Linear Operator Rules for Ker T and Ker S: Solving for T and S in R^3

AI Thread Summary
The discussion centers on finding a linear operator S such that its kernel is equal to the image of T, and its image is equal to the kernel of T. Participants express uncertainty about starting the problem and whether S can be assumed to be the inverse of T. It is clarified that if the kernels are not trivial, T does not have an inverse. Despite finding the kernels, the participants remain stuck on determining the rule for S, suggesting that a geometric description might help clarify the problem. The conversation highlights the complexity of linear transformations and their relationships in R^3.
Cyannaca
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Let T: R^3 -> R^3 be a linear operator and
T(x;y;z)= (x -y + 3z;2x-y+8z;3x -5y+5z).

Find the rule, for the linear operator S: R^3 -> R^3 such that ker S= I am T and I am S=Ker T.

I'm not really sure how I should start this problem. Also I would like to know if I can assume S is the inverse operator of T.
 
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Seems obvious to me that the first thing to do is to find ker T and I am T. *shrug*

Also I would like to know if I can assume S is the inverse operator of T.

What do you think?
 
" Also I would like to know if I can assume S is the inverse operator of T."

If the two kernals are not trivial, then T does not HAVE an inverse!
 
I found ker T and I am T and I'm still stuck. I know it's not the inverse operator but I still don't know how to find the rule for the linear operator S.
 
Maybe if you described them geometrically, it would spark an idea?
 

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