Linear Algebra

This is linear transformation stuff and just something I'm confused with.

Suppose imT=kerT, is there a standard matrix for a transformation for which this holds?
Basically, if T:V->V and imT=kerT, show that T^2=0.

And then I am stuck with this.

Let T:R^3->R^3 be a linear transformation. Show that imT^2 is a subset of imT and that kerT is a subset of T^2. Prove the equivilance of the following statements.

a. R^3=kerTΦimT
b. kerT=kerT^2
c. imT=imT^2

Cheers :smile:

Reply 1

lgs98jonee

OP

2

Can anyone offer any assistance with the first bit? I cant workout why kerT=imT?

Reply 2

apd35

9

lgs98jonee

This is linear transformation stuff and just something I'm confused with.

Suppose imT=kerT, is there a standard matrix for a transformation for which this holds?
Basically, if T:V->V and imT=kerT, show that T^2=0.

Can help you with this bit but not the rest (brain has stopped working for today). imT=kerT => T²(x) = T(imT) = T(kerT) = 0
If T maps to it's own kernel, applying T to this image is applying T to its kernel, any map applied to its kernel is 0. Meaning for all x:
T²(x) = 0

Reply 3

Neapolitan

2

apd35

Can help you with this bit but not the rest (brain has stopped working for today). imT=kerT => T²(x) = T(imT) = T(kerT) = 0
If T maps to it's own kernel, applying T to this image is applying T to its kernel, any map applied to its kernel is 0. Meaning for all x:
T²(x) = 0

That bit though needs rewriting as x is an element and ImT, kerT are spaces.

Something like:

"Take x in V so that Tx is in both ImT and kerT. Then

T²(x) = T(Tx) = 0 as Tx is in KerT"

Reply 4

evariste_3086

1

lgs98jonee

This is linear transformation stuff and just something I'm confused with.

Suppose imT=kerT, is there a standard matrix for a transformation for which this holds?
Basically, if T:V->V and imT=kerT, show that T^2=0.

And then I am stuck with this.

Let T:R^3->R^3 be a linear transformation. Show that imT^2 is a subset of imT and that kerT is a subset of T^2. Prove the equivilance of the following statements.

a. R^3=kerTΦimT
b. kerT=kerT^2
c. imT=imT^2

Cheers :smile:

assume (a)
clearly if T(x)=0 T^2(x)=0 so Ker(T^2) contained in Ker T
assume we have v in V st v is not in ker T but T^2(v)=0
then T(T(v))=0 but T(v) is in im T and Ker T which contradicts direct sum assumption. (since this implies T(v)=0 giving v in Ker T which we chose it not to be)
so we have (a)=>(b)

assume (b)
by the isomorphism theorem
V/ker T=im T
V/ker T^2=im T^2
by assumption kerT=kerT^2
hence
im T=V/ker T=V/ker T^2=im T^2

giving (b)=>(C)
note if we assume (c) by same argument we have (c)=>(b)

now assume (c)
T^2(x)=0 implies T(x)=0..
this follows because (c)=>(b) so Ker T^2=ker T
if T^2(x)=0 =>x is in ker T^2=> x is Ker T=>T(x)=0
so if x is in im T and Ker T
then there is v st T(v)=x and T(x)=0
since T(v)=x
T^2(v)=T(x)=0
so T(v)=0
hence x=0
so im T n Ker T= 0
by assumption imt=im T^2
so there exist y such that T^2(y)=Tx
write
x=T(y)+x-T(y)
T(x-T(y))=T(x)-T^2(y)=0
hence x-T(y) is in Ker T
clearly T(y) is in Im T
and the result follows.

Reply 5

Neapolitan

2

evariste

assume (b)
by the isomorphism theorem
V/ker T=im T
V/ker T^2=im T^2
by assumption kerT=kerT^2
hence
im T=V/ker T=V/ker T^2=im T^2

The OP won't have met quotient vector spaces so I'm not sure this will help.

But also these (in bold) are canonical isomorphisms, not equalities, so all you've proved is that ImT is isomorphic to imT^2 - of course we know one is contained in the other by the first part, so a dimension argument means they are equal.

Reply 6

El Stevo

11

looks remarkably like dr. stoy's problem sheet :smile:

Linear Algebra

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