Matrices: Technical Questions

[xenogizmo]

Hey Guys,
I was solving some practice questions on linear algebra and I came across this one. I've solved many of its kind, but this one has a variation I'm not so sure about.. Here's the question:

Let A =
| 0 3 -6 3 |
| 0 -3 6 -3 |

Find a basis of nullspace (A). (a set of basic solutions for the homogeneous system)

(That's a matrix, sorry about not using latex, I'm kinda new here)

Now, I know how to do a normal question, by doing elementary row operations and then factoring the parameters.. But what do I do when the first column is all zeroes?? do I make the first basic solution a column of 4 zeroes? or what?
Your help would be greatly appreciated, thx..

 

[HallsofIvy]

The problem asks you to find a basis for the null space of A. A itself, in this example, is from R⁴ to R². If we represent a vector in R⁴ as < x, y, z, u> then your two equations are 0x+ 3y- 6z+ 3u= 0 and 0x- 3y+ 6z- 3u= 0.

The first column being 0 means that x disappears from the equations! Okay, that's easy- that just means that x can be anything and that any vector that is a multiple of <1, 0, 0, 0> is in the null space: one basis vector is <1, 0, 0, 0>.

The two equations are 3y- 6z+ 3u= 0 and -3y+ 6z- 3u= 0. It's easy to see that the second equation is just -1 times the first: you really don't have 2 independent equation. (Normally, you would try to "solve" the equations by eliminating unknowns until you don't have enough equations left. IF you had enough independent equations to completely solve for each unknown, the null space would be just {0}.)
Since you really have just the one equation in 3 unknowns, solve for one of them in terms of the other 2: 3y= 6z- 3u so y= 2z- u. Now, to be sure you get independent vectors, take one of the remaining unknowns to be 1 and the other 0:
If z= 1 and u= 0, then y= 2(1)- 0= 2. One basis vector for the null space is <0,2,1,0>.
If z= 0 and u= 1, then y= 2(0)- 1= -1. Another basis vector for the null space is <0,-1,0, 1>.

The null space of A is 3 dimensional and has basis <1, 0, 0, 0>, <0, 2, 1, 0> and
<0, -1, 0, 1>.

You understand, of course, that the basis of the null space is not unique: we could have solved for z or u instead of y as well as choosing numbers other than 0 and 1 to try and arrived at a different answer.

 

[M98Ranger]

Hi there,

Thank you for reaching out with your question about matrices. I understand that you are having trouble finding a basis for the nullspace of the given matrix A. Let me try to explain the process to you.

To find a basis for the nullspace of a matrix, we need to solve the equation Ax = 0, where A is the given matrix and x is a vector of variables. In this case, A is a 2x4 matrix, so x will be a vector with 4 variables.

Now, when the first column of A is all zeroes, it means that the first variable in x does not affect the solution. So, when you perform elementary row operations, you will end up with an equation that looks like 0x1 + 3x2 - 6x3 + 3x4 = 0. We can ignore the first variable, x1, and only focus on the remaining three variables.

Next, you can solve for the remaining variables using elementary row operations and factoring the parameters. Once you have solved for x2, x3, and x4, you can set x1 to any value (since it does not affect the solution) and form a vector with the values of x2, x3, and x4. This vector will be your first basic solution.

You can repeat this process to find the other basic solutions, keeping in mind that the first variable will always be 0. Once you have a set of basic solutions, you can form a basis for the nullspace by taking the linear combinations of these basic solutions.

I hope this helps clarify the process for you. Let me know if you have any other questions or need further clarification. Best of luck with your practice questions!