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It's only semi autobiographical
Wednesday, January 26, 2005
Also, see below for badger fun. Especially you Emma.
Comments:
Why I'll have you know that give f=f^2, if we let U be the Image of the function, and W the Kernel, we can suppose v is a member of V, and then v=f(v)+(v-f(v))
Now f(v) is in U and
f(v-f(v))
=f(v)-f^2(v)
=f(v)-f(v)
=0
This v-f(v) is a member of W, which is the kernel of the function. We conclude V=U+W.
Suppose, x is in both, and thus in W, so f(x)=0
Since x is in U also, then it is f(y) for a y in V.
Now f(x)=0=f^2(y)
=>f(y)=f^2(y)
and x=0
Thus V is the direct sum of U and W, U=Im(f) W=Ker(f)
The reverse argument is trivial in that each v can be described as a unique u+v, and through linearity, the v dissapears and the u is unnafected.
The reason Kieran thinks the above is not complicated is the same as the reason he gets better marks than me in maths.
Now f(v) is in U and
f(v-f(v))
=f(v)-f^2(v)
=f(v)-f(v)
=0
This v-f(v) is a member of W, which is the kernel of the function. We conclude V=U+W.
Suppose, x is in both, and thus in W, so f(x)=0
Since x is in U also, then it is f(y) for a y in V.
Now f(x)=0=f^2(y)
=>f(y)=f^2(y)
and x=0
Thus V is the direct sum of U and W, U=Im(f) W=Ker(f)
The reverse argument is trivial in that each v can be described as a unique u+v, and through linearity, the v dissapears and the u is unnafected.
The reason Kieran thinks the above is not complicated is the same as the reason he gets better marks than me in maths.
Aw come on, piece of cake. A real hard proof is showing if there is a Vector space V=Span(y) then there exists a set of vectors X st X contained in Y and X a basis of V. THATS tough.
I know what you are up to! You're trying to trick me into revision! You're just like Steve! DAAAMNNNN YOUUUUUUUU
You know, you're both quite incredibly sad... now I'm going to bed so I can get in some extra revision in in the morning, just for fun...
RECTIFY
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RECTIFY
# posted by 
: 12:09 am
: 12:09 am
