Find a kernel and image basis of a linear transformation

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Homework Statement



Find a kernel and image basis of the linear transformation having:

[tex] \displaystyle T:{{\mathbb{R}}^{3}}\to {{\mathbb{R}}^{3}}[/tex] so that

[tex] \displaystyle _{B}{{\left( T \right)}_{B}}=\left( \begin{matrix}
1 & 2 & 1 \\
2 & 4 & 2 \\
0 & 0 & 0 \\
\end{matrix} \right)[/tex]

[tex] \displaystyle B=\left\{ \left( 1,1,0 \right),\left( 0,2,0 \right),\left( 2,0,-1 \right) \right\}[/tex]

Homework Equations




The Attempt at a Solution



For the image basis it is easy given the fact that the rank of the associated matrix is 1 so, the image is generated by one column.

The problem comes when finding the Kernel basis. My idea is to save the general fromula of the linear map which would work for sure but I wanted to know if there's a quicker way of doing it without finding the general formula of the linear map.


Thanks!
 

Answers and Replies

  • #2
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To find the kernel, you'll need to find the (x,y,z) such that

[tex]\left(\begin{array}{ccc} 1 & 2 & 1\\ 2 & 4 & 2\\ 0 & 0 & 0\end{array}\right)\left(\begin{array}{c} x\\ y\\ z\\\end{array}\right)=\left(\begin{array}{c} 0\\ 0\\ 0\\\end{array}\right)[/tex]

This is a system of equations whose solution is not that hard to find.

(note: you know the rank of the matrix, so the dimension of the kernel is easy to find out)
 
  • #3
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Yeah, the dimension of the kernel should be 2. I also thought about that system, but I've got some doubts with it.

We have:

$$\left(\begin{array}{ccc} 1 & 2 & 1\\ 2 & 4 & 2\\ 0 & 0 & 0\end{array}\right)\left(\begin{array}{c} x\\ y\\ z\\\end{array}\right)=\left(\begin{array}{c} 0\\ 0\\ 0\\\end{array}\right)$$

But there, what is (x,y,z)? Are they coordinates or is it a vector? What I understand is that if I multiply the associated matrix with the COORDINATES of a vector in base B then the result is the coordinates of the vector TRANSFORMATION. Is that right?

Thanks!
 
  • #4
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The (x,y,z) are coordinates of a vector with respect to base B. If you want to find the corresponding vector, this can easily be done.
 
  • #5
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Ahhm, I see. So we get that:

[tex]Ker\left( T \right) = \left\{ {\left( {x,y, - x - 2y} \right)} \right\}[/tex]

Right? So there we have the set of the coordinates of the vectors in basis B whose transformation is 0. Or are they vectors?

Thanks
 
  • #6
22,129
3,298
Ahhm, I see. So we get that:

[tex]Ker\left( T \right) = \left\{ {\left( {x,y, - x - 2y} \right)} \right\}[/tex]

Right? So there we have the set of the coordinates of the vectors in basis B whose transformation is 0. Or are they vectors?

Thanks

They are coordinates for vectors. You can first take special values for x and y to find two special coordinates in the kernel. Then you have to find the corresponding vector.
 
  • #7
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Alright so, we have the set of the coordinates of the vectors whose images are the null vector.
To find the generator of the coordinates:

[tex]\left( {x,y, - x - 2y} \right) = x\left( {1, - 1,0} \right) + y\left( {0,1, - 2} \right)[/tex]

So: [tex]\left\{ {\left( {1, - 1,0} \right),\left( {0,1, - 2} \right)} \right\}[/tex] is a generator (basis) of the coordinates. Now, how can I get the basis of the actual vectors but not in base B?


Thanks!
 
  • #8
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Do you know what coordinates are?? How are they defined??
 
  • #9
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Yes, they are the coefficients of the linear combination of the base. I got that

$$\left\{ {\left( {1, - 1,0} \right),\left( {0,1, - 2} \right)} \right\}$$

is a basis of the kernel. But those elements are vectors or what?
 
  • #10
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Ok, so take the linear combination.

For example, the coordinates (1,2,3) would correspond to the vector

[tex](1,1,0)+2(0,2,0)+3(2,0,-1)[/tex]
 
  • #11
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Ahhh I get it now by that way! So:

[tex]x\left( {1,1,0} \right) + y\left( {0,2,0} \right) + \left( { - x - 2y} \right)\left( {2,0, - 1} \right) = \left( { - x - 4y,x + 2y,x + 2y} \right)[/tex]

And:

[tex]\left( { - x - 4y,x + 2y,x + 2y} \right) = x\left( { - 1,1,1} \right) + y\left( { - 4,2,2} \right)[/tex]

So a basis of the kernel is:
[tex]{\left( { - 1,1,1} \right),\left( { - 4,2,2} \right)}[/tex]

Now just a last question. What was wrong about:

$$\left( {x,y, - x - 2y} \right) = x\left( {1, - 1,0} \right) + y\left( {0,1, - 2} \right)$$

How could I solve the problem following that way? It would be the same, wouldnt it?


Thanks!!
 
  • #12
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3,298
Now just a last question. What was wrong about:

$$\left( {x,y, - x - 2y} \right) = x\left( {1, - 1,0} \right) + y\left( {0,1, - 2} \right)$$

How could I solve the problem following that way?

Nothing was wrong about that. It just gives you coordinates of a vector. That is, (1,-1,0) and (0,1,-2) are coordinates of a certain vector. Take linear combinations to find the actual vector.
 
  • #13
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Thank you very much for your efficiency and quickness for helping (the best combination) :)
 

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