Linear algebra, linear independance

mikau · 2007-07-23 09:40:31

for the matrix

|1 2 0 3|
|1 2 3 3|
|1 0 1 1|
|1 1 1 2|

determine the maximum number of linearly independant rows.

My book says the answer is 3 which must mean the 4 rows are dependant. But I can't see how they found this.

For a set of rows to be linnearly independant, we have to show that the only way for all the rows to add to zero, is to multiply each row by zero.

Obviously the system of equations required is

a + b + c + d = 0
2a + 2b + 0 + d = 0
0 + 3b + c + d = 0
3a + 3b + c + 2d = 0

now if we can find a solution to this equation such that a, b, c and d are not all equal to zero, it follows that its dependant. But every time I solve it using traditional methods, it only gives me a = b = c = d = 0.

Now it makes sense that if one of the rows is derived from the other rows via elementary row operations, that it is infact dependant. But i have no idea to go about doing this.

Overal question is, when you solve a system of equations, and you find a solution, how do you know if thats the only solution? How do I find all the solutions?

Frankly, i was hoping linear algebra would show me this, but right now it seems to be asking me before telling me.

mikau · 2007-07-23 09:46:57

to confuse me even more, in an earlier example my book said

in the matrix A =

|1 2 0 2|
|0 1 1 1|
|1 0 1 0|

the columns are linearly dependant. Indeed the second and fourth columns are the same, so that a2 = a4 which we can write in the form 1a2 - 1a4 = 0. However, the first three columns of A are linearly independant. To see this, suppose k1a1 + k2a2 + k3a3 = 0 then we have

k1 + 2 k2 = 0
k2 + k3 = 0
k1 + k3 = 0

from which it is easily seen that k1 = k2 = k3 = 0.

It can be easily seen? sure. But can it be easily seen that this is the only solution?

I'm confused.

Last edited by mikau (2007-07-23 09:47:12)

Ricky · 2007-07-23 12:56:39

Edited to add: I think the thing you're missing is that by doing row operations, you are guaranteed not to alter the system in any way. Thus, if you come up with a requirement after doing row operations, this same exact requirement was in the original system. It was just a bit hidden.

Overal question is, when you solve a system of equations, and you find a solution, how do you know if thats the only solution? How do I find all the solutions?
Frankly, i was hoping linear algebra would show me this, but right now it seems to be asking me before telling me.

When you solve a system of equations, if there are multiple solutions, you will be shown this automatically. This is a great example for it.

For sake of writing (and I suggest you do this in your own work as well, makes it easier), we write:

a + b + c + d = 0
2a + 2b + 0 + d = 0
0 + 3b + c + d = 0
3a + 3b + c + 2d = 0

as:

1 1 1 1 0
2 2 0 1 0
0 3 1 1 0
3 3 1 2 0

By doing row operations, I turn this matrix first into:

1 1 1 1 0
0 0 -2 -1 0
0 3 1 1 0
0 0 -2 -1 0

Simply multiply and add two rows making the left column all 0's (except the top). Then switch the middle rows:

1 1 1 1 0
0 3 1 1 0
0 0 -2 -1 0
0 0 -2 -1 0

Right now you should stop and say to yourself, "These aren't linearly independant." But to find solutions, you must keep going. I'm leaving off the last row now because you can easily make it all 0's. Divide the middle row by 3:

1 1 1 1 0
0 1 1/3 1/3 0
0 0 -2 -1 0

Now make the 2nd column have all 0's but the middle row:

1 0 2/3 2/3 0
0 1 1/3 1/3 0
0 0 1 1/2 0

Finally, do the same for the third column

1 0 0 1/3 0
0 1 0 1/6 0
0 0 1 1/2 0

And for the sake of "integerness":

3 0 0 1 0
0 6 0 1 0
0 0 2 1 0

Now we get our variable back out.

3a + d = 0
6b + d = 0
2c + d = 0

Now rewrite the equations:

a = -d/3
b = -d/6
c = -d/2

Now note that because we started with all of our restrictions, these are all of the restrictions on the system. So pick d to be anything, solve for a, b, and c, and you will get a solution. For example, if d = -6, a = 2, b = 1, c = 3 is a solution.

Ricky · 2007-07-23 13:02:34

It can be easily seen? sure. But can it be easily seen that this is the only solution?

Same method.

|1 2 0 2|
|0 1 1 1|
|1 0 1 0|

1 2 0 0
0 1 1 0
1 0 1 0

Remove the 1 in the bottom left corner:

1 2 0 0
0 1 1 0
0 -2 1 0

Do the same for the 2nd column:

1 0 -2 0
0 1 1 0
0 0 3 0

And stop there. We see that there is a requirement for 3c = 0 from that last line. Thus, it must be that c = 0. Now from the 2nd line, b + c = 0. But we already know that c must be 0. So b = 0. And the same goes for the top. It must be that a = 0. Thus, each coefficient must be 0, and so the first three columns are linearly independant.

mikau · 2007-07-23 13:06:00

it looks like in the top example you reduced it to hermite form, which is what my book hasn't gotten to yet. I think i can see where they're going but they asked me to solve that system WITHOUT matrices. So i thought..maybe i'm expected to know how.

Ricky · 2007-07-23 13:43:03

*shrug* I certainly don't know of any other way to tell dependence.

mikau · 2007-07-23 13:46:31

why is it my books always pose questions when introducing a concept that people years beyond my level can't solve?

Ricky · 2007-07-23 13:58:19

What's the books title/author?

mikau · 2007-07-23 15:23:38

Basic Linear Algebra. Second edition. Springer Undergraduate Mathematics Series
authors, Professor T.S Blyth, Professor E.F. Robertson

This is the book used at my school.
its an okay book overal. But its only a 180 pages so i think its condensed. (i HATE condensed books!)

Last edited by mikau (2007-07-23 15:24:08)

Ricky · 2007-07-23 15:45:06

The first few pages of chapter 3 go over how to solve the matrix in the way I did. Are you past that point yet?

mikau · 2007-07-23 15:49:26

thats where i am. Chapter 3. What, do you have the book?

Ricky · 2007-07-23 15:52:54

Google Rocks

mikau · 2007-07-23 15:59:57

hehehe. I almost had a tantrum until i realized its just a preview. Anyway, I'm not sure but i thiiink if you'll look closer, you'll see that the few pages they have there don't actually explain how to solve it in the way you did. They're undoubtebly heading there but i don't think they reached it in whats displayed there. (the reason i'm saying that is i past those pages already)

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#1 2007-07-23 09:40:31

Linear algebra, linear independance

#2 2007-07-23 09:46:57

Re: Linear algebra, linear independance

#3 2007-07-23 12:56:39

Re: Linear algebra, linear independance

#4 2007-07-23 13:02:34

Re: Linear algebra, linear independance

#5 2007-07-23 13:06:00

Re: Linear algebra, linear independance

#6 2007-07-23 13:43:03

Re: Linear algebra, linear independance

#7 2007-07-23 13:46:31

Re: Linear algebra, linear independance

#8 2007-07-23 13:58:19

Re: Linear algebra, linear independance

#9 2007-07-23 15:23:38

Re: Linear algebra, linear independance

#10 2007-07-23 15:45:06

Re: Linear algebra, linear independance

#11 2007-07-23 15:49:26

Re: Linear algebra, linear independance

#12 2007-07-23 15:52:54

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#13 2007-07-23 15:59:57

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