Finding Basis for Kernel of Linear Transformations

[andrey21]

Identify the Hermite form of the following linear transformations and the basis for its kernel

(x,y,z) = (x-y+2z,2x+y-z,-3x-6y+9z)

So when finding basis for kernel we have to set equal to 0, giving:

x-y+2z=0 (1)
2x+y-z=0 (2)
-3x-6y+9z=0 (3)

So from (1) we can see:

x=y-2z

Therefore (2) becomes:

2y-4z+y-z = 0

3y=5z

So (3) becomes:

-5z+6z -10z +9z=0

15z = 15z
z=z

This is as far as I can get, not sure of the next step, any help be fantastic:

 

[micromass]

Identify the Hermite form of the following linear transformations and the basis for its kernel

(x,y,z) = (x-y+2z,2x+y-z,-3x-6y+9z)

So when finding basis for kernel we have to set equal to 0, giving:

x-y+2z=0 (1)
2x+y-z=0 (2)
-3x-6y+9z=0 (3)

So from (1) we can see:

x=y-2z

Therefore (2) becomes:

2y-4z+y-z = 0

3y=5z

So (3) becomes:

-5z+6z -10z +9z=0

15z = 15z
z=z

This is as far as I can get, not sure of the next step, any help be fantastic:

OK, so you've reduced your system to

x=y-2z
3y=5z
z=z

So, you can even reduce this further to

x=3z
y=(5/3)z

so, the kernel is \{(3z,5z/3,z)~\vert~z\in \mathbb{R}\}. Now, can you find a basis for this?

 

[andrey21]

Sorry I'm a little confused how you got x=3z??

 

[micromass]

Oh, sorry, that was a mistake.

Basically, you have
x=y-2z
y=(5/3)z

Thus
x=(5/3)z-2z=-z/3

I hope I didn't do any wrong calculations this time.

 

[andrey21]

Ok so from what you have said:

=z(-1/3, 5/3, 1)

=(-1/3, 5/3, 1)

 

[micromass]

Yes, that is indeed a basis!

 

[andrey21]

Thank you micromass

 

[HallsofIvy]

Ok so from what you have said:

=z(-1/3, 5/3, 1)

=(-1/3, 5/3, 1)

Just a comment on notation- what you have written is incorrrect, unless you are assuming z= 1. Better would be to say that any vector in the kernel is of the form (-z/3, 5z/3, z)= z(-1/3, 5/3, 1) so that if z= 1 we have (-1/3, 5/3, 1) showing that the dimension of the kernel is 1 and that a basis (which is a set of vectors) is {(-1/3, 5/3, 1)}.

If you are like me and don't like fractions, you could take z= 3 and say that {(-1, 5, 3)} is a basis.