The vector [69, 69, 69] can moved to any point in R3 as vectors do not have a fixed position. Therefor, [69, 69, 69] can trace out all of R3. Therefor, [69, 69, 69] spans R3.
>>8004100
show me how to write an arbitrary vector in [math]\mathbb{R}^3[/math] in terms of a scalar multiplied by the 69 vector and I'll give you $100000000000000000000000000000000
>>8004110
[1, 1, 1] is an arbitrary vector
[69, 69, 69] / 69 = [1, 1, 1]
Can you set up a coefficient matrix to prove this?
>>8004100
The fact you can move it through the R3 doesn't imply it spans the R3.
And if something spans a given space, it has to contain the 0 vector as well... { [x, x, x] | x = 69 } doesn't contain the 0 vector
Nor is it closed under addition and multiplication
>>8004100
R3 doesn't all point in the same direction as [69,69,69] figlord
>>8004374
No, you're fucking retarded.
the span of [69, 69, 69] does contain the 0 vector. However, the rank of the matrix
[69, 69, 69] (i.e. the dimension of the space spanned by the vector) is 1.
>>8004147
>[1, 1, 1] is an arbitrary vector
No it isn't.
>>8004390
Why does it contain the zero vector? [0, 0, 0] isn't in OPs space, because his vectors can only consist of 69s
Or OP should've given a better definition of his space...
>>8004397
"span" means "take all the linear combination of the basis vectors". So he implied linear combinations (notably c = 0) when he said [69, 69, 69] spans R3.
Shit just spans a 1D line in R3 bro, are you seriously this stupid?
>>8004397
Span must include [math]\overrightarrow{0}[/math], as it can be formed as a linear combination of [math][69,69,69][/math].
