1
$\begingroup$

Let $T$ be a linear operator on a vector space $V\rightarrow V$. Let $K_r$ and $W_r$ denote the kernel and image , respectively, of $T^r$.

(b) The following conditions might or might not hold for a particular value of $r$:

(1) $K_r=K_{r+1}$, (2)$W_r=W_{r+1}$, (3) $W_r\cap K_1=\{0\}$, (4) $W_1+K_r=V$. Find all implications among the conditions (1)-(4) when V is finite dimensional.

I know 1, 2, and 3 are equivalent, but 4 is giving me a lot of trouble.

I know that if 1 holds, $K_r\cap W_r=\{0\}$, so $K_r\oplus W_r =V$, and since $W_r\subset W_1$, $K_r + W_1 =V$. I also know the 4 implies that there exists some subspace $W_1'$ such that $K_r\oplus W_1'=V$, but I don't know how to prove $W_1'=W_r$.

Note: there is a solution to this on this site here, but it uses notation that I don't understand.

$\endgroup$
  • $\begingroup$ @Drike just an integer, that was supposed to be $T^r$, just edited it $\endgroup$ – Miles Johnson Oct 13 '18 at 16:23
  • $\begingroup$ @MorganRodgers my bad, those were supposed to be subscripts $\endgroup$ – Miles Johnson Oct 13 '18 at 16:32
  • $\begingroup$ Nice, looks good now. One last question: are you allowed to assume that $V$ is finite dimensional? $\endgroup$ – Morgan Rodgers Oct 13 '18 at 18:57
  • $\begingroup$ @MorganRodgers yes you are $\endgroup$ – Miles Johnson Oct 13 '18 at 19:03
1
$\begingroup$

As long as you are allowed to assume that $V$ is finite dimensional, here is something that may help for (4):

Notice that for any fixed $r > 1$,

  1. $\dim{W_{r}}+\dim{K_{r}} = \dim{V}$ (by rank-nullity)
  2. $\dim{W_{r+1}} \leq \dim{W_{r}}$, since $W_{r+1}\subseteq W_{r}$.
  3. $\dim{K_{r+1}} \geq \dim{K_{r}}$, since $K_{r} \subseteq K_{r+1}$.

Also, remember that that for any subspaces $A$,$B$, $$\dim{(A+B)} = \dim{A}+\dim{B} - \dim{(A \cap B)}.$$

Since you know that (1),(2), and (3) are equivalent, see if you can show that $\{(1),(2),(3)\} \Rightarrow (4)$, or that $(4) \Rightarrow \{(1),(2),(3)\}$.

If you assume (4), then every vector $\vec{v} \in V$ can be written as a linear combination of a vector in $W_{1}$ and a vector in $K_{r}$. So if we take a vector $\vec{u} \in W_{r}$ then $\vec{u} = T^{r}(\vec{w} + \vec{k})$ where $\vec{w} \in W_{1}$ and $\vec{k} \in K_{r}$. Thus $$\vec{u} = T^{r}(\vec{w} + \vec{k}) = T^{r}(\vec{w}) + T^{r}(\vec{k}) = T^{r}(\vec{w}) + 0 = T^{r}(\vec{w}).$$ Now since $\vec{w} \in W_{1}$, $\vec{w} = T(\vec{v})$ for some $v \in V$, and so $$T^{r}(\vec{w}) = T^{r}(T(\vec{v})) = T^{r+1}(\vec{v}) \in W_{r+1}.$$ This shows that $W_{r} \subseteq W_{r+1}$, and combined with point 2. above, shows that $W_{r}=W_{r+1}$ (and so $(4) \Rightarrow \{(1),(2),(3)\}$).

I'll leave the other direction for you. This proof really is a lot easier if you understand quotient spaces and can understand the other version; I'm not sure where you found this exercise, but you should make sure you work through all of the material leading up to it so you know you have all of the tools needed.

$\endgroup$
  • $\begingroup$ I'm still not getting it. I feel like I have to prove that intersection is null, but I keep just getting it equal to something that is not necessarily zero. Is there some property or something I should know about that intersection? $\endgroup$ – Miles Johnson Oct 13 '18 at 19:40
  • $\begingroup$ But if you plug that in you get dim$V$-dim$(W_1\cap K_r)\leq$ dim$V=>0\leq$ dim$(W_1\cap K_r)$, which doesn't help $\endgroup$ – Miles Johnson Oct 13 '18 at 22:19
  • $\begingroup$ @MilesJohnson I flipped my inequality somehow, sorry about that. Have edited my answer above. $\endgroup$ – Morgan Rodgers Oct 13 '18 at 23:16
  • $\begingroup$ $W_{r+1}\subset W_r$ is the same as point 2. But ya you're right I should probably learn to use quotient spaces. Probably the point of this problem was to show how useful they can be. $\endgroup$ – Miles Johnson Oct 13 '18 at 23:28
  • 1
    $\begingroup$ That was a typo, the argument in my proof shows $W_{r} \subseteq W_{r+1}$ (I start with $\vec{u} \in W_{r}$ and show also $\vec{u} \in W_{r+1}$), and along with point 2. shows they are equal. $\endgroup$ – Morgan Rodgers Oct 14 '18 at 0:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.