Proof - linear isomorphism

Question

I have the following question: Let $X$ be a finite-dimensional linear space, and let $T \in L(X)$ such that $T^{k} = 0$ for some $k \ge 1$. $\DeclareMathOperator{\id}{id}\DeclareMathOperator{\Ker}{Ker}$ Show that $\id_{x} + T$ is a linear isomorphism.

My idea:

We have to show that $\id_{x} + T$ is linear and $\id_{x} + T$ is an isomorphism which holds if and only if $\Ker((\id_{x} + T))= $ {0}

Let $x_{1},x_{2} \in X$ then $(\id_{x} + T)(x_{1}+x_{2}) = \id_{x}(x_{1}+x_{2}) + T(x_{1}+x_{2}) = \id_{x}(x_{1}) + \id_{x}(x_{2}) + T(x_{1}) + T(x_{2}) = (\id_{x} + T)(x_{1}) + (\id_{x} + T)(x_{2})$

Let $x \in X, a \in \Re $ then $(\id_{x} + T)(a\cdot x) = \id_{x}(a\cdot x) + T(a\cdot x) = a \cdot \id_{x}(x) + a \cdot T(x) = a \cdot (\id_{x}(x) + T(x)) = a \cdot ((\id_{x} + T)(x)) $

So $\id_{x} + T$ is linear

Still to prove: $\Ker((\id_{x} + T)) := \{x \in X : (\id_{x} + T)(x) = 0 \} = \{0\}$

So let $x \in \Ker((\id_{x} + T))$ then

$(\id_{x} + T)(x) = \id_{x}(x) + T(x) = x + 0 = x = 0 $

q.e.d.

My question: Does this make any sense? I'm very unsure about the given $T^{k}=0$ and how to handle it correctly.

Thanks for your help :)

Answer 1

It is $T(x)+Id(x)=0$ implies that $T^{k-1}(T(x)+x)=T^k(x)+T^{k-1}(x)=T^{k-1}(x)=0$. Recursively, you obtain $x=0$ as follows:

Suppose that $T^{k-i}(x)=0$, $T(x)+x=0$ implies that $T^{k-i-1}(T(x)+x)=T^{k-i}(x)+T^{k-i-1}(x)=T^{k-i-1}(x)=0$.

Answer 2

First of all, you are not required to verify that kernel of $id+T$ is a subspace. It always is.

Secondly if $x $ is in the kernel you got $x+Tx=0$. You are not given that $Tx=0$ so you proof does not work. Now start with $Tx=-x$ and apply $T$ repeatedly to get $x=0$. [You get $T^{i}(x)=(-1)^{i}x$. Put $i=k$]. That will complete the proof.

Answer 3

In view of the last part, $(id+T)x=0$ iff $id(x) + T(x)=0$ iff $T(x)=-x$. Thus $T^2(x)=x$, $T^3(x)=-x$ and so on. But $T^k(x)=\pm x =0$ by hypothesis and so $x=0$.