Nullspace Matrix Homework: Struggling to Find an Example Vector

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Homework Help Overview

The discussion revolves around finding example vectors in the nullspace of a given matrix A, specifically addressing the conditions for vectors to belong to the nullspace and the dimensionality of the space involved.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the original poster's attempts to identify vectors in the nullspace, questioning the validity of the answer sheet's feedback regarding dimensionality. There is also exploration of what constitutes a vector not in the nullspace.

Discussion Status

The discussion is ongoing, with participants providing insights into the dimensionality of the vectors and suggesting potential issues with how the answers are being submitted to the online portal. There is no explicit consensus on the resolution of the original poster's issue.

Contextual Notes

Participants note that the original poster's vectors are 1x4, while there is a suggestion that they should be input as 4x1 vectors. There is also mention of the online portal's feedback being potentially misleading or erroneous.

negation
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Homework Statement



Let A be the matrix:

[3,3,-2,0;-3,-3,3,-2]


a) An example of a vector in the nullspace of A is
b) An example of a vector NOT in the nullspace of A is


Sorry guy but I'm really STRUGGLING

The Attempt at a Solution




a) I found x1 ,x2,x3,x4 = -x2+4/3x4, x2, 2x4, x4

=x2( -1,1,0,0) + x4(4/3,0,2,1)

For more than an hour, the answer sheet is telling that my vectors are wrong.
I don't see how it can be wrong.

Ax = 0
I reduced A to row echelon form, then solve in terms of the free variables x2 and x4.
 
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That answer is correct as you can see by apply A to each of (-1,-, 0, 0) and (4/3, 0, 2, 1). The nullspace is two dimensional and there are an infinite number of bases for it. How is the answer sheet telling you that you are wrong? If it gives a different pair of vectors, that may just be a different basis for the same space. For example, if (4/3, 0, 2, 1) is a basis vector, so is (4, 0, 6, 3).
 
HallsofIvy said:
That answer is correct as you can see by apply A to each of (-1,-, 0, 0) and (4/3, 0, 2, 1). The nullspace is two dimensional and there are an infinite number of bases for it. How is the answer sheet telling you that you are wrong? If it gives a different pair of vectors, that may just be a different basis for the same space. For example, if (4/3, 0, 2, 1) is a basis vector, so is (4, 0, 6, 3).

The answer sheet is an online portal prompting me of my error whenever I submit the answer.
I know my answer has to be correct. I submit ( -1,1,0,0),(4/3,0,2,1) in vector form. and I know my answer cannot be wrong because 1) I perform RREF on the matrix A then, 2) express the solution as free variables.

I do not know what is going on with the portal prompting me of my error.

edit: Message: "What you have given as vector in the nullspace of A is not of the correct dimension."
 
I'll solve part (b) first:

so, a vector that is not in the nullspace implies ax =/= 0

[3,3,-2,0;-3,-3,3,-2] [x1,x2,x3,x4] = [0;1]

I get

x1 = 2/3 - x2 + 4/3x4
x3 = 1 + 2x4
free : x2 , x4

x1,x2,x3,x4 = ( 2/3 - x2 + 4/3x4, x2, 1+2x4, x4)
 
A has four columns so acts on four dimensional vectors. If you gave (-1, 1, 0, 0) and (4/3, 0, 2, 1) they have the correct dimension. How did you enter the fraction 4/3? The answer "sheet" might be interpreting that as (4, 3, 0, 2, 1). Try (4, 0, 6, 3), as I suggested before, instead.
 
HallsofIvy said:
A has four columns so acts on four dimensional vectors. If you gave (-1, 1, 0, 0) and (4/3, 0, 2, 1) they have the correct dimension. How did you enter the fraction 4/3? The answer "sheet" might be interpreting that as (4, 3, 0, 2, 1). Try (4, 0, 6, 3), as I suggested before, instead.

I entered the fraction as it is, that is- 4/3. There is a button that shows my answer in some kind of tex format and it corresponds to 4/3 as how one would normally write a value as a fraction form.


Untitled.png
 
HallsofIvy said:
A has four columns so acts on four dimensional vectors. If you gave (-1, 1, 0, 0) and (4/3, 0, 2, 1) they have the correct dimension. How did you enter the fraction 4/3? The answer "sheet" might be interpreting that as (4, 3, 0, 2, 1). Try (4, 0, 6, 3), as I suggested before, instead.

It isn't working. There isn't any reason for me to enter (4,0,..) because the system does in fact recognize 4/3 as an input language for the answer.
 
negation said:
It isn't working. There isn't any reason for me to enter (4,0,..) because the system does in fact recognize 4/3 as an input language for the answer.

Hey!

I might be wrong, but the way you are inputting the answer might be incorrect.

Notice the vectors you are putting in are 1x4, and you know you should be inputting 4x1.

Maybe try putting a semicolon after your numbers instead?

If that doesn't work.. well, maybe there is an error in the program.
 

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