Basis for a Set of Vectors 

did you?

@@mnirapp4801 yeah he did

@@RubLox_Live how do you know?

@@gracehardo7436 they married each other and one is wife and the other is husband. I am their son, my names is PatrickJMT.

@@Fire1nside you are 6 years old mate, wth are you studyin basis for?

Congratulations Graduation

damn how did your midterm go, its been 8 years now lol

@@asfarsarfaraz3510 Bro got ignored after 8 years... still counting

I need to understand linear algebra can you hook me up with some videos?

Spanning means at least one solution or consistent and not spanning means no solution or inconsistent. A equation that's Inconsistent means 0 = nonzero.

To add onto to what the other guy say, when you have infinitely many solutions, that means one of your equations is a multiple of the other.

Lots of classes don't teach determinants until the end for this reason. It makes it all too easy and doesn't teach what is exactly going on like going through the tedious row reduction algebra does.

+Robert Wilson III thank you

response to 2), that's exactly what I was thinking. Not sure why he did that. Seems redundant.

a bit late i suppose but might help others.. 1) It does not span R3. The explanation is simple: if y!=0, then you cannot get *every* vector in R3 using those vectors.

+✰Leanz✰ I think it is because if you show that the matrix has only the trivial solution then it is linearly independent

He shows the algebraic definition of linear independence. It's entirely possible to work with kernels, images, rank dimension, basis, null space, column space, linear independence and dependence and spans without using matrices at all. It's just that matrix notation was supposed to make manipulating complicated systems of linear equations simpler. This isn't to say that it's easy though. I personally find this stuff to be extremely tedious and challenging. I prefer analysis!

You can set up a matrix and get to RREF too, he's just doing another method :)

You should try linear subspace by khan academy...he is also very good....

+Cole Calendar for the last part yes you can. then, you don't have to do any back substitution. just use caution when manipulating the rows (meaning "keep track" of what you have done to b_1, b_2, b_3; or as he called it x,y,z).

did you pass your finals?

+jazzj2 if linearly dependent, the set of vectors needs to be simplified to be a basis. A basis is a minimally spanning set!

+User name I'm kind of confused.. so all vectors can be linearly independent ?

To determine if a set of vectors are L. I. you have to row reduce the matrix. If you get the identity matrix after reducing, then that set I'd linearly independent

User name If you row reduce and the matrix is not an identity matrix, then theres a vector in a set that is not linearly independent ? so how would you find the vectors in that set that are L.I ? would you look at the leading entrys ?

The columns of the pivots are the L.I. vectors (each vector that is L.I will have a 1 in its collumn). So Yes; this L.I. set is also the basis

Hey man, I need to learn linear algebra can you hook me up with some channels or linsk?

no, you can have a linearly independent set that is not in R3 and you can have a linearly dependent set in R3.

bro, how. determinants are only for square matrices

what does this mean sir