Extending to a basis using row operations

Conversation started Oct 19, 2015 at 15:32.

Tien-Cheng Huang

Oct 19, 2015 15:32

Given an arbitrary independent set of a finite-dimensional vector space, how to expand the set to obtain a basis?

robjohn

@r9m Numerically integrating and summing the series I got both agree, so I am inclined to believe those results.

Tien-Cheng Huang

(oh... possibly a newbie question...)

r9m

@robjohn I am probably making some kind of mistake and I need to identify it ,, thanks!

Martin Sleziak

Oct 19, 2015 15:50

@Tien-ChengHuang Have you learned Gauss-Jordan elimination?

I.e. finding row reduced echelon form of a matrix.

Tien-Cheng Huang

well sorry i haven't work through the whole LA book...

Martin Sleziak

For example if you get rref of this form from the original vectors: $\begin{pmatrix}
1 & 0 & 1 & 1 \\
0 & 1 & 3 & 1 \\
\end{pmatrix}$

Then you know for sure that you can add (0,0,1,0) and (0,0,0,1).

I.e., the ones are in the positions where the rref does not have pivots.

$\begin{pmatrix}
1 & 0 & 1 & 1 \\
0 & 1 & 3 & 1 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}$

@Tien-ChengHuang Could you give an example which you are trying to solve?

Or some example similar to your problem.

Tien-Cheng Huang

I have two exercises giving spanning set and requiring trim them to obtain a basis. but have no exercise giving independent set and requiring expand them to obtain a basis.

so i can just learn form your example now

Martin Sleziak

And you are working in $R^3$ or $R^4$ or some similar space?

Tien-Cheng Huang

$R^5$

Martin Sleziak

Oct 19, 2015 16:00

Ok, let's do that then.

Suppose we are given vectors $(1,2,-1,3,2)$ and $(1,1,0,2,1)$.

We check whether they are linearly independent. (That should be easy right?)

We want to add some other vectors - we need to add 3 vectors.

Let us denote them by $\vec a-(1,2,-1,3,2)$ and $\vec b=(1,1,0,2,1)$.

Notice that $\vec a-\vec b=(0,1,-1,1,1)$

Would you agree that $[\vec a,\vec b]=[\vec a-\vec b,\vec b]$?

I.e. that $a-b$ and $b$ span the same subspace as $a$ and $b$?

@Tien-ChengHuang

Tien-Cheng Huang

you mean the same span? yes

Martin Sleziak

So we need some three vectors which are not in their span.

I write them under each other so thet we can see better what ve are doing.

$\begin{pmatrix}
1 & 1 & 0 & 2 & 1 \\
0 & 1 &-1 & 1 & 1 \\
\end{pmatrix}$

That should be b and a-b, unless I have made a mistake.

Maybe I could use b-a instead (it does not matter that much).

$\begin{pmatrix}
1 & 1 & 0 & 2 & 1 \\
0 &-1 & 1 &-1 &-1 \\
\end{pmatrix}$

Notice the first and the third column.

Those are columns where I have number one and all other entries in that column are zeroes.

This is the important thing.

Now if I ask whether I can get (0,1,0,0,0) as a linear combination of these two vectors, could you answer that?

I will write them again into a matrix, so that we can see them better.

$\begin{pmatrix}
1 & 1 & 0 & 2 & 1 \\
0 &-1 & 1 &-1 &-1 \\
0 & 1 & 0 & 0 & 0 \\
\end{pmatrix}$

We are asking whether the third row can be obtained as linear combination of the first two rows.

Can you answer this question @Tien-ChengHuang?

Tien-Cheng Huang

i think no

Martin Sleziak

That's correct. They are linearly independent. Is there some easy argument how to see that the third row is not linear combination of the first two?

As I said, the first and the third column might be useful to find such an argument.

Tien-Cheng Huang

i am listening :)

Martin Sleziak

Oct 19, 2015 16:12

To make this shorter I will denote the first two rows by $\vec r_1$ and $\vec r_2$.

If I write a linear combination $c\vec r_1+d\vec r_2$ then I can see what will be on the first and the third coordinate.

The first coordinate must be equal to $c$. The third coordinate must be equal to $d$.

So if I want to get $(0,1,0,0,0)$, I must have $c=0$ and $d=0$, because of the first and third coordinate.

But this does not work on the second coordinate.

So I can't get the third row as linear combination of the first two.

Does this argument make sense?

Tien-Cheng Huang

yes i can follow you

Martin Sleziak

Great.

And for the same reason I can add two more vectors.

$\begin{pmatrix}
1 & 1 & 0 & 2 & 1 \\
0 &-1 & 1 &-1 &-1 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1
\end{pmatrix}$

Maybe this is not the only possibility, but for these vectors I can be sure that they are independent without checking it by manual computation.

Would you agree that in this way we extended {a,b-a} to the basis of $R^5$?

Tien-Cheng Huang

ok i see the second column can be made containing only one 1 in the 3rd row

the same as the 3rd and 4th column

4th an 5th*

Martin Sleziak

Yes. Using row operations. So you have learned a bit about elementary row operations right?

Tien-Cheng Huang

yes so that is a use of row operation

Martin Sleziak

Oct 19, 2015 16:19

So we extented {a,b-a} by these three vectors.

This is the same as extending {a,b}.

But b-a was better, because I had a column containing zeroes on all places with one exception.

I simply guessed that I can use b-a. (For two vectors this was easy.)

But no guess work is needed.

Tien-Cheng Huang

so {a,b}U{the three vectors} are maximal independent set

which is spanning

Martin Sleziak

Yes.

To avoid guesswork, I could have taken the original two vectors:

$\begin{pmatrix}
1 & 2 &-1 & 3 & 2 \\
1 & 1 & 0 & 2 & 1
\end{pmatrix}$

Use row operations until I get matrix in such form that some column has all zeroes with one exception.

This is what I meant by Gauss-Jordan elimination and rref. (I am not sure whether you have learned about this yet.)

If I get such matrix, I can add the vectors as described above. I.e., I take vectors from the standard basis, but avoid these columns.

Tien-Cheng Huang

thank you!! you really teach me something :)

Martin Sleziak

I could also have done this.

$\begin{pmatrix}
1 & 2 &-1 & 3 & 2 \\
1 & 1 & 0 & 2 & 1
\end{pmatrix}
\sim
\begin{pmatrix}
0 & 1 &-1 & 1 & 1 \\
1 & 1 & 0 & 2 & 1
\end{pmatrix}\sim
\begin{pmatrix}
1 & 1 & 0 & 2 & 1 \\
0 & 1 &-1 & 1 & 1
\end{pmatrix}\sim
\begin{pmatrix}
1 & 0 & 1 & 1 & 0 \\
0 & 1 &-1 & 1 & 1
\end{pmatrix}$

Now I got pivots in different place (the first and the third columns).

So in this case I see I can take e_3, e_4, e_5 from the standard basis.

I have just added this to show that e_2, e_4, e_5 is not the only possibility. (Those were the vectors we took before.)

ok, I hope this helped a bit and that you would be able to solve something similar with three vectors

Conversation ended Oct 19, 2015 at 16:27.

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