Does Linear Algebra Prove Mutual Exclusivity of Transformation Properties?

[stunner5000pt]

Let T: V->V be a linear transformation where V is finite dimensional. Show ath exactly one of (i) an (ii) holds
i) T(v) = 0 for some v not zero in V
ii) T(x) = v has a solution x in V for every v in V

do they mean that if i holds then ii cannot hold?
Ok suppose i holds
T(v) = 0 for some v in V, v not zero
then T(T(v)) = T(0) = 0
let T(v) = x
then T(x) = 0
only solution here is x = v
So T(x) = 0 for all x. ANd thus is not possible for T(x) = v if T(v) = 0

 

[Hurkyl]

do they mean that if i holds then ii cannot hold?

There's more to it than that...


Anyways, your proof doesn't make any sense to me: here are the problems I see:

T(v) = 0 for some v in V, v not zero
then T(T(v)) = T(0) = 0
let T(v) = x

Okay. I don't see the point of defining x to be zero, though.

then T(x) = 0
only solution here is x = v

(1) Solution to what?
(2) For what are you solving? You haven't written down any unknowns!
(3) Why is it a solution?
(4) Why is it the only solution?

So T(x) = 0 for all x.

You've defined x to be T(v) (which is zero) -- so it doesn't make sense to say "for all x".

 

[stunner5000pt]

Clearly i missed the point of the question

what are they asking for actually?
i imples that ii cannot h0old and vice versa?? Is that what I am aiming for here?

 

[HallsofIvy]

Let T: V->V be a linear transformation where V is finite dimensional. Show that exactly one of (i) an (ii) holds
i) T(v) = 0 for some v not zero in V
ii) T(x) = v has a solution x in V for every v in V

"Show that exactly one holds" means "one of these and only one" You must show that for any vector v either (i) is true or (ii) is true and also show that they can't both be true.

What do you know about T(0)? If (i) is true, if T(v)= 0 for non-zero v, then what is the solution to T(x)= 0?

 

[stunner5000pt]

Let T: V->V be a linear transformation where V is finite dimensional. Show that exactly one of (i) an (ii) holds
i) T(v) = 0 for some v not zero in V
ii) T(x) = v has a solution x in V for every v in V

"Show that exactly one holds" means "one of these and only one" You must show that for any vector v either (i) is true or (ii) is true and also show that they can't both be true.

What do you know about T(0)? If (i) is true, if T(v)= 0 for non-zero v, then what is the solution to T(x)= 0?

'
isnt T(0) = 0
suppose T(v) = 0, then if T(x) = 0, then x = v

 

[HallsofIvy]

'
isnt T(0) = 0
suppose T(v) = 0, then if T(x) = 0, then x = v

Did you notice the emphasis in the solution?

Yes, T(0)= 0. Don't you see a problem with that and "T(v)= 0 for some no-zero v"?

 

[stunner5000pt]

Did you notice the emphasis in the solution?

Yes, T(0)= 0. Don't you see a problem with that and "T(v)= 0 for some no-zero v"?

so
suppose T(v) = 0 for v not zero
then suppose T(x) = 0, for some x in V then x =v.
But T(0) = 0 , so x must be zero? But v is non zero. SO the second one cannot hold? Is it like that?

 

[Hurkyl]

In general, T(x) = T(y) does not imply x = y.

 

[stunner5000pt]

In general, T(x) = T(y) does not imply x = y.

right, T in this case is not specified to be one to one
so for T(x) = 0 ,x MUST be v because of the assertion of the first condition.

 

[Hurkyl]

so for T(x) = 0 ,x MUST be v because of the assertion of the first condition.

Why do you think (i) tells you x = v?