. The reason both d=b+c+α(a+b+c) and d=-a+α(a+b+c) are valid is because -a=b+c, and this last claim is true because a+b+c=0 is true. By subtracting a on both sides of the equation, the latter becomes the former.

Additionally, if we take both forms of expressing and decomposing d and we write the set of possible linear combinations in terms of explicit coefficients, both expressions will get the exact same coefficients and are thus exactly equal.

Here is my conjecture: both decompositions of d can only be valid if the expressions that express the set of possible linear combinations are equivalent. The reason is that, if they aren't, then there is at least one value of α for which b+c+α(a+b+c) does not =-a+α(a+b+c), which means that neither expression contains all of the possible linear combinations in the set. But that is a contradiction because we're assuming both expressions do capture the entire set of possible linear combinations.

[smalljbug]

I'm not sure that's a proper use of the equal sign. by themselves, each expression captures the entire set, but the equal sign suggests that alpha refers to the same constant in both. why not just use beta in the second expression? then alpha and beta are both arbitrary constants, and the equal sign is valid for some appropriate choices of alpha and beta. Great series!

[graceliu]

Hi Dr. Grinfeld, thank you for your videos on Linear Algebra. I have been struggling learning it -- I feel everything is talking about the same thing and when I am given a problem I never know how to approach it. I think your approach is filling in the holes I have. And you passion about is contagious!
I have a question about this first deposition question: are vector a, b, c are in a plane? visually they look like the unit vector in 3D.

[alojzybabel]

I think I have a better name for your "fancy zero": let's call it *"composite zero"* :)
Why? Because it is a zero vector expressed as a composition of other non-zero vectors, and if you use it in an expression, it "contaminates" it with its components.