I\'m having issues with a dimension and basis finding problem. The mapping is F:

Question

I'm having issues with a dimension and basis finding problem.
The mapping is F: R^4 --> R^2 and F(x,y,z,t)=(2x+y,x-y-t)

The Ker F={((x,y,z,t) s.t. F(x,y,z,t)=0}
So we have 2x+y=0 and x-y-t=0 which yields x=-1/3t and y=2/3t.
Then the Ker F = span{(-1/3, 2/3, 1)} and consequently dim Ker F = 1

The dim Im(F) must = 3 then. I applied F to the standard bases to get
(2,1),(1,-1),(0,0), and (0,-1).
I inserted the vectors as columns of a matrix and row-reduced.
The problem is I'm getting the Dimension of the Im F to be 2 instead of 3.

Where am I going wrong?
Thanks for your help.

Explanation / Answer

SEE MY COMMENTS IN CAPITALS I'm having issues with a dimension and basis finding problem. The mapping is F: R^4 --> R^2 and F(x,y,z,t)=(2x+y,x-y-t).....OK So we have 2x+y=0 and x-y-t=0 which yields x=-1/3t and y=2/3t.......OK...GOOD... Then the Ker F = span{(-1/3, 2/3,1)} ........NO..YOU HAVE DONE WELL SO FAR ,BUT SLIPPED HERE.. YOU MISSED Z ...Z CAN HAVE ANY VALUE..SO TAKING Z=0, WE GET [-1/3 , 2/3 , 0 , 1] IS ONE VECTOR AND [-1/3 , 2/3 , 1 , 1]...SO DIMENSION IS 2.. ACTUALLY YOU CAN GAUGE IT FROM THE GIVEN MAPPING...F: R^4 --> R^2 HOPE NOW YOU CAN CONTINUE ? IF NOT PLEASE COME BACK. I SHALL GIVE THE COMPLETE SOLUTION AS YOU SHOW AN INCLINATION TO DO IT YOUR SELF ,WHICH IS VERY GOOD TRAIT, I AM NOT SOLVING IT FOR THE PRESENT..HOPE YOU WOULD AGREE... and consequently dim Ker F = 1.............. The dim Im(F) must = 3 then. I applied F to the standard bases to get (2,1),(1,-1),(0,0), and (0,-1). I inserted the vectors as columns of a matrix and row-reduced. The problem is I'm getting the Dimension of the Im F to be 2 instead of 3. Where am I going wrong? Thanks for your help.

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