null

• May 8th 2011, 11:31 AM
alexandrabel90
null
given that f: V->W and g: W->Z. to show that null (gf) less than or equal to null (g) + null(f),

can i say that null(gf) = null(g) intersect rank (f) < null(g) < null(g) +null(f)?
• May 8th 2011, 01:07 PM
Deveno
by null(g) do you mean the nullspace, or the nullity of g (because one is a vector space, the other is an integer)?

the dimension of ker(gf) might not be dim(ker(g)∩f(V)). for example suppose V = W = Z = R^2,

and g(x,y) = (x,y), and f(x,y) = (x,0). then ker(gf) = ker(f) = {(0,y) : y in R}, which has dimension 1, but ker(g)∩f(V)

= {(0,0)} ∩ {(0,y): y in R} = {(0,0)}, which has dimension 0.

use the rank-nullity therorem instead. it may be helpful to consider the linear transformation g*:f(V)-->Z g*(w) = g(w),

(g restricted to the range of f) and to prove first that null(g*) ≤ null(g).
• May 8th 2011, 01:11 PM
alexandrabel90
I mean nullity
• May 8th 2011, 01:16 PM
alexandrabel90
Is this working for nullity or nullspace? By proving that null(g*)<null(g), we are showing that null(gf) <null(g) right?

Sorry but im quite confused about how to go about using linear transformation here
• May 8th 2011, 01:39 PM
Deveno
yes but it really is ≤. for all we know, f might be onto, in which case g* = g.

sure, null(gf) ≤ null(g), but that is taking "too big a bite" all at once. we want to get some idea of how much smaller null(gf) is than null(g).

null(h), for any linear transformation h is dim(ker(h)), where ker(h) is the nullspace.

now f may "shrink down the size" of the space of vectors v available for gf to annihilate, whereas g has the whole of W to look at.

gf will take to 0 anything in ker(g)∩f(V), but g may take even more vectors of W to 0.

the only thing we have to guage "how much smaller" null(gf) is than null(g) is the behavior of f.

unfortunately, g is defined on all of W, not just on f(V)∩W = f(V).

which amongst f,g and gf have common domains? gf, and f. the rank-nullity theorem says:

rank(f) + null(f) = rank(gf) + null(gf) = dim(V).

now, we want to keep the "null" parts of this equation, but we need to somehow get rid of the "rank" parts.

we know that rank(g) + null(g) = dim(W), but that's not very helpful.

but if we consider g*, we have: rank(g*) + null(g*) = dim(f(V)) = rank(f).

note that rank(g*) = rank(gf) (because g* only looks at w's with w = f(v)).

so rank(gf) + null(g*) = rank(f). now, if we combine our two equations we have:

rank(gf) + null(g*) + null(f) = rank(gf) + null(gf)...can you see what comes next?
• May 8th 2011, 01:53 PM
alexandrabel90
Since null(g*) is less than or equal to null(g), we get null(gf) less than or equal to null(g) + null (f).

But what about when it means nullity. Isit the same method?
• May 8th 2011, 04:19 PM
Deveno
i am using null(f) to mean the nullity of f, i am referring to the nullspace as ker(f) to emphasize the distinction.
• Dec 16th 2013, 02:51 PM
sipnayan
Re: null
how to show null(g*)<= null g?
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