# [SOLVED]Finding Basis of the Quotient Space

#### Sudharaka

##### Well-known member
MHB Math Helper
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

##### Member
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

##### Well-known member
MHB Math Helper
Thank you so much for your answer. I learnt 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

##### Member
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

##### Well-known member
MHB Math Scholar
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

##### Well-known member
MHB Math Helper
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

##### Well-known member
MHB Math Helper
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

##### Well-known member
MHB Math Scholar
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

##### Member
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

##### Well-known member
MHB Math Helper
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.

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