Finding a basis

**paupsers** · June 1st 2009, 08:35 PM

Can anyone verify that I am doing this correctly?

I have to find a basis for

$W=\{(a_{1}, a_{2}, a_{3}, a_{4}, a_{5} \in F^{5} | a_{1}-a_{3}-a_{4}=0\}$

So, does this mean that $a_{2}$ and $a_{5}$ can be any numbers?

I began by setting them both equal to one, giving me (0, 1, 0, 0 ,0) and (0, 0, 0, 0, 1)

Then, by letting $a_{1}=1$ , this gave me $1=a_{3}+a_{4}$ . Then choosing $a_{3}=0$ or $a_{3}=1$ , I obtain $a_{4}=1$ or $a_{4}=0$ , respectively.

So that gave me the vectors (1, 0, 0, 1, 0) and (1, 0, 1, 0, 0).

Repeating this process for a different $a$ gave me
(0, 0, 1, 1, 0).

That's a total of 5 vectors, which is how many the basis should have, correct? Can anyone verify this solution? Is this a correct method to use? I just feel weird "choosing" the a's to be 1 or 0 and stuff.

**NonCommAlg** · June 2nd 2009, 09:13 AM

Originally Posted by paupsers

Can anyone verify that I am doing this correctly?

I have to find a basis for

$W=\{(a_{1}, a_{2}, a_{3}, a_{4}, a_{5} \in F^{5} | a_{1}-a_{3}-a_{4}=0\}$

So, does this mean that $a_{2}$ and $a_{5}$ can be any numbers? yes!

I began by setting them both equal to one, giving me (0, 1, 0, 0 ,0) and (0, 0, 0, 0, 1)

Then, by letting $a_{1}=1$ , this gave me $1=a_{3}+a_{4}$ . Then choosing $a_{3}=0$ or $a_{3}=1$ , I obtain $a_{4}=1$ or $a_{4}=0$ , respectively.

So that gave me the vectors (1, 0, 0, 1, 0) and (1, 0, 1, 0, 0).

Repeating this process for a different $a$ gave me
(0, 0, 1, 1, 0).

That's a total of 5 vectors, which is how many the basis should have, correct? Can anyone verify this solution? Is this a correct method to use? I just feel weird "choosing" the a's to be 1 or 0 and stuff.

no, it's not quite correct. from the relation $a_1-a_3-a_4=0$ you get $a_1=a_3+a_4.$ so an element of $W$ is in the form:

$(a_3+a_4,a_2,a_3,a_4,a_5)=a_2e_1+a_3e_2+a_4e_3+a_5 e_4,$ where $e_1=(0,1,0,0,0), \ e_2=(1,0,1,0,0), \ e_3=(1,0,0,1,0),$ and $e_4=(0,0,0,0,1).$

the set $\{e_1,e_2,e_3,e_4 \}$ is a basis for $W.$

**HallsofIvy** · June 2nd 2009, 09:45 AM

Originally Posted by paupsers

Can anyone verify that I am doing this correctly?

I have to find a basis for

$W=\{(a_{1}, a_{2}, a_{3}, a_{4}, a_{5} \in F^{5} | a_{1}-a_{3}-a_{4}=0\}$

So, does this mean that $a_{2}$ and $a_{5}$ can be any numbers?

I began by setting them both equal to one, giving me (0, 1, 0, 0 ,0) and (0, 0, 0, 0, 1)

Then, by letting $a_{1}=1$ , this gave me $1=a_{3}+a_{4}$ . Then choosing $a_{3}=0$ or $a_{3}=1$ , I obtain $a_{4}=1$ or $a_{4}=0$ , respectively.

So that gave me the vectors (1, 0, 0, 1, 0) and (1, 0, 1, 0, 0).

Repeating this process for a different $a$ gave me
(0, 0, 1, 1, 0).

That's a total of 5 vectors, which is how many the basis should have, correct? Can anyone verify this solution? Is this a correct method to use? I just feel weird "choosing" the a's to be 1 or 0 and stuff.

No. A basis for all of $R^5$ must have 5 vectors but this is a subspace, not all of $R^5$ . Typically each equation restricting values reduces the dimension by 1. Since this space is restricted by one equation, we would expect the dimension to be 4 and expect a basis to contain four vectors.
Yes, $a_2$ and $a_5$ have no restrictions so (0, 1, 0, 0, 0) and (0, 0, 0, 0, 1) are basis vectors for it. NonCommAlg solved the single equation for $a_1$ . You could, of course, solve for either $a_3$ or $a_4$ as functions of the other two.

For example, solving for $a_3= a_1+ a_4$ , we can see that if $a_1= 1$ and $a_4= 0$ , then $a_3= 1+ 0= 1$ so (1, 0, 1, 0, 0) is another basis vector. If $a_1= 0$ and $a_4= 0$ , then $a_3= 0+ 1= 1$ so (0, 0, 1, 1, 0) is a fourth basis vector. Those four vectors, {(0, 1, 0, 0, 0), (0, 0, 0, 0, 1), (1, 0, 1, 0, 0), (0, 0, 1, 1, 0)} form a basis for this subspace slightly different from the one NonCommAlgebra gave.

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