Question

Let V be a finite dimensional vector space and A \(\in L(V)\). Suppose \(N(A)=N(A^2)\)
a) Prove \(V=N(A)\oblus R(A)\)
b) Prove that \(\bar A\in L(V/N(A))\) , defined by A[v]=[Av]is well-defined

Answer 1

I got part a
not part b

Answer 2

someone gives me the proof:
\[\bar A :V/N(A)\rightarrow V/N(A)\]
if \(\vec u'\in [\vec u] \) then \(\vec u' -\vec u\in N(A)\) --> Au'=Au
-->A[u']=[Au]
so, \(\bar A\) is well defined
I don't understand a word. :(
The argue start from A and the conclusion is on \(\bar A\)

Answer 3

So let's go through the argument step by step.
1. If \(\vec{u}'\in[\vec{u}]\), then \(A\vec{u}'=A\vec{u}\), so \(A\vec{u}'-A\vec{u}=\vec{0}\). By factoring \(A\) out, this means that \(A(\vec{u}'-\vec{u})=\vec{0}\). So by definition, \(\vec{u}'-\vec{u}\in N(A)\).

Answer 4

yes.

Answer 5

To get that \(A[\vec{u}']=[A\vec{u}]\), in set notation we have that\[A[\vec{u}']=\{\vec{v}\in V:\vec{v}=A\vec{u}''\text{ for some }\vec{u}''\in[\vec{u}']\}\]\[[A\vec{u}]=\{\vec{v}\in V:A(A\vec{u})=A\vec{v}\}.\]

Answer 6

Sorry, having some internet problems on my side. I've nearly convinced myself this is true, but there's one step I'm not sure of yet.

Answer 7

The net is on and off here, too. If you don't see me, please, leave the instruction here. I will get right after I can access the net.

Answer 8

We're definitely going to have to use the fact that \(N(A)=N(A^2)\).

Answer 9

Now we have that since \([\vec{u}']=[\vec{u}]\), \[\begin{aligned}
A[\vec{u}']&=\{\vec{v}\in V:\vec{v}=A\vec{u}''\text{ for some }\vec{u}''\in[\vec{u}']\}\\
&=\{\vec{v}\in V:\vec{v}=A\vec{u}''\text{ for some }\vec{u}''\in[\vec{u}]\}\\
&=\{\vec{v}\in V:\vec{v}=A\vec{u}''\text{ s.t. }A\vec{u}''=A\vec{u}\}\\
&=\{\vec{v}\in V:\vec{v}=A\vec{u}\}\\
&=\{\vec{v}\in V:A\vec{v}=A^2\vec{u}\}\quad\leftarrow\text{This step doesn't seem right to me}\\
&=[A\vec{u}].
\end{aligned}\]

Answer 10

where do you want me to go to? hihihihi

Answer 11

This argument shows that we have one containment (\(A[\vec{u}']\subseteq[A\vec{u}]\)). It's only other one that's giving me some trouble.

Answer 12

Well, if \(\vec{v}\in [A\vec{u}]\), then \(A^2\vec{u}-A\vec{v}=\vec{0}\), so \(A\vec{u}-\vec{v}\in N(A)=N(A^2)\). But I don't immediately see how that helps.

Answer 13

Ah. I think I can convince myself now. Since \(N(A)=N(A^2)\), we have that if \(A\vec{v}=\vec{0}\) iff \(A(A\vec{v})=\vec{0}\) as well. So restricting our attention to the subset \[V\supseteq S=\{\vec{w}\in V:\vec{w}=A\vec{v}\text{ for some }\vec{v}\in V\},\]we see that \(N(A)\cap S=\vec{0}\). Furthermore, since \(A[\vec{u}']\subseteq S\), we do in fact have that \[\{\vec{v}\in V:\vec{v}=A\vec{u}\}=\{\vec{v}\in V:A\vec{v}=A^2\vec{u}\}.\]The conclusion I had above then follows.

Answer 14

what do you think?