# Finding basis problem

• Dec 11th 2009, 05:27 PM
firebio
Finding basis problem
Given S= [a,b,c,d] belongs to R4, a not = 0. Find a basis that is a subset of s.

I am unsure what the a not = 0 means.

Would it just be a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)+d(0,0,0,1) a not = 0.

If not, can someone give me an example of the basis.

• Dec 11th 2009, 05:43 PM
dedust
Quote:

Originally Posted by firebio
Given S= [a,b,c,d] belongs to R4, a not = 0. Find a basis that is a subset of s.

I am unsure what the a not = 0 means.

Would it just be a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)+d(0,0,0,1) a not = 0.

If not, can someone give me an example of the basis.

it is clear that $v_{1}=[a,0,0,0],v_{2}=[a,1,0,0],v_{3}=[a,0,1,0],v_{4}=[a,0,0,1]$ are subset of $S$.

now, we need to prove that $v_{1},v_{2},v_{3},v_{4}$ are linearly independent.
let $\lambda_{1}v_{1}+\lambda_{2}v_{2}+\lambda_{3}v_{3} +\lambda_{4}v_{4}=0$,
then we get $\lambda_{2}=\lambda_{3}=\lambda_{4}=0$
and $a\lambda_{1} +a\lambda_{2}+a\lambda_{3}+a\lambda_{4}=0$
because $\lambda_{2}=\lambda_{3}=\lambda_{4}=0$, then $a\lambda_{1}=0$.
because $a \not= 0$, then $\lambda_{1}=0$
hence, $v_{1},v_{2},v_{3},v_{4}$ are linearly independent.
• Dec 11th 2009, 11:26 PM
Raoh
Quote:

Originally Posted by dedust
it is clear that $v_{1}=[a,0,0,0],v_{2}=[a,1,0,0],v_{3}=[a,0,1,0],v_{4}=[a,0,0,1]$ are subset of $S$.

now, we need to prove that $v_{1},v_{2},v_{3},v_{4}$ are linearly independent.
let $\lambda_{1}v_{1}+\lambda_{2}v_{2}+\lambda_{3}v_{3} +\lambda_{4}v_{4}=0$,
then we get $\lambda_{2}=\lambda_{3}=\lambda_{4}=0$
and $a\lambda_{1} +a\lambda_{2}+a\lambda_{3}+a\lambda_{4}=0$
because $\lambda_{2}=\lambda_{3}=\lambda_{4}=0$, then $a\lambda_{1}=0$.
because $a \not= 0$, then $\lambda_{1}=0$
hence, $v_{1},v_{2},v_{3},v_{4}$ are linearly independent.

probably you should check your vectors(Wondering)
• Dec 12th 2009, 04:49 AM
HallsofIvy
Quote:

Originally Posted by firebio
Given S= [a,b,c,d] belongs to R4, a not = 0. Find a basis that is a subset of s.

I am unsure what the a not = 0 means.

Would it just be a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)+d(0,0,0,1) a not = 0.

If not, can someone give me an example of the basis.

That simply means vectors of the form [a, b, c, d] where a is not 0. But that also means that [0, 0, 0, 0] is not in the set nor is the scalar product of 0 with any member of the set. This is NOT a subspace of $R^4$ and so does not have a basis!
• Dec 12th 2009, 05:51 AM
dedust
Quote:

Originally Posted by HallsofIvy
That simply means vectors of the form [a, b, c, d] where a is not 0. But that also means that [0, 0, 0, 0] is not in the set nor is the scalar product of 0 with any member of the set. This is NOT a subspace of $R^4$ and so does not have a basis!

i thought the problem said that $S$ is a subset of $R^4$ and we need to find the basis for vector space $S$. I also didn't check whether it is closed under scalar multiplication or not.

(Whew)
• Dec 12th 2009, 06:45 AM
Raoh
Quote:

Originally Posted by HallsofIvy
That simply means vectors of the form [a, b, c, d] where a is not 0. But that also means that [0, 0, 0, 0] is not in the set nor is the scalar product of 0 with any member of the set. This is NOT a subspace of $R^4$ and so does not have a basis!

if $a$ is any real,what would be the basis ?
• Dec 12th 2009, 07:46 AM
dedust
Quote:

Originally Posted by Raoh
if $a$ is any real,what would be the basis ?

if $a$ is any real, then $S = R^4$ and any basis for $R^4$ is also basis for $S$ (Rofl)(Rofl)
• Dec 12th 2009, 07:52 AM
Raoh
Quote:

Originally Posted by dedust
if $a$ is any real, then $S = R^4$ and any basis for $R^4$ is also basis for $S$ (Rofl)(Rofl)

but when you answered his question you assumed that $a$ is any real,right ?
• Dec 12th 2009, 10:32 AM
firebio
Quote:

Originally Posted by HallsofIvy
That simply means vectors of the form [a, b, c, d] where a is not 0. But that also means that [0, 0, 0, 0] is not in the set nor is the scalar product of 0 with any member of the set. This is NOT a subspace of $R^4$ and so does not have a basis!

That was what i thought at first too... but i didnt think it was suppose to be the kind of question with a strange twist to it..

And a is any real number not = to 0?
• Dec 13th 2009, 06:43 AM
HallsofIvy
Someone corrected my interpretation of this question. I though it was asking for a basis for the "subspace" S but in fact it is asking for a basis for $R^4$ consisting of vectors in the set S.

Try this: the standard basis for $R^4$ is <1, 0, 0, 0>, <0, 1, 0, 0>, <0, 0, 1, 0>, and <0, 0, 0, 1> where the last three are not from set S. Okay, just change the first component: try <1, 0, 0, 0>, <1, 1, 0, 0>, <1, 0, 1, 0>, and <1, 0, 0, 1>. Is that a basis for $R^4$? There are four vectors in it; are they independent?
• Dec 13th 2009, 07:13 AM
Raoh
Quote:

Originally Posted by HallsofIvy
Someone corrected my interpretation of this question. I though it was asking for a basis for the "subspace" S but in fact it is asking for a basis for $R^4$ consisting of vectors in the set S.

Try this: the standard basis for $R^4$ is <1, 0, 0, 0>, <0, 1, 0, 0>, <0, 0, 1, 0>, and <0, 0, 0, 1> where the last three are not from set S. Okay, just change the first component: try <1, 0, 0, 0>, <1, 1, 0, 0>, <1, 0, 1, 0>, and <1, 0, 0, 1>. Is that a basis for $R^4$? There are four vectors in it; are they independent?

i think the standard basis of $R^4$ is also a basis of $S$.
no ?
• Dec 13th 2009, 02:12 PM
dedust
Quote:

Originally Posted by Raoh
but when you answered his question you assumed that $a$ is any real,right ?

any real but not 0
Quote:

Originally Posted by HallsofIvy
Someone corrected my interpretation of this question. I though it was asking for a basis for the "subspace" S but in fact it is asking for a basis for $R^4$ consisting of vectors in the set S.

Try this: the standard basis for $R^4$ is <1, 0, 0, 0>, <0, 1, 0, 0>, <0, 0, 1, 0>, and <0, 0, 0, 1> where the last three are not from set S. Okay, just change the first component: try <1, 0, 0, 0>, <1, 1, 0, 0>, <1, 0, 1, 0>, and <1, 0, 0, 1>. Is that a basis for $R^4$? There are four vectors in it; are they independent?

yes,they are linearly independent, as I proved before