0
$\begingroup$

Let $T:V\to V$ a linear transformation above $\mathbb{F}$ (s.t $1+1\neq 0$) and $T^2 = \operatorname{Id}$.

Prove:

  1. $T$ is invertible.

  2. $\operatorname{Ker}(I-T)=\operatorname{Im}(I+T)$.

  3. $\operatorname{Ker}(I+T)=\operatorname{Im}(I-T)$.

  4. $V=\operatorname{Ker}(I+T)\oplus \operatorname{Ker}(I-T)$.

  1. Let assume that $T$ is not invertible, namely there is no $S$ such that $S\circ T = T\circ S = \operatorname{Id}$, in contradiction to the fact that $T\circ T = \operatorname{Id}$.

2 and 3. I have tried to take an element from $\operatorname{Ker}(I-T)$ and prove it is contained in $\operatorname{Im}(I+T)$.

Let $v\in \operatorname{Ker}(I-T)$ so $(I-T)(v)=0$ from linearity $I(v)-T(v)=0$ which is $v-T(v)=0$ adding $T(v)$ to both sides we get $T(v)=v$ which is a. incorrect as $T$ can be $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$, so which step was incorrect? b. It does not prove that $v\in \operatorname{Im}(I+T)$.

  1. Assume I have proved sections 2 and 3, we can say that $V=\operatorname{Ker}(I+T)\oplus \operatorname{Ker}(I-T)=\operatorname{Ker}(I+T)\oplus \operatorname{Im}(I+T)$ but else from the fact that $\dim V =\dim(\operatorname{Ker} T)+\dim(\operatorname{Im} T)$ how can we conclude on something on the space itself (and not the dimension)?
$\endgroup$
2
  • $\begingroup$ Sorry, what is $I $? I suppose it is the identity... $\endgroup$ Jan 24, 2018 at 23:42
  • $\begingroup$ @LucioTanzini $I$ Identity matrix or the identity transformation $\endgroup$
    – gbox
    Jan 24, 2018 at 23:43

3 Answers 3

1
$\begingroup$

Hints: $\DeclareMathOperator{\Id}{Id}\DeclareMathOperator{\Ker}{Ker}\DeclareMathOperator{\Im}{Im}$

  1. Why not just observe the $T^2=\Id$ precisely means $T$ is its own inverse?
  2. Rewrite $T^2=\Id$ as $\;\Id-T^2=0\iff (\Id-T)\circ(\Id+T)=0$. This implies that $$\Im(\Id+T)\subset\Ker(\Id-T).$$ For the reverse inclusion, consider a vector $v\in \Ker(\Id-T)$: this means $v=T(v)$ (in other words, $v$ is an eigenvector of $T$ for the eigenvalue $1$. Then $$2v=v+T(v),\enspace\text{so }\enspace v=\tfrac12\bigl(v+T(v))\bigr)=(\Id+T)\bigl(\tfrac12v\bigr).$$
  3. Rewrite as $\;\Id-T^2=0\iff (\Id+T)\circ(\Id-T)=0$ (these linear transformations commute).
  4. All you have to prove is that $\;\Ker(\Id-T)\cap\Ker(\Id+T)=\{0\}$.
$\endgroup$
1
$\begingroup$

Note that for question 2 (and similarly for question 3), to establish set equality, you must show two things:

(i) $Ker(I - T) \subseteq Im(I + T)$

(ii) $Im(I + T) \subseteq Ker(I - T)$

In addition, you should expect to use the hypothesis $T^2 = I$. To get you started, I'll outline the proof of (ii).

Take an arbitrary element $v \in Im(I+T)$ so that, by definition, there exists $x \in V$ such that $(I + T)x = v$. By linearity (of the vector space in which $I$ and $T$ reside, not the linearity of $T$ nor $I$ individually), this expands to $I(x) + T(x) = v$. Operating on each side by $T$, we see that $T\big(I(x) + T(x)\big) = T(v)$. Now, by the linearity of $T$, this is equivalent to $T(I(x)) + T^{2}(x) = T(v)$. Because $T(I(x)) = T(x)$ and $T^{2} = I$ by hypothesis, we have $T(x) + I(x) = T(v)$. This, together with the fact that $I(x) + T(x) = v$ from earlier, implies that $T(v) = v$. Thus $v - T(v) = (I - T)v = 0$, so that $v \in ker(I - T)$ by definition of kernel. Because $v \in Im(I+T)$ was chosen arbitrarily, we conclude that $Im(I + T) \subseteq Ker(I - T)$.

I'll leave (i), as well as question 3, as exercises to you.

$\endgroup$
0
$\begingroup$

So in the point 2 you have actually proved a right thing and interpreted it (I guess) in the wrong way, $v=T (v) $ means that $Ker (I-T)=\{v\in V|v=T (v)\} $.

Then you say that $(I+T)(v)\in Im (I+T) $, and you just plug the last expression (the element of the image) into the first equation: $v+T(v)=T(v+T(v))=T(v)+I(v)=v+T(v)$ which is true.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .