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Find the bases for the kernal

  • Thread starter andrey21
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  • #1
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Find bases for the kernal of the following:


T(x1,x2,x3,x4) = (x1+x2+x3,-x3,x1+x2)


Any help would be great thank you
 

Answers and Replies

  • #2
HallsofIvy
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You have been told this several times now- do NOT just post a problem for other people to do. Show what you have attempted and what you do understand about the problem.

What is the definition of "kernel"of a linear tranformation?

What is the kernel of this particular linear transformation?

Do you know how to find a basis for a given vector space?

(I note you titled this "find the bases for the kernel". There are an infinite number of bases for any vector space. I suspect you really just want to find a basis.)
 
  • #3
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Ok so here is what I know:

T(x1,x2,x3,x4) = (x1+x2+x3,-x3,x1+x2)

Becomes:

1, 1, 1, 0
0, 0,-1, 0
1, 1, 0, 0

Next I reduced it to HERMITE form:

1,1,0,0
0,0,1,0
0,0,0,0

Set equal to zero

1,1,0,0 0
0,0,1,0 0
0,0,0,0 0

Im nt sure where to go from here??
 
  • #4
HallsofIvy
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Yes, that would be a matrix representation of this linear transformation but is irrelevant to answering this question.

I will ask again- What is the definition of "kernel of a linear transformation"?

What is the kernel of this particular linear transformation?
(One simple variation, what is [itex]x_3[/itex] if [itex](x_1, x_2, x_3, x_4)[/itex] is in the kernel? What can you say about [itex]x_4[/itex]?)
 
  • #5
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It is the set of vectors in the domain for which T(v) = 0 correct???
 
  • #6
HallsofIvy
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It is the set of vectors in the domain for which T(v) = 0 correct???
Yes, good! So you have
[tex]T(x_1, x_2, x_3, x_4)= (x_1+x_2+ x_3, x_3, x_1+ x2)= (0, 0, 0)[/tex]
which gives the three equations
[itex]x_1+ x_2+ x_3= 0[/itex]
[itex]x_3= 0[/itex]
[itex]x_1+ x_2= 0[/itex]

Since, by the second equation, [itex]x_3= 0[/itex], the first equation becomes [itex]x_1+ x_2= 0[/itex], the same as the third equation. We cannot solve a single equation in two unknowns-the best we do is solve for one in terms of the other: [itex]x_2= -x_1[/itex]. Since [itex]x_4[/itex] does not appear in any of these equations, there is no restriction on it- it can be any number.

With the conditions that [itex]x_2= -x_1[/itex] and [itex]x_3= 0[/itex] we can write any such vector as
[tex](x_1, -x_1, 0, x_4)= (x_1, -x_1, 0, 0)+ (0, 0, 0, x_4)= x_1(1, -1, 0, 0)+ x_4(0, 0, 0, 1)[/tex].

Now, do you see a basis for the kernel?

You could have done this from your reduced form of the matrix by arguing that
[tex]\begin{bmatrix}1 & 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x_1 & x_2 & x_3 & x_4\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}[/tex]
results in the same equations: [itex]x_1+ x_2= 0[/itex] and [itex]x_3= 0[/itex] but I think it is more instructive to go back to the original definitions than to rely on partially-remembered and not very well understood formulas.
 
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  • #7
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Rite I think I understand now. I have done another example if u wouldnt mind checking if it is correct?

(2x1 -x2 + 4x3, x1+x2-x3 ,x1-x3) = 0
so

2x1 -x2 + 4x3 = 0
x1+x2-x3 = 0
x1-x3=0

Therefore x1 = x3

Therefore second equation becomes:

x2 = 0

first equation becomes

6x1

I think I may have made an error?
 
  • #8
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So I think the basis for kernal is:

x1(1,0,1)

Can someone confirm if this is correct thank you!!
 
  • #9
Oh, you were so close!

Your last equation you put was

6x1

What you should have done is

6x1 = 0

This should help make clear what you did wrong.
 
  • #10
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Ah ok so that would be x1(0,0,1) correct???
 
  • #11
Not quite. you have x3=x1 as well.
 
  • #12
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Ah is it x1(000)??
 
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  • #13
Ah is it x1(000)??
Yup. To be more precise, the only coordinate that is in the kernal is (0,0,0). You could even loosely say that there is no basis for the kernal.
 
  • #14
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Ah brilliant with there being no basis for the kernal is tht like saying it has a dimension of 0??
 
  • #15
chiro
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Ah brilliant with there being no basis for the kernal is tht like saying it has a dimension of 0??
Anything with no variation will be dimensionless.
 
  • #16
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I know this is an old thread but I have a similar question I wish to ask.

Find the basis for the kernel of the following:

T3(x,y,z) = (x-2y,3x-6y)

so

x-2y=0 (1)
x=2y

3x-6y=0 (2)

Therefore (2) becomes:

3(2y)-6y=0

6y=6y
y=y

So the basis for kernel is:

y(2,1,0)+z(0,0,1)

is this correct??
 
  • #17
HallsofIvy
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I know this is an old thread but I have a similar question I wish to ask.

Find the basis for the kernel of the following:

T3(x,y,z) = (x-2y,3x-6y)

so

x-2y=0 (1)
x=2y

3x-6y=0 (2)

Therefore (2) becomes:

3(2y)-6y=0

6y=6y
y=y

So the basis for kernel is:

y(2,1,0)+z(0,0,1)

is this correct??
Any vector in the kernel is of that form. Strictly speaking a 'basis" is a set of vectors: {(2, 1, 0), (0, 0, 1)}.
 

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