What Is the Basis for the Kernel of These Linear Transformations?

[andrey21]

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

 

[HallsofIvy]

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.)

 

[andrey21]

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??

 

[HallsofIvy]

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 x_3 if (x_1, x_2, x_3, x_4) is in the kernel? What can you say about x_4?)

 

[andrey21]

It is the set of vectors in the domain for which T(v) = 0 correct?

 

[HallsofIvy]

It is the set of vectors in the domain for which T(v) = 0 correct?

Yes, good! So you have
T(x_1, x_2, x_3, x_4)= (x_1+x_2+ x_3, x_3, x_1+ x2)= (0, 0, 0)
which gives the three equations
x_1+ x_2+ x_3= 0
x_3= 0
x_1+ x_2= 0

Since, by the second equation, x_3= 0, the first equation becomes x_1+ x_2= 0, 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: x_2= -x_1. Since x_4 does not appear in any of these equations, there is no restriction on it- it can be any number.

With the conditions that x_2= -x_1 and x_3= 0 we can write any such vector as
(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).

Now, do you see a basis for the kernel?

You could have done this from your reduced form of the matrix by arguing that
\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}
results in the same equations: x_1+ x_2= 0 and x_3= 0 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.

 

[andrey21]

Rite I think I understand now. I have done another example if u wouldn't 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?

 

[andrey21]

So I think the basis for kernal is:

x1(1,0,1)

Can someone confirm if this is correct thank you!

 

[transphenomen]

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.

 

[andrey21]

Ah ok so that would be x1(0,0,1) correct?

 

[transphenomen]

Not quite. you have x3=x1 as well.

 

[andrey21]

Ah is it x1(000)??

 

[transphenomen]

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.

 

[andrey21]

Ah brilliant with there being no basis for the kernal is tht like saying it has a dimension of 0??

 

[chiro]

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.

 

[andrey21]

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??

 

[HallsofIvy]

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)}.