# Find the bases for the kernal

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

Homework Helper
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??

Homework Helper
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???

Homework Helper
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.

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andrey21
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?

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)??

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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??

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??

Homework Helper
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)}.