Question about the Null Space for this Zero Matrix

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Homework Help Overview

The discussion revolves around determining the null space of a 2 x 6 zero matrix. Participants are exploring the implications of the definition of the null space in the context of a matrix with all zero entries.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants discuss the definition of the null space and question whether any vector in R^6 would suffice as a solution. Others suggest that the problem may not be as straightforward as it seems, considering it is labeled as a "conceptual" problem.

Discussion Status

The conversation is ongoing, with participants sharing their thoughts on the nature of the problem. Some have provided insights into the implications of the zero matrix, while others are questioning the depth of the problem and whether additional considerations are necessary.

Contextual Notes

There is a sense of uncertainty regarding the expectations of the homework assignment, as participants wonder if the simplicity of the zero matrix problem aligns with the intended challenge of the bonus problem.

Theelectricchild
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How can I determine the null space for the 2 x 6 zero matrix as precisely as I can?

Clearly N(A) = {x: Ax = 0, x in R^n},

So if A is this 2x6 matrix, wouldn't virtually any vector x that is in R^6 work?

This is supposed to be a "conceptual" problem, and I KNOW it can't be this easy for the bonus problem on the HW assignment!

Can anyone tell me what I am missing? THANKS A LOT!
 
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it's been a long time but check out

http://cnx.rice.edu/content/m10368/latest/

Basically you need to get A into row reduced echelon form. You have 2 equations, 6 unknowns (so you have 4 free parameters.)

x1 and x2 will be your piviots and your equation for x (your general vector which describes the set of all vectors in your null space is:


x = x3*a+x4*b+x5*c+x6*d

where a,b,c,d are your column vectors which give the coefficents of your x3,x4,x5,x6 when you solve for these variables. So for example, you should be able to get it down to where x1 is out of your system of equations and x2 is solved by all the other variables. So we can take the coefficient of this to also be zero. So we should have a vector in terms of the free variables only. Now solve this equation in terms of the remaining variables. X3=x3(x4,x5,x6), x4=x4(x3,x5,x6), and so on. So youll have zeros for x1,x2 in all a,b,c,d vectors, and 1s in the values for the variables you solve for, for "a" above, the 3rd element will be 1, because you're dealing with x3 here.

I'm too tired, and I'm sure there are some errors in this bad explaniation. But I hope it helps.
 
Last edited by a moderator:
If A is the zero matrix (the matrix with all zero entries), then every vector x in R^6 will give Ax=0.
It's pretty trivial.
You could also use the rank equation.
 
Did they really say zero matrix? That doesn't even seem like a problem.
 
Theelectricchild said:
How can I determine the null space for the 2 x 6 zero matrix as precisely as I can?

Clearly N(A) = {x: Ax = 0, x in R^n},

So if A is this 2x6 matrix, wouldn't virtually any vector x that is in R^6 work?

This is supposed to be a "conceptual" problem, and I KNOW it can't be this easy for the bonus problem on the HW assignment!

Can anyone tell me what I am missing? THANKS A LOT!

Yes, this is a "conceptual" problem. What is the definition of "null space"??
 
but did he not define it ?
 

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