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

[maximus101]

Is this statement true or false

if false a counterexample is needed

if true then an explanation

If T : U [tex]\rightarrow[/tex] V is a linear map, then Rank(T) [tex]\leq[/tex] (dim(U) + dim(V ))/2

 

[spamiam]

Is this a homework problem? If so, it should be posted in the homework section of the forum.

In any case, this looks true to me. Here's my proof.

Suppose U and V are finite-dimensional vector spaces and [itex]T \colon U \to V[/itex] is linear. Since [itex] \text{ran}(T) \subseteq V [/itex] (where ran(T) is the range of T), then rank(T) [itex] \leq [/itex] dimV. Also, by the dimension theorem, dimU = rank(T) + nullity(T). Putting these two facts together, we have

[tex]
\frac{dimU + dimV}{2} \geq \frac{dimU + rank(T)}{2} = \frac{rank(T) + nullity(T) + rank(T)}{2} = rank(T) + \frac{nullity(T)}{2} \geq rank(T)
[/tex]

 

[brydustin]

Well strickly speaking, spamiam's equation should be read Right to Left, but otherwise this is absolutely true.

 

[maximus101]

Is this a homework problem? If so, it should be posted in the homework section of the forum.

In any case, this looks true to me. Here's my proof.

Suppose U and V are finite-dimensional vector spaces and [itex]T \colon U \to V[/itex] is linear. Since [itex] \text{ran}(T) \subseteq V [/itex] (where ran(T) is the range of T), then rank(T) [itex] \leq [/itex] dimV. Also, by the dimension theorem, dimU = rank(T) + nullity(T). Putting these two facts together, we have

[tex]
\frac{dimU + dimV}{2} \geq \frac{dimU + rank(T)}{2} = \frac{rank(T) + nullity(T) + rank(T)}{2} = rank(T) + \frac{nullity(T)}{2} \geq rank(T)
[/tex]

Hey, thank you I understand now. Could you give me an example where [tex](dimU + dimV)/2[/tex] > [tex]rankT[/tex]

 

[HallsofIvy]

It is not at all difficult to make up examples.
Suppose U and V have the same dimension but T in not invertible. Then (dim U+ dim V)/2= dim U> rank of T

If T is invertible, then V must have dimension at least equal to the dimension of U. If T is invertible and dim(U)= dim(V), then (dim(U)+ dim(V))/2= dim(U)= rank of T but if dim(V)> dim(U), (dim(U)+ dim(V))/2> dim(U)= rank(T).

 

[maximus101]

It is not at all difficult to make up examples.
Suppose U and V have the same dimension but T in not invertible. Then (dim U+ dim V)/2= dim U> rank of T

okay, in this case, why is dim U > Rank T ?

I'm not sure how to use the fact that it is non-singular

 

[HallsofIvy]

If T is not invertible, then it kernel is not trivial (there exist non-zero v such that T(v)= 0) and its nullity (dimension of the kernel) is greater than 0. rank(T)+ nullity(T)= dim(V) (which equals dim(U) in this example) so that rank(T)= dim(U)- nullity(T)< dim(U)/

 

[maximus101]

If T is not invertible, then it kernel is not trivial (there exist non-zero v such that T(v)= 0) and its nullity (dimension of the kernel) is greater than 0. rank(T)+ nullity(T)= dim(V) (which equals dim(U) in this example) so that rank(T)= dim(U)- nullity(T)< dim(U)/

great got it, thanks

 

[mathwonk]

this follows from the meaning of dimension and rank.

i.e. if v1,...vn is a basis for U, then the rank of T is the dimension of the span of Tv1,...,.Tvn.thus rankT ≤ min{dimU, dimV} ≤ (dim(U) + dim(V ))/2.

 

[mathwonk]

another proof uses the notion of rank of a matrix as the number of pivot columns in a reduced form. this is obviously no greater than the number of all columns, and since there is at most one pivot per row, also no greater than the number of rows.

try cooking up a matrix defining a map R-->R of rank less than one.