Is Rank(TS) Always Less Than or Equal to Rank(T)?

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Discussion Overview

The discussion revolves around the relationship between the ranks of two linear maps, S and T, specifically exploring whether Rank(TS) is always less than or equal to Rank(T). The scope includes mathematical reasoning and exploration of linear algebra concepts related to vector spaces and linear transformations.

Discussion Character

  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant suggests that since im(TS) is a subspace of W and im(T) is also a subspace of W, it follows that Rank(TS) should be less than or equal to Rank(T).
  • Another participant refines this by stating that im(TS) is a subspace of im(T), leading to the conclusion that Rank(TS) is less than or equal to Rank(T), but questions whether im(T) equals W.
  • A third participant clarifies that im(T) equals W only when T is onto, indicating a condition for the previous statements.
  • A later reply emphasizes that the reasoning is correct, noting that for finite-dimensional subspaces, the rank condition holds true when im(TS) is a subset of im(T).

Areas of Agreement / Disagreement

Participants generally agree on the direction of the reasoning regarding the ranks of the linear maps, but there is uncertainty about the conditions under which im(T) equals W, indicating that multiple views remain on this aspect.

Contextual Notes

There is an assumption that the subspaces involved are finite-dimensional, which may limit the generality of the discussion. The dependence on the properties of the linear map T (specifically whether it is onto) is also noted but not resolved.

garyljc
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I came across a question saying
Let S: U -> V and T:V -> W be linear maps, where U V and W are vectors spaces over the same field K
Show that Rank(TS) =< Rank(T)

This is my attempt
the im(TS) is a subspace of W
and so is the im(T)

am I missing out something ?
 
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This is my second attempot oon it
im(TS) is a subspace of im(T)
and im(T) is a subspace of W
therefore rank(TS) =< rank (T)
is it correct ?

but isn't im(T) = W ?
 
[tex]Im(T) = W[/tex] when [tex]T[/tex] is onto.

Your second attempt is the right direction, you might want to explain more of the details.
 
You were pretty much done, but you did some extra stuff that wasn't necessary.

Assuming the subspaces involved are finite-dimensional, [itex]\operatorname{rank} TS \le \operatorname{rank} T[/itex] means, by definition, [itex]\dim(\operatorname{im} TS) \le \dim(\operatorname{im} T)[/itex]. This happens exactly when [itex]\operatorname{im} TS \subseteq \operatorname{im} T[/itex], which you know is true.
 

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