hey quick question for part d) of this question..... i am not sure of the answer but i think i know the rest of it...

hey quick question for part d) of this question..... i am not sure of the answer but i think i know the rest of it...

The columns of [math]A[/math] span [math]\mathbb{R} ^3[/math] if and only if they are linearly independent. The columns of [math]A[/math] are linearly independent if and only if, when you create a matrix, call it [math]B[/math], whose rows are the columns of [math]A[/math], and compute the reduced row echelon form of [math]B[/math], you get that the reduced row echelon form of [math]B[/math] is the 3 x 3 identity matrix.

 

Using the above information, you should be able to determine whether or not part d.) is true.

The columns of [math]A[/math] span [math]\mathbb{R} ^3[/math] if and only if they are linearly independent. The columns of [math]A[/math] are linearly independent if and only if' date=' when you create a matrix, call it [math']B[/math], whose rows are the columns of [math]A[/math], and compute the reduced row echelon form of [math]B[/math], you get that the reduced row echelon form of [math]B[/math] is the 3 x 3 identity matrix.

 

Using the above information, you should be able to determine whether or not part d.) is true.

 

 

well i found this answer, but i don't really understand what it is saying....

[1 0 0]

[0 1 0.5]

[0 0 0 ]

 

thats what i get for B in "rref", so therefore the columns of A do no span R^3 ? yes? lol

 

cheers dap

The columns of [math]A[/math] span [math]\mathbb{R} ^3[/math'] if and only if they are linearly independent.

Hmm... I am not sure about that... I think it spans iff every element of R^3 can be written as a linear combination of the column vectors. A set of vectors which span a space and on top of that are linearly independent is the basis set.

?

I was just saying if some vectors span R^3 they dont have to be necessarily linearly independent.

I was just saying if some vectors span R^3 they dont have to be necessarily linearly independent.

Yup. However, this is only possible if there are more than 3 vectors that we are dealing with. I made a minor simplification since the question only deals with 3 vectors.

ok then, lol another question to do with axioms.....

 

is say

a x 0 = 0

an axiom or something? because if a x 1 = 1 x a = a is one, i figure a x 0 = 0 has to be some kind of axiom???

ok then' date=' lol another question to do with axioms.....

 

is say

a x 0 = 0

an axiom or something?[/quote']Nope. It's a result that can be derived from the axioms of the real numbers.

 

because if a x 1 = 1 x a = a is one

This is one of the axioms of the real numbers.

 

i figure a x 0 = 0 has to be some kind of axiom???

Nope. I can provide a list of the axioms for the real numbers if you care to know what they are.

thanks for the offer but i've already got a list... how do you derive it? or more importantly, if i am trying to prove ac=bc or something like that using the axioms, can i just say at a step that a x 0 = 0? like i don't have to give a reason, eg. because of axoim * or something like that?

thanks for the offer but i've already got a list... how do you derive it?

Derivation:

 

Consider [math]0 \cdot x + 0 \cdot x[/math]

 

By the distributive axiom, [math]x(y+z)=xy+xz, \forall x,y,z \in \mathbb{R}[/math].

 

Therefore, [math]0 \cdot x + 0 \cdot x = (0+0)\cdot x = 0 \cdot x[/math].

 

One of the consequences of the axioms is that [math]x+y=x \implies y = 0[/math], therefore [math]0 \cdot x = 0[/math]

or more importantly, if i am trying to prove ac=bc or something like that using the axioms, can i just say at a step that a x 0 = 0? like i don't have to give a reason, eg. because of axoim * or something like that?

Yes, as long as whomever is grading your assignment has access to the list as well, otherwise they won't know which axiom you are referring to.

allright cheers Dapthar, that makes sense to me

 

thanks a million (yet again!)