MHB Finding Basis of the Quotient Space

[Sudharaka]

Hi everyone, :)

This seems like a pretty simple question, but up to now I haven't found a method to solve it. Hope you can provide me a hint. :)

Problem:

Let \(V\) be a space with basis \(B=\{b_1,\,b_2,\,b_3,\,b_4,\,b_5\},\,U\) the subspace spanned by \(u_1=b_1+b_2+b_3+b_4+b_5\), \(u_2=b_2-b_3+b_4-b_5\). Find a basis of \(V/U\).

 

[Turgul]

You want to find a generating set for a quotient. It suffices to find a generating set for the original space. That is, if you have a generating set for $V$, the image of that set under the quotient map will be a generating set for $V/U$.

The problem is that you don't just want a generating set, you want a basis, and the image of a basis of $V$ in $V/U$ might have dependence relations (actually it surely will since a basis of $V$ requires 5 elements and you are quotienting by a 2D space, so the quotient will be 3 dimensional).

But suppose you picked a basis of $V$ which included a basis for $U$. Then what could you say about the image of the elements of such a basis?

 

[Sudharaka]

Thank you so much for your answer. I learned a lot from it. :)

Sorry, but I thought \(V/U\) stands for the relative compliment, not the quotient group. Anyway now I see that I have interpreted the problem incorrectly. The notation \(V/U\) denotes the factor group (quotient group). I will change the title accordingly.

You want to find a generating set for a quotient. It suffices to find a generating set for the original space. That is, if you have a generating set for $V$, the image of that set under the quotient map will be a generating set for $V/U$.

The problem is that you don't just want a generating set, you want a basis, and the image of a basis of $V$ in $V/U$ might have dependence relations (actually it surely will since a basis of $V$ requires 5 elements and you are quotienting by a 2D space, so the quotient will be 3 dimensional).

But suppose you picked a basis of $V$ which included a basis for $U$. Then what could you say about the image of the elements of such a basis?

Is it that the images of the three elements of our basis of \(V\) that are not in the basis of \(U\) become the basis for \(V/U\). Sorry if this looks stupid. I have little background knowledge about these stuff. :)

 

[Turgul]

No worries, glad I was helpful. It is worth noting that "/" is almost always used for quotient in this context. Either "\" or "-" is typically used for set compliments, but beware because the compliment of a vector space in another is NEVER a vector space (can't have 0, for example), so looking for a basis would be a nonsensical task.

Indeed you are correct, the vectors which are not collapsed to 0 will form a new basis for the quotient. Certainly you will have the correct number of vectors (the dimension of $U$ is given by the number of basis elements generating it, as with $V$) and they must span the quotient since the original basis spanned $V$ and the quotient map is surjective, so they must form a basis of the quotient.

It is a worthwhile exercise to decide why they are linearly independent directly (without knowing the dimension of the quotient space, as I have done).

 

[Deveno]

Re: Finding Basis of the Relative Compliment

There are a couple of ways to approach this, Turgul's post suggest the following:

Let $B' = \{u_1,u_2,b_3,b_4,b_5\}$.

Since $b_1 = u_1 - u_2 - 2b_3 - 2b_5$ and:

$b_2 = u_2 + b_3 - b_4 + b_5$.

It should be clear this set spans $V$, and since it has 5 elements must therefore be a basis for $V$ (this saves us the rigamarole of showing linear independence).

Applying the quotient map to this set gives us the set:

$\{b_3 + U, b_4 + U, b_5 + U\}$ and it suffices to show this is a basis for $V/U$ (the first two elements map to $u_1 + U = u_2 + U = 0 + U = U$, the 0-vector of $V/U$).

Since $\text{dim}(V/U) = \text{dim}(V) - \text{dim}(U) = 5 - 2 = 3$ it suffices to show that this set is LI.

So suppose that (for $c_i \in F$) we have:

$c_1(b_3 + U) + c_2(b_4 + U) + c_3(b_5 + U) = U$

This is equivalent to saying:

$c_1b_3 + c_2b_4 + c_3b_5 \in U$, that is:

$c_1b_3 + c_2b_4 + c_3b_5 = a_1(b_1 + b_2 + b_3 + b_4 + b_5) + a_2(b_2 - b_3 + b_4 - b_5)$, or equivalently:

$-a_1b_1 - (a_1 + a_2)b_2 + (c_1 - a_1 + a_2)b_3 + (c_2 - a_1 - a_2)b_4 + (c_3 - a_1 + a_2)b_5 = 0$.

By the LI of $B$, we have:

$a_1 = 0$
$a_1 + a_2 = 0$
$c_1 - a_1 + a_2 = 0$
$c_2 - a_1 - a_2 = 0$
$c_3 - a_1 + a_2 = 0$

From the first and second equation, we see that $a_1 = a_2 = 0$, which then shows that all the $c_i = 0$.

********

One could just form the set $\{b_1 + U, b_2 + U,b _3 + U, b_4 + U, b_5 + U\}$ which clearly spans $V/U$, and try to form a basis out of it, somehow.

Since $b_1 + b_2 + b_3 + b_4 + b_5 \in U$, this means:

$b_1 + U = -b_2 + U - (b_3 + U) - (b_4 + U) - (b_5 + U)$ so we can safely remove $b_1 + U$ from the set above without changing the span.

Similarly, since:

$b_2 - b_3 + b_4 - b_5 \in U$, we have:

$b_2 + U = b_3 + U - (b_4 + U) + b_5 + U$, so we may also likewise remove $b_2 + U$ without affecting the span.

Since $\{b_3 + U,b_4 + U,b_5 + U\}$ spans $V/U$ and we have shown LI above, we therefore have a basis.

 

[Sudharaka]

No worries, glad I was helpful. It is worth noting that "/" is almost always used for quotient in this context. Either "\" or "-" is typically used for set compliments, but beware because the compliment of a vector space in another is NEVER a vector space (can't have 0, for example), so looking for a basis would be a nonsensical task.

Indeed you are correct, the vectors which are not collapsed to 0 will form a new basis for the quotient. Certainly you will have the correct number of vectors (the dimension of $U$ is given by the number of basis elements generating it, as with $V$) and they must span the quotient since the original basis spanned $V$ and the quotient map is surjective, so they must form a basis of the quotient.

It is a worthwhile exercise to decide why they are linearly independent directly (without knowing the dimension of the quotient space, as I have done).

Thanks again, I think I am getting the idea now. Let me write down my answer. Correct me if I am wrong. :)

We shall try to construct a basis of \(V\) that includes \(u_1\mbox{ and }u_2\). Take and \(u\in U\). Then,

\[u=\alpha u_1+\beta u_2=\alpha b_1+(\alpha+\beta)b_2+(\alpha-\beta)b_3+(\alpha+\beta)b_4+(\alpha-\beta)b_5\]

Now if \(b_1\in U\) then \(\alpha=1,\,\alpha+\beta=0\mbox{ and }\alpha-\beta=0\) which is impossible. Therefore \(b_1\notin U=\left<u_1,\,u_2\right>\). So we extend the basis by adding \(b_1\). If \(b_2\in \left\{u_1,\,u_2,\,b_1\right\}\) then the coefficient of \(b_2\) is equal to \(1\) and the coefficient of \(b_4\) is equal to zero, which is again impossible. Therefore we also add \(b_2\) to the basis. Similarly if \(b_3\in\left\{u_1,\,u_2,\,b_1,\,b_2\right\}\) then then the coefficient of \(b_3\) is equal to \(1\) and the coefficient of \(b_5\) is equal to zero; impossible. Therefore finally we obtain the basis \(\left\{u_1,\,u_2,\,b_1,\,b_2,\,b_3\right\}\).

Hence a basis for \(V/U\) is \(\left\{b_1+U,\,b_2+U,\,b_3+U\right\}\).

 

[Sudharaka]

Re: Finding Basis of the Relative Compliment

There are a couple of ways to approach this, Turgul's post suggest the following:

Let $B' = \{u_1,u_2,b_3,b_4,b_5\}$.

Since $b_1 = u_1 - u_2 - 2b_3 - 2b_5$ and:

$b_2 = u_2 + b_3 - b_4 + b_5$.

It should be clear this set spans $V$, and since it has 5 elements must therefore be a basis for $V$ (this saves us the rigamarole of showing linear independence).

Applying the quotient map to this set gives us the set:

$\{b_3 + U, b_4 + U, b_5 + U\}$ and it suffices to show this is a basis for $V/U$ (the first two elements map to $u_1 + U = u_2 + U = 0 + U = U$, the 0-vector of $V/U$).

Since $\text{dim}(V/U) = \text{dim}(V) - \text{dim}(U) = 5 - 2 = 3$ it suffices to show that this set is LI.

So suppose that (for $c_i \in F$) we have:

$c_1(b_3 + U) + c_2(b_4 + U) + c_3(b_5 + U) = U$

This is equivalent to saying:

$c_1b_3 + c_2b_4 + c_3b_5 \in U$, that is:

$c_1b_3 + c_2b_4 + c_3b_5 = a_1(b_1 + b_2 + b_3 + b_4 + b_5) + a_2(b_2 - b_3 + b_4 - b_5)$, or equivalently:

$-a_1b_1 - (a_1 + a_2)b_2 + (c_1 - a_1 + a_2)b_3 + (c_2 - a_1 - a_2)b_4 + (c_3 - a_1 + a_2)b_5 = 0$.

By the LI of $B$, we have:

$a_1 = 0$
$a_1 + a_2 = 0$
$c_1 - a_1 + a_2 = 0$
$c_2 - a_1 - a_2 = 0$
$c_3 - a_1 + a_2 = 0$

From the first and second equation, we see that $a_1 = a_2 = 0$, which then shows that all the $c_i = 0$.

********

One could just form the set $\{b_1 + U, b_2 + U,b _3 + U, b_4 + U, b_5 + U\}$ which clearly spans $V/U$, and try to form a basis out of it, somehow.

Since $b_1 + b_2 + b_3 + b_4 + b_5 \in U$, this means:

$b_1 + U = -b_2 + U - (b_3 + U) - (b_4 + U) - (b_5 + U)$ so we can safely remove $b_1 + U$ from the set above without changing the span.

Similarly, since:

$b_2 - b_3 + b_4 - b_5 \in U$, we have:

$b_2 + U = b_3 + U - (b_4 + U) + b_5 + U$, so we may also likewise remove $b_2 + U$ without affecting the span.

Since $\{b_3 + U,b_4 + U,b_5 + U\}$ spans $V/U$ and we have shown LI above, we therefore have a basis.

Thanks so much for your detailed reply. This gives the complete picture of the question and how to think about it. :)

 

[Deveno]

Yes, that will work as well...recall that the basis elements themselves are not invariants of a given space, only the number of basis elements.

In fact:

$b_5 + U = -(b_1 + U) - (b_2 + U) - (b_3 + U) + (b_4 + U)$

which shows that:

$\text{span}(b_1+U,b_2+U,b_3+U,b_4+U,b_5+U) = \text{span}(b_1+U,b_2+U,b_3+U,b_4+U)$

However, showing $b_4 + U \in \text{span}(b_1+U,b_2+U,b_3+U)$ is not as straight-forward, which is why I elected not to go that route.

 

[Turgul]

You have to be a little bit careful. The argument you give actually just shows that each $b_1,b_2,b_3$ are each individually not in the span of $u_1,u_2$, but it does not show that $b_2$ is not in the span of $u_1,u_2,b_1$. Using your method, to show $b_2$ is independent of $u_1,u_2,b_1$, you need to show that $b_2$ cannot be written as $\alpha u_1 + \beta u_2 + \gamma b_1$. Then you would have to add the next step to show $b_3$ cannot be written in terms of $u_1,u_2,b_1,b_2$.

A very similar task is to show all at once that $\{u_1,u_2,b_1,b_2,b_2\}$ is a linearly independent set, which just requires check that if $\alpha u_1 + \beta u_2 + \gamma b_1 + \delta b_2 + \sigma b_3 = 0$, then $\alpha = \beta = \gamma = \delta = \sigma = 0$. Once you plug in the expressions for $u_1$ and $u_2$, you will get a pretty easy system of linear equations in the Greek letters to solve, and it goes very similar to showing $b_1$ is not in the span of $u_1$ and $u_2$.

But Deveno has provided a pretty comprehensive solution as well.

 

[Sudharaka]

You have to be a little bit careful. The argument you give actually just shows that each $b_1,b_2,b_3$ are each individually not in the span of $u_1,u_2$, but it does not show that $b_2$ is not in the span of $u_1,u_2,b_1$. Using your method, to show $b_2$ is independent of $u_1,u_2,b_1$, you need to show that $b_2$ cannot be written as $\alpha u_1 + \beta u_2 + \gamma b_1$. Then you would have to add the next step to show $b_3$ cannot be written in terms of $u_1,u_2,b_1,b_2$.

A very similar task is to show all at once that $\{u_1,u_2,b_1,b_2,b_2\}$ is a linearly independent set, which just requires check that if $\alpha u_1 + \beta u_2 + \gamma b_1 + \delta b_2 + \sigma b_3 = 0$, then $\alpha = \beta = \gamma = \delta = \sigma = 0$. Once you plug in the expressions for $u_1$ and $u_2$, you will get a pretty easy system of linear equations in the Greek letters to solve, and it goes very similar to showing $b_1$ is not in the span of $u_1$ and $u_2$.

But Deveno has provided a pretty comprehensive solution as well.

I understand you perfectly now. Well I thought of the whole summation,

\[\alpha u_1+\beta u_2=\alpha b_1+(\alpha+\beta)b_2+(\alpha-\beta)b_3+(\alpha+\beta)b_4+(\alpha-\beta)b_5+\gamma b_1\]

when I wrote, "If \(b_2\in \left\{u_1,\,u_2,\,b_1\right\}\) then the coefficient of \(b_2\) is equal to \(1\) and the coefficient of \(b_4\) is equal to zero, which is again impossible". But due to being lazy and not writing down every detail, you might have misunderstood it. :p

Thanks very much for all your help. You have been wonderful in providing insight to this question. I learned a great deal from it. :)