Linear Algebra - [ 07. Solving Ax = 0: Pivot Variables, Special Solutions ]

date

Jun 15, 2023

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linear-algebra-07

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Public

Solving Ax = 0

So can I take a particular matrix A and describe the natural algorithm, and I’ll execute it for that matrix.

So let me take the matrix as an example.

OK, so let me execute elimination for that matrix. My goal is always, while I’m doing elimination - I’m not changing the null space. That’s very Important. What does elimination leave unchanged? The answer is the solutions to the system are not changed because I’m doing a legitimate operations on the equations.

first column to zero

I look in that red position, I see a zero. I look below it, hoping for a non-zero that I can do a row exchange. But it’s zero below. What it’s really going to be telling me is that that column is a combination of earlier columns . It’s that second column is dependent on the earlier columns. But I don’t stop to think here. In that column there’s nothing go do. I go on the next.

second column to zero

(사실 stair case 는 upper triangular 이 아닌 echelon form 라고 부른다.)

So actually we’ve already discovered the most important number about this matrix. The number of pivots is two. That number we will call the rank of the matrix.

I’m always solving Ax = 0, bout now I can solve ux = 0. Same solution, same null space. So now I have to ask you, how do I describe the solutions?

Column with pivot is called pivot column and without pivot is called free column. Why do I use those words? Why do I use that word ‘free’? Because I want to find the solutions to ux = 0. These free columns I can assign any number freely to those to the variables x2 and x4, the ones that multiply columns two and four. So I can assign anything I like x2 and x4. And then I can solve the equations for x1 and x3.

There is a vector in the null space. There is a solution to Ax = 0. In fact, what solution is that? The second column is twice as big as the first column.

OK, tell me some more vectors in the null space. Tell me, how to get a bunch more immediately out of that one. Just take multiples of it.

So x could be any multiple of x.

OK, that describes now a line in four-dim space. But which is in the null space. Is that the whole null space? NO. I’ve got two free variables here. I made x3 = 0, x1 = -2 but I could have made another choice. Let me make the other choice.

This is the algorithm that you just learned to do.

Do elimination.

Decide which are pivot columns and which are the free columns.

Then those free variables, you assign them - you give one of them the value.

OK, now I’ve found another vector in the null space. Now we’re ready to tell me the whole null space. What are all the solutions to Ax = 0?

These are my two special solutions. (I just invented that name) What’s special about them is the numbers I gave to the free variables. I could have given the free variables any values and got vectors in the null space.

I left a little space for what? A plus sign. I can take any combination. I’m taking all the combinations of my two special solutions. That’s my conclusion there. The null space contains exactly all the combinations of the special solutions. And how many special solutions are there? SO MANY

I would like to clean up this matrix even more.

OK, so now I’m going to speak about the reduced row echelon form.

R = reduced row echelon form.

How can I clean that up further? I can do, elimination upwards.

This is what I’m interested here. Let’s suppose we’re got this system already in rref.

I want to solve Rx = 0 and I’ll have some special solutions. Can I do them all at once? I’m going to create a null space matrix.(cols=special solutions)

RN = 0

So what N will do the job?

It does back substitution, but it’s totally simple. What is this system?

Rx = 0

So, what does that mean? That means that the pivot variables plus F times the free variables give zero.

There’s my equation as simple as it can be. That’s what back substitution comes to when I’ve reduced and reduced and reduced this system to the best form.

Another Example

I’m totally expecting that these will be pivot columns co1, col2 because they’re independent, but the third column, which is dependent on the first two, is going to be a free column.

r (Rank) = 2 again !

What about the null space? How many special solutions for this matrix?

OK, so how do I start? I give the free variable some convenient value. And what’s that? I set the free variable to one. (If I set the free variable to zero and solve for the pivot variables, I’ll get all zeros : no progress. But by setting the free variable to one - you see my two equations now are)

And now you can tell me what else is in the null space. What’s the whole null space now? I multiply by c.

OK, now one more natural thing to do with this example is keep going to the reduced matrix, R.

The x has the identity. (Well, it’s only a single number but it’s the identity matrix in the free part.) And what does it have in the pivot variables? What did back substitution give? -F

So do you see, though, how the minus F just automatically shows up in the special solutions.

Summary

Converting Ax=0 to the rref form gives Rx=0, where R is,

And, the solution of Rx=0, the whole null space, is as follows.

Next lecture

Ax = b