lineaar algebra question

[Sarahisme]

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...

[Dapthar]

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.

[Sarahisme]

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....

[Sarahisme]

[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

[bloodhound]

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.

[Sarahisme]

?

[bloodhound]

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

[Dapthar]

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.

[Sarahisme]

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???

[Dapthar]

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.

[Sarahisme]

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?

[Dapthar]

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.

[Sarahisme]

allright cheers Dapthar, that makes sense to me

 

thanks a million (yet again!)