Nullspace of Orthogonal Vectors: Example Matrix

[negation]

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?

 

[ehild]

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

ehild

 

[negation]

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?

 

[ehild]

[ 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

 

[negation]

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


ehild

I'm still unclear. Could you explain?

 

[ehild]

There is the system of equations:

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

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

What is the general solution?

ehild

 

[negation]

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

 

[ehild]

Well, solve for the variables...

ehild

 

[negation]

Well, solve for the variables...

ehild

Post #7

 

[ehild]

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

 

[negation]

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?

 

[Mark44]

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?

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.

 

[negation]

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?

 

[ehild]

S is the nullspace of a linear transformation T, that is, Tv =0 for all vectors v$\in$S. Show a matrix of that transformation. (You have shown one in Post #3) .

ehild

 

[Mark44]

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.

 

[negation]

S is the nullspace of a linear transformation T, that is, Tv =0 for all vectors v$\in$S. 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.

 

[ehild]

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

 

[negation]

The subspace S contains infinite number of vectors. You have found these vectors already.

The next question is

ehild

any matrix with coefficient $\in R$?

 

[negation]

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)

 

[HallsofIvy]

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.

 

[Mark44]

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.

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.

 

[negation]

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.

 

[Mark44]

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.

 

[ehild]

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

 

[negation]

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!

 

[HallsofIvy]

What are you talking about? Have you found the matrix yet?

 

[HallsofIvy]

Similar to what I said in post #12.

Yes, it was. But negation had then asked the same question again.

 

[Mark44]

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.

 

[negation]

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

 

[negation]

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