Linear Algebra, kerT subseteq kerT^2

let T:V -> V be a linear transformation

claim: kerT subseteq kerT^2
proof: let v in kerT, now
T(v) = 0 and by the properties of linear transformation
T(0) = 0, so T^2(v) = (TT)(v) = T(T(v)) = T(0) = 0
∴ kerT subseteq kerT^2

claim2: kerT = kerT^2
now let w in imT
clearly, imT subseteq V, so
kerT with restriction to imT subseteq kerT

or its not? imT is only a part of V so it should generate only part of kerT;
in other words, if imT is a subset or equal to V theres nothing new in kerT that
it can generate, because I've already considered kerT to be part of kerT^2

and so kerT^2 = kerT

im preety sure that it is not the case that kerT^2 = kerT in general...

but when I think about it, I get a little confused...

can someone clarify this?

Thanks ^_^

Reply 1

RichE

[QUOTE='[maciek]']
and so kerT^2 = kerT

Where does this line come from? I can't see how you arrive at it.

Re other points:

\ker T_{|ImT} = \ker T \cap Im T[br]

and why do you start with w from ImT?

Reply 2

[maciek]

consider T(T(v))

now if v is in kerT T(T(v)) = 0 so its in kerT^2

if its not in kerT, its in imT

so T(v) = w and T(T(v)) = T(w)

and T(w) = 0 iff w is in kerT

so take any v in kerT^2 and it must be in kerT

and take any v in kerT and it must be in kerT^2

so kerT^2 should be equal to kerT

Reply 3

Neapolitan

[QUOTE='[maciek]']consider T(T(v))

now if v is in kerT T(T(v)) = 0 so its in kerT^2

if its not in kerT, its in imT

so T(v) = w and T(T(v)) = T(w)

and T(w) = 0 iff w is in kerT

so take any v in kerT^2 and it must be in kerT

and take any v in kerT and it must be in kerT^2

so kerT^2 should be equal to kerT

The line in bold isn't true.

It's not even generally true that V = kerT + ImT (let alone their union).