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

1. ## 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 $\displaystyle \rightarrow$ V is a linear map, then Rank(T) $\displaystyle \le$(dim(U) + dim(V ))/2

2. Well, you know that

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

$\displaystyle \text{rank}(T)\le\dim(V),$ right? Can you see where to go from here?

3. 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) $\displaystyle \leq$ dim (U)

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

4. 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) $\displaystyle \leq$ dim (U)

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

Tonio

5. 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) $\displaystyle \leq$ 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).

rank (T) $\displaystyle \leq$ dim (V)
"rank T" is the dimension of T(U) as above and, of course, T(U) is a subspace of V.