- #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

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- Thread starter andrey21
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- #1

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

Any help would be great thank you

- #2

HallsofIvy

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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

- #3

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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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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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Yes, good! So you haveIt is the set of vectors in the domain for which T(v) = 0 correct???

[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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(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!!

x1(1,0,1)

Can someone confirm if this is correct thank you!!

- #9

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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

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Not quite. you have x3=x1 as well.

- #12

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Ah is it x1(000)??

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- #13

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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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Find the basis for the kernel of the following:

T

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

Find the basis for the kernel of the following:

T_{3}(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??

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