Show that if T^2 = 0 then I - T is one-to-one and onto

[AzwinAwin]

Let V be a vector space and T:V->V a linear transformation. Show that if T^2 = 0 then I - T is one-to-one and onto. As I denoted as the identity transformation.

can i relate it with (1-t)(1+t)=(1-t^2) when i do the proving ? how?

 

[fresh_42]

You've found the basic idea behind it. But is it also true, that (A-B)^2 = A^2 - 2AB +B^2 for any linear A,B : V → V? When and why can you apply the binomial formula?

 

[Bipolarity]

Try using this with what you have: If the product of two linear maps is invertible, then each of those maps is invertible.

BiP

 

[AzwinAwin]

your hints re very helpful. thanks a lot.

 

[Abhineshwary]

I'm sorry.
I don't think so the hint help me to solve the question.
Can someone explain more clearly?

 

[Samy_A]

I'm sorry.
I don't think so the hint help me to solve the question.
Can someone explain more clearly?

$(I-T)(I+T)=I²-T²=I-0=I$
and now use @Bipolarity 's remark.

 

[HallsofIvy]

You've found the basic idea behind it. But is it also true, that (A-B)^2 = A^2 - 2AB +B^2 for any linear A,B : V → V?

(A- B)^2= (A- B)(A- B)= A^2- AB- BA+ B^2. That is NOT generally A^2- 2AB+ B^2 because, for linear transformations, it is NOT general y true that AB= BA.

When and why can you apply the binomial formula?

The binomial formula assumes that the values are commutative under multiplication so that the product of the same number of terms in any order can be combined.

 

[fresh_42]

The binomial formula assumes that the values are commutative under multiplication so that the product of the same number of terms in any order can be combined.

... which for linear functions in general is only true to multiples of 1.

 

[mathwonk]

a function is 1-1 and onto iff it has an inverse. you have proved that 1-t has as inverse 1+t. qed. what can you say if T^3 =0? or if T^n = 0?

 

[Bill_Nye_Fan]

I am kind of interested in what the answer to this would be.

I would have said that because T^2 = 0, T must also equal 0. Therefore, 1 - T = 1 - 0 = 1.

x/y = x:y
1/1 = 1:1

Is this process a valid proof for this kind of question? Or am I completely off?

 

[Samy_A]

I am kind of interested in what the answer to this would be.

I would have said that because T^2 = 0, T must also equal 0. Therefore, 1 - T = 1 - 0 = 1.

x/y = x:y
1/1 = 1:1

Is this process a valid proof for this kind of question? Or am I completely off?

No, for operators, T^2 = 0 doesn't imply that T must also equal 0

An example, take the following operator $T: \mathbb R² \to \mathbb R²: (x,y) \mapsto (0,x)$
$T(x,y)=(0,x)$, $T²(x,y)=T(0,x)=(0,0)$
So $T \neq 0$, but $T²=0$.

Also note: $(I-T)(x,y)=I(x,y)-T(x,y)=(x,y)-(0,x)=(x,y-x) \neq I(x,y)$

 

[Bill_Nye_Fan]

No, for operators, T^2 = 0 doesn't imply that T must also equal 0

An example, take the following operator $T: \mathbb R² \to \mathbb R²: (x,y) \mapsto (0,x)$
$T(x,y)=(0,x)$, $T²(x,y)=T(0,x)=(0,0)$
So $T \neq 0$, but $T²=0$.

Also note: $(I-T)(x,y)=I(x,y)-T(x,y)=(x,y)-(0,x)=(x,y-x) \neq I(x,y)$

Okay, thank you. I am unfamiliar with what operators are or how they work, so I sort of just tried to solve it as a regular equation. I suppose I will need to go a research these operators further now.

 

[Abhineshwary]

Okay, thanks a lot everyone!