Linear transformation that is like projection

Question

I am given a linear transformation $\;T:V\to V\;$, with $\;V\;$ linear space over field $\;F\;$ , and with $\;\dim\text{Im}\,T=1\;$ .

I am asked to prove that there exists scalar $\;c\in F\;$ such that $\;T\circ T=cT\;$ .

This is what I did so far: Since

$$\;\dim\text{Im}\,T=1\implies \text{Im}\,T=\text{Span}\,\{w\}=:W\;\;,\;\;0\neq w\in V\;$$ This for me is like projection of $\;V\;$ onto one dimensional space, but I can't find that $\;c\;$. I thought

$$v\in V\implies Tv\in W\implies \exists\; c_v\in F\;\;s.t.\;\;Tv=c_vw$$

but then I am in problems when I do

$$\color{red}{T\circ T(v)}=T(c_vv)=c_vTv=c_v^2w\stackrel{??}=\color{red}{cTv}=cc_vw$$

how can I go to get that fixed scalar $\;c\;$ for all vectors in $\;V\;$ ?

Any help will be greatly appreciated.

Answer 1

Use that $w$ is an eigenvector for$~T$ (it's span is a $T$-invariant subspace, namely the image of $T$). The corresponding eigenvalue gives you your $c$.

If you don't know the term eigenvector, that $w$ is an eigenvector for$~T$ just means that $T(w)$ is in the span of $\{w\}$. This is obvious, because that span is the image of$~T$. But the consequence is that there is some scalar $c$ such that $T(w)=cw$. Since $w\neq0$ there can be only one such$~c$. That's your $c$.

Now consider any other vector. Again $T(v)$ is a scalar multiple of$~w$, though probably a different one; say $T(v)=dw$. Now try to prove that $T(T(v))=cT(v)$ for the scalar $c$ obtained from$~w$, and independently of $d$. It is quite easy. This shows that $T\circ T=cT$.

See also this answer.

Answer 2

Take a basis for the kernel of $T$, and extend to a basis of $V$. Now argue that the extension only required 1 additional vector (if $V$ is finite dimensional this follows from rank nullity, but even if $V$ is infinite dimensional this still holds, proof left to the end).

Now to see that $T\circ T = c T$ for some $c$, note we have $V= \text{Ker } T \oplus v$ where wlog we may assume $Tv=w$ (since $T\neq 0$). For $k\in \text{Ker T}$ we have $T(k+av)=aw$. Write $w=k'+bv$ where $k'\in \text{Ker } T$. Then $T(aw)=abw$. Hence $T\circ T= b T$, since we may divide by $a$ if $a\neq 0$, and when $a$ is zero, $T\circ T$ and $T$ applied to $k+0v$ are both zero, so the equation still holds. This completes the proof.

ADDENDUM: A proof of the claim in the the first paragraph in the infinite dimensional case would be to show that any two vectors which are not in the kernel have some linear combinatino in the kernel; suppose $v,u$ are such that $Tv=aw$ and $Tu=bw$ where $a,b\neq 0$. Then $Tv/a-Tu/b=0$ and hence $v/a-u/b$ is in the kernel of $T$.

Answer 3

$T:Im(T)\to Im(T)$ is linear and since $\dim(Im(T))=1$ it is equal to multiplication by a scalar ($T(x)=cx$, for some scalar $c$ and all $x\in Im(T)$).

In fact, if $Im(T)=span(w)$ then for each $x\in Im(T)$, there is a scalar $a_x$ such that $x=a_xw$. By linearity $T(x)=T(a_xw)=a_xT(w)=a_xcw=cx$, where $c$ is the unique scalar such that $T(w)=cw$.

Then, in particular, since $\forall x\in V,\ T(x)\in Im(T)$, we get $\forall x\in V,\ T(T(x))=cT(x)$.

Answer 4

Since $V/\ker T\simeq\text{Im}T$，we have $\dim(V/\ker T)=1$ so there is a $u\in V$ such that $V=\langle u\rangle+\ker T=V$. Since $u\notin\ker T$, $Tu\neq0$. Because $\dim\text{Im}T=1$, there exist a $c\in F$ Such that $T(Tu)=cTu$. Then $\;T\circ T=cT\;$ (This equation holds for $v\in\langle u\rangle$ and $v\in\ker T$, thus holds for $v\in V$).

Answer 5

You were nearly there, but as Marc van Leeuwen said in the comments you wrote $T\circ T(v)=T(c_vv)$ when you needed $T(c_vw)$ instead. Fixing this makes your method work immediately: $$T\circ T(v)=T(c_vw)=c_vTw=c_vc_ww=c_wc_vw=c_wTv$$ and $c_w$ is constant as you vary $v$.