Nullspace of Orthogonal Vectors: Example Matrix

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Homework Help Overview

The problem involves finding a matrix whose nullspace consists of all vectors in R4 that are orthogonal to the vectors (0, 4, 4, 2) and (3, 4, -2, -4). Participants are exploring the implications of orthogonality and the relationship between the given vectors and the nullspace.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the interpretation of the problem, particularly the relationship between the orthogonal vectors and the nullspace. There are attempts to set up equations based on the dot product and to express the solution in terms of free parameters. Questions arise about the correct form of the matrix and how to represent the solution set.

Discussion Status

There is ongoing exploration of the problem, with some participants suggesting that the original poster is on the right track. However, confusion remains regarding the specific requirements of the question, particularly in distinguishing between finding a matrix and expressing the solution set. Guidance has been offered regarding the setup of the equations and the nature of the solutions.

Contextual Notes

Participants are grappling with the definitions and properties of nullspaces and orthogonality, as well as the implications of free parameters in their solutions. There is a noted lack of consensus on the interpretation of the problem statement and the expected format of the answer.

negation
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Homework Statement



Let S be the subspace of all vectors in R4 that are orthogonal to each of the vectors
(0, 4, 4, 2), (3, 4, -2, -4)
What is an example of a matrix for which S is the nullspace?

The Attempt at a Solution



I'm not sure how I should be intepreting the question:

[ 0 ,4 ,4 ,2 ;3, 4, -2, -4] = [ x , y ,z , t] = [ 0 , 0 ]

0x + 4y + 4z + 2t = 0

3x + 4y -2 z -4t = 0

from here I set up an augmented matrix and solve for the set of vectors x ,y ,z ,t? By definition if x, y, z, t results in each linear equation = 0, then, x ,y ,z ,t are vectors with properties of orthogonality?
 
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There is multiplication instead of equality in the first line? If yes, it looks all right, do it.

ehild
 
ehild said:
There is multiplication instead of equality in the first line? If yes, it looks all right, do it.

ehild

[ 0 ,4 ,4 ,2 ;3, 4, -2, -4] .[ x , y ,z , t] = [ 0 , 0 ]

Yes, it should be a multiplicative operation-dot product.
But I aren't exactly sure what should I be doing with the 2 linear equation. Do I perform reduce row echelon?
 
negation said:
[ 0 ,4 ,4 ,2 ;3, 4, -2, -4] .[ x , y ,z , t] = [ 0 , 0 ]


But I aren't exactly sure what should I be doing with the 2 linear equation. Do I perform reduce row echelon?
Yes. But it is of that form already... You will have two free parameters.


ehild
 
ehild said:
Yes. But it is of that form already... You will have two free parameters.


ehild

I'm still unclear. Could you explain?
 
There is the system of equations:

0x + 4y + 4z + 2t = 0

3x + 4y -2 z -4t = 0

What is the general solution?

ehild
 
ehild said:
There is the system of equations:

0x + 4y + 4z + 2t = 0

3x + 4y -2 z -4t = 0

What is the general solution?

ehild

It is done by setting up an augmented matrix:

0 4 4 2 | 0
3 4 -2 -4 | 0

Perform REF(Not RREF), then solve for the variables.


Edit: x1 = 2x3 + 2x4, x2 = -x3-0.5x4, x3, x4
 
Last edited:
Well, solve for the variables...

ehild
 
ehild said:
Well, solve for the variables...

ehild

Post #7
 
  • #10
X3=a, x4=b a,b free parameters. So the S subspace consists of vectors ((2a+2b); (-a-0.5b); a; b).
And a matrix for which S is the nullspace?

ehild
 
  • #11
ehild said:
Well, solve for the variables...

ehild

Relooking at the question, I'm not sure if we interpreted it correctly.

I think the question is asking for the {SX=0|x1=0,4,4,2, x2=(3,4,-2,-4),S\in R4}

Edit: I think we were on the right track.

But this brings up a question. What if we're asked to solve S, given X where SX = 0?
 
Last edited:
  • #12
negation said:
Let S be the subspace of all vectors in R4 that are orthogonal to each of the vectors
(0, 4, 4, 2), (3, 4, -2, -4)
What is an example of a matrix for which S is the nullspace?

negation said:
Relooking at the question, I'm not sure if we interpreted it correctly.

I think the question is asking for the {SX=0|x1=0,4,4,2, x2=(3,4,-2,-4),S\in R4}
Let's untangle your notation. S is a subspace of R4 such that if v ##\in## S, then v ##\cdot## <0, 4, 4, 2> = 0 and v ##\cdot## <3, 4, -2, -4> = 0.

The problem asks that you find a matrix (call it A) for which S is the nullspace of A. IOW, if v ##\in## S, then Av = 0.
 
  • #13
ehild said:
X3=a, x4=b a,b free parameters. So the S subspace consists of vectors ((2a+2b); (-a-0.5b); a; b).
And a matrix for which S is the nullspace?

ehild
Isn't the vector [2z+2t; -z-0.5t;z;t] the answer?
 
  • #14
S is the nullspace of a linear transformation T, that is, Tv =0 for all vectors v\inS. Show a matrix of that transformation. (You have shown one in Post #3) .

ehild
 
  • #15
negation said:
Isn't the vector [2z+2t; -z-0.5t;z;t] the answer?
See post #12. The "answer" should be a matrix, not a vector.

In addition, what you have above would be better written as two vectors - a basis for S.
 
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  • #16
ehild said:
S is the nullspace of a linear transformation T, that is, Tv =0 for all vectors v\inS. Show a matrix of that transformation. (You have shown one in Post #3) .

ehild

Am I suppose to perform another one? I thought we have had all the solutions already. 2 are free variables and the corollary is that there are infinite solutions.
 
  • #17
The subspace S contains infinite number of vectors. You have found these vectors already.

The next question is
What is an example of a matrix for which S is the nullspace?

ehild
 
  • #18
ehild said:
The subspace S contains infinite number of vectors. You have found these vectors already.

The next question is

ehild

any matrix with coefficient \in R?
 
  • #19
Mark44 said:
See post #12. The "answer" should be a matrix, not a vector.

In addition, what you have above would be better written as two vectors - a basis for S.

basis for S : z (2,-1,1,0) + t (2,-0.5,0,1)
 
  • #20
Your wording is peculiar. z (2,-1,1,0) + t (2,-0.5,0,1) is a general vector in S. The set of vectors {(2, -1, 1, 0), (2, -0.5, 0, 1)} is a basis.

The answer to this question is any matrix, A, such that A applied to each of those two vectors is 0.
 
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  • #21
HallsofIvy said:
The answer to this question is any matrix, A, such that A applied to each of those two vectors is 0.
Similar to what I said in post #12.
Mark44 said:
The problem asks that you find a matrix (call it A) for which S is the nullspace of A. IOW, if v ∈ S, then Av = 0.
 
  • #22
HallsofIvy said:
Your wording is peculiar. z (2,-1,1,0) + t (2,-0.5,0,1) is a general vector in S. The set of vectors {(2, -1, 1, 0), (2, -0.5, 0, 1)} is a basis.

The answer to this question is any matrix, A, such that A applied to each of those two vectors is 0.


I think I got it. Set any value for the free variables vand work from there.
 
  • #23
negation said:
I think I got it. Set any value for the free variables vand work from there.
I'm not sure you get it. As has been said multiple times, you need to find a matrix.
 
  • #24
negation said:
any matrix with coefficient \in R?

NO!

You have written in post #3:

[ 0 ,4 ,4 ,2 ;3, 4, -2, -4] .[ x , y ,z , t] = [ 0 , 0 ]

When you perform matrix multiplication, you do the dot product of the vector v=(x,y,z,t) with the first line of the matrix as row vector and get zero, as the are normal to each other, and the same when multiply the second row of the matrix with the vector v. So one matrix A that fulfils equation Av=0 is ?

ehild
 
  • #25
ehild said:
no!

You have written in post #3:
When you perform matrix multiplication, you do the dot product of the vector v=(x,y,z,t) with the first line of the matrix as row vector and get zero, as the are normal to each other, and the same when multiply the second row of the matrix with the vector v. So one matrix a that fulfils equation av=0 is ?

Ehild
v1 = 0x + 4y + 4z +2t = 0
v2 = 3x + 4y -2z -4t = 0

for v1 and v2 = 0; (x,y,z,t) = (0,0,0,0)

zero vector!
 
  • #26
What are you talking about? Have you found the matrix yet?
 
  • #27
Mark44 said:
Similar to what I said in post #12.
Yes, it was. But negation had then asked the same question again.
 
  • #28
negation,
This is becoming very frustrating for us. We have said in posts 10, 12, 14, 15, 17, 20, 21, 23, 24, and 26 that this problem is asking you to find a matrix. We can't say it more plainly than that.
 
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  • #29
HallsofIvy said:
What are you talking about? Have you found the matrix yet?

I have. It's (0,0,0,0),(0,0,0,0) which was correct.

Edit: Too many question to keep track of.

0,4,4,2
3,4,-2,-4
 
Last edited:
  • #30
Mark44 said:
negation,
This is becoming very frustrating for us. We have said in posts 10, 12, 14, 15, 17, 20, 21, 23, 24, and 26 that this problem is asking you to find a matrix. We can't say it more plainly than that.

I apologized. I misread the question. But I found the answer to be (0,0,0,0),(0,0,0,0)

Edit: Opps, answer to wrong question. Too many question to keep track.

Ans:

0,4,4,2
3,4,-2,-4
 
Last edited:

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