Is this statement about the rank of a linear map true or false?

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February 17th 2011, 05:41 AM
maximus101

Is this statement about the rank of a linear map true or false?

Is this statement true or false

if false a counterexample is needed

if true then an explanation

If T : U $\rightarrow$ V is a linear map, then Rank(T) $\le$ (dim(U) + dim(V ))/2
February 17th 2011, 06:22 AM
Ackbeet

Well, you know that

$\text{rank}(T)\le\dim(U)$ and

$\text{rank}(T)\le\dim(V),$ right? Can you see where to go from here?
February 21st 2011, 03:49 AM
maximus101

hey, I think I will use inequalities, I have an idea using the fact that

dimU = rank(T) + nullity(T) and the inequalities you provided?

but I do not quite understand yet even from the link why exactly

rank (T) $\leq$ dim (U)

rank (T) $\leq$ dim (V)
February 21st 2011, 04:09 AM
tonio

Quote:

Originally Posted by maximus101

hey, I think I will use inequalities, I have an idea using the fact that

dimU = rank(T) + nullity(T) and the inequalities you provided?

but I do not quite understand yet even from the link why exactly

rank (T) $\leq$ dim (U)

rank (T) $\leq$ dim (V)

Add sidewise both given inequalities...(Wink)

Tonio
February 21st 2011, 04:37 AM
HallsofIvy

Quote:

Originally Posted by maximus101

hey, I think I will use inequalities, I have an idea using the fact that

dimU = rank(T) + nullity(T) and the inequalities you provided?

but I do not quite understand yet even from the link why exactly

rank (T) $\leq$ dim (U)

Any linear transformation from U to V maps a basis for U onto a basis for T(U).
If T is "one to one", then it maps the set of basis vectors of U onto an independent set and so maps U onto a subspace of V having exactly the same dimension as U. If not, then it maps U onto a subset of V having smaller dimension. "rank T" is the dimension of T(U).

Quote:

rank (T) $\leq$ dim (V)

"rank T" is the dimension of T(U) as above and, of course, T(U) is a subspace of V.