# Basis of ker L --> Basis of vector space?

• Sep 16th 2011, 02:14 PM
mathminor827
Basis of ker L --> Basis of vector space?
Hi all ---

I'm trying to prove the rank-nullity theorem and am just stuck on the first line of the proof. I'm using the first proof on Wikipedia --- Rank ---

If $L : V \to W$ is a linear transformation from an $n$-dimensional vector space $V$ to a vector space $W$, then

$\dim \ker L + \dim \ker L = \dim V$.

First proof on Wikipedia --- I've changed it a little ---

Assume $\dim \ker L = m$ and $\dim V = n$

Also assume { $u_1, u_2, ..., u_m$} forms a basis for $\ker L.$

We can extend this to form a basis for $V$: { $u_1, u_2, ..., u_m, w_1, .., w_n$}

etc...

My problem ---

If { $u_1, u_2, ..., u_m$} forms a basis for $\ker L,$, I think I see why { $u_1, u_2, ..., u_m, w_1, .., w_n$} is a basis for $\ker L,$ <-- not a typo.

The last word isn't a typo. I'm not sure if this is right ---

If { $u_1, u_2, ..., u_m$} is a basis for $\ker L$, I ??can?? extend this to

{ $u_1, u_2, ..., u_m, w_1, .., w_n$} --- a basis for $\ker L$???

This is why I think this --- Because the { $u_i$}s form a basis already, then adding { $w_1, .., w_n$} doesn't change that because these { $w_i$}s are kind of unnecessary.

Original question --- But how does { $u_1, u_2, ..., u_m, w_1, .., w_n$} as a basis for $\ker L$ suddenly turn into a basis for $V$? I can't figure out this jump? How can something represent $\ker L$ now represent vector space $V$?

Thanks a lot ---
• Sep 16th 2011, 07:20 PM
FernandoRevilla
Re: Basis of ker L --> Basis of vector space?
Please, read it again. We have a basis $\{u_1,\ldots ,u_m, w_1,\ldots ,w_n \}$ of $V$ (not of $\ker L$) and we need to prove that $\{L(w_1),\ldots,L(w_n)\}$ is a basis of $\textrm{Im}L$ .
• Sep 16th 2011, 07:53 PM
mathminor827
Re: Basis of ker L --> Basis of vector space?
Quote:

Originally Posted by FernandoRevilla
Please, read it again. We have a basis $\{u_1,\ldots ,u_m, w_1,\ldots ,w_n \}$ of $V$ (not of $\ker L$) and we need to prove that $\{L(w_1),\ldots,L(w_n)\}$ is a basis of $\textrm{Im}L$ .

Hi Professor ---

But I'm stuck on exactly why $\{u_1,\ldots ,u_m, w_1,\ldots ,w_n \}$ is a basis for $V$?

The proof assumed ---
$\{u_1,\ldots ,u_m \}$ is a basis for $\ker L$.

How did they get from this basis to $\{u_1,\ldots ,u_m, w_1,\ldots ,w_n \}$ --- a basis for $V$?

What I meant in my post is --- I edited out the confusion now --- is that if $\{u_1,\ldots ,u_m\}$ is a basis for $\ker L$, then it is possible? to extend this to ---
$\{u_1,\ldots ,u_m, w_1,\ldots ,w_n \}$ is a basis for $\ker L$. But I'm not sure about this.

But I DON'T see how $\{u_1,\ldots ,u_m, w_1,\ldots ,w_n \}$ is a basis for $V$?

Thanks ---
• Sep 16th 2011, 10:47 PM
FernandoRevilla
Re: Basis of ker L --> Basis of vector space?
Quote:

Originally Posted by mathminor827
But I'm stuck on exactly why $\{u_1,\ldots ,u_m, w_1,\ldots ,w_n \}$ is a basis for $V$? The proof assumed ---
$\{u_1,\ldots ,u_m \}$ is a basis for $\ker L$. How did they get from this basis to $\{u_1,\ldots ,u_m, w_1,\ldots ,w_n \}$ --- a basis for $V$?

They are using the Incomplete basis theorem. Look Theorem 5.4.5 here: Basis of a vector space.
• Sep 17th 2011, 06:27 AM
mathminor827
Re: Basis of ker L --> Basis of vector space?
Quote:

Originally Posted by FernandoRevilla
They are using the Incomplete basis theorem. Look Theorem 5.4.5 here: Basis of a vector space.

Hi Professor ---

Thanks for answering. I copy the theorem here ---

http://i51.tinypic.com/24xpa8h.jpg

Because I've been stuck on this, I just want to make sure ---
The given linearly independent set of vectors ---

ie $\{u_1,\ldots ,u_r\}$ --- does NOT have to be in $V$?

And why does it NOT have to be in $V$? What's the intuition? I'm not after the formal proof.

I still don't see why a basis for $\ker L$ can be a basis for $V$?

Thanks ---
• Sep 17th 2011, 07:59 AM
HallsofIvy
Re: Basis of ker L --> Basis of vector space?
You said "Also assume { $u_1, u_2, ..., u_m$} forms a basis for ker L.

We can extend this to form a basis for V: { $u_1, u_2,..., u_m, w_1, w_2, ..., w_n$}".

Do you not understand what "extend" means? We have a basis for ker L, a subspace of V and we add new vectors from V, that are not in ker L to fill out a full basis for V.

You also say, in your first post, "I think I see why { $u_1, u_2,..., u_m, w_1, w_2, ..., w_n$} is a basis for ker L'. No, that's impossible (unless ker L= V, of course) because we have extended adding vectors $w_1$, etc. vectors that are NOT in ker L.

Here's how you do that: if ker L= V, then we are done. A basis for one is a basis for the other.

If ker L is not V, then there exist vectors in V that are not in ker L. Pick any one and add it to the basis of ker L. If we call that new vector $w_1$, then we now have { $u_1, u_2, ..., u_m, w_1$}. Since $w_1$ was not in L, it cannot be written as a linear combination of the u vectors so this set is still a set of independent vectors. If it spans V, we are done. If it does not, by definition of "span", there must exist a vector, $w_2$ in V that is not in the span. Add it to the set. Again, because it was not in the span of the previous vectors, it is independent of them. See it this new set spans V. If it does, we are done- we have a set that is both independent and spans V- a basis for V. If not, there exist some other vector, $w_3$ that is not in its span. Add that vector.

That process must eventually end because we are given that V is a finite dimensional vector space. There must be some number N such that no set of more than N vectors can be independent.
• Sep 17th 2011, 08:57 AM
mathminor827
Re: Basis of ker L --> Basis of vector space?
Quote:

Originally Posted by HallsofIvy
You said "Also assume { $u_1, u_2, ..., u_m$} forms a basis for ker L.

We can extend this to form a basis for V: { $u_1, u_2,..., u_m, w_1, w_2, ..., w_n$}".

Do you not understand what "extend" means? We have a basis for ker L, a subspace of V and we add new vectors from V, that are not in ker L to fill out a full basis for V.

You also say, in your first post, "I think I see why { $u_1, u_2,..., u_m, w_1, w_2, ..., w_n$} is a basis for ker L'. No, that's impossible (unless ker L= V, of course) because we have extended adding vectors $w_1$, etc. vectors that are NOT in ker L.

Here's how you do that: if ker L= V, then we are done. A basis for one is a basis for the other.

If ker L is not V, then there exist vectors in V that are not in ker L. Pick any one and add it to the basis of ker L. If we call that new vector $w_1$, then we now have { $u_1, u_2, ..., u_m, w_1$}. Since $w_1$ was not in L, it cannot be written as a linear combination of the u vectors so this set is still a set of independent vectors. If it spans V, we are done. If it does not, by definition of "span", there must exist a vector, $w_2$ in V that is not in the span. Add it to the set. Again, because it was not in the span of the previous vectors, it is independent of them. See it this new set spans V. If it does, we are done- we have a set that is both independent and spans V- a basis for V. If not, there exist some other vector, $w_3$ that is not in its span. Add that vector.

That process must eventually end because we are given that V is a finite dimensional vector space. There must be some number N such that no set of more than N vectors can be independent.

Hi HallsofIvy ---

Thank you so much for answering. I get it now!

But for other readers - I just want to point out a typo. I hope you don't mind this at all. I just want other new to linear algebra --- like me! --- to know about it.

I put it in red --- it should be $\ker L$.

Thanks a lot again ---