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Let $V$ be a finite dimensional vector-space over $mathbbK$ and $phi:Vto V$ a linear transformation with $phi circ phi = phi$

1. Show that $phi$ can have only the eigenvalues $0$ and $1$.


2. Describe all endomorphisms $phi : V to V$ with $phi circ phi = phi$ that have only the eigenvalue $0$ and prove your claim.


3. Describe all endomorphisms $phi : V to V$ with $phi circ phi = phi$ that have only the eigenvalue $1$ and prove your claim.


4. Specify a vector space $V$ and a linear transformation $phi : V to V$ with $phi circ phi = phi$ that has eigenvalues $0$ and $1$.

1.) Let $vin V$ and $vneq 0$ with $ phi(v)=lambda v$ and $ lambda in mathbbK$

Since $phi circ phi = phi$, we have $(phi circ phi)(v) = lambdalambda viff phi(v) = lambda^2v$

How do I show, that this will only hold for eigenvalues $1$ and $0$?

2.) With $kerphi$ we only map those vectors that are zero, that's what happens, then $lambda = 0$

(How do I prove this?)

3.) Not really sure, (maybe the image of the linear transformation?)

4.)

Could I just use the given general vector-space and linear transformation as an example?

I guess $V$ is a finite vector space (say $n$)

Hint

1) You started well. Let $lambda $ an eigenvalue and $v$ s.t. $varphi (v)=lambda v$. Then $$lambda v=varphi (v)=varphi circ varphi (v)=lambda ^2v.$$
Therefore $lambda (lambda -1)v=0$. I let you conclude.

2) $varphi $ has only $0$ as eigenvalue $iff$ $varphi $ is Nilpotent, i.e. there is $k$ s.t. $varphi ^k=0$. This follow from the fact that the characteristic polynomial of such an application is $p(x)=x^n$. So, there is only one choice for $varphi $.

3) $varphi $ has only $1$ as eigenvalue $iff$ $varphi =alpha cdot textid_X$ for a certain $alpha >0$. This follow from the fact that the characteristic polynomial is $p(x)=(x-1)^n$. So, there is only one choice for $varphi $.

4) Think to an application that has a minimal polynomial of the from $m(x)=x(x-1)$.

Hints:

1. If $phi(v) = lambda v$ then $$lambda v =phi(v) = (phicircphi)(v) = phi(lambda v) = lambda^2 v$$
   so $lambda^2 = lambda$ if $v ne 0$.




2. The polynomial $x^2-x = x(x-1)$ annihilates $phi$ so $phi$ is diagonalizable. The only eigenvalue is $0$ so $phi$ diagonalizes to the zero matrix.



3. The same reasoning as above implies $phi$ diagonalizes to the identity matrix.


4. Consider $phi : mathbbR^2 to mathbbR^2$, $phi(x,y) = (x,0)$.

1. 
   $lambda v = phi(v) = phi^2(v) = lambda^2v$. Since $v neq 0$ (because $v$ is an eigenvector) $Rightarrow lambda = lambda^2$. This equation can only be true for $lambda = 0, 1$.


2. -


3. 
   $forall v in V: : phi(phi(v)) = phi(v)$. This means that $Im(phi)$ is a subspace of the eigenspace of the eigenvalue $lambda = 1$. Since $Ker(phi) = 0 $ (because $phi$ does not have $0$ as an eigenvalue) and since $phi$ is an endomorphism, it follows that $Im(phi) = V$, which means that the eigenspace of the eigenvalue $lambda = 1$ is the whole space $V$ $Leftrightarrow phi = id_V$.


4. Let $V = mathbbR^2$, $phi = left( beginmatrix 0 & 0 \ 0 & 1 endmatrix right)$. One can easily prove that $phi$ satisfies $phi^2 = phi$ and that the eigenvalues of $phi$ are $1$ and $0$.

It turns out that we don't need $V$ to be finite-dimensional over $Bbb K$ to prove the first one. (The other answers more than cover the other three.)

For the first, take any eigenvector $vec vin V$ of $phi,$ say with eigenvalue $lambda.$ Then by definition, we have that $vec vnevec 0,$ and $phileft(vec vright)=lambdavec v.$ Since $phi$ is a linear transformation, then we have $$phileft(lambdavec vright)=lambdaphileft(vec vright)=lambda^2vec v,$$ but on the other hand, $$phileft(lambdavec vright)=phibigl(phileft(vec vright)bigr)=(phicircphi)left(vec vright)=phileft(vec vright)=lambdavec v.$$ Thus, $$lambda^2vec v=lambdavec v,$$ so $$left(lambda^2-lambdaright)vec v=vec 0,$$ and since $vec vnevec 0$ and $Bbb K$ is a field, we must have $lambda^2-lambda=0,$ or $lambda(lambda-1)=0.$ Since $Bbb K$ is a field, then it follows that $lambda=0$ or $lambda=1.$

If $lambda$ denote the eigenvalue of $phi$ then $lambda^2=lambdaimplieslambda(lambda-1)=0$ i.e. $lambda(lambda-1)$ is the anihilating polynomial for $phi$.

So you have three possibilities for minimal polynomial of $phi$ viz. $m_1(lambda)=lambda$ or $m_2(lambda)=(lambda-1)$ or $m_3(lambda)=lambda(lambda-1)$,

• $1$. Each of these possibilities thereby giving $0$ or $1$ or $0, 1$ respectively as the eigenvalues set.




• $2$. From the choice $m_1(lambda)=lambda$, you can see the only possibility is $phi=O$




• $3$. From the choice $m_2(lambda)=lambda-1$, you can see the only possibility is $phi-I=Oimplies phi=I$




• $4$. From the choice $m_3(lambda)=lambda(lambda-1)$, you can see there are infinite possibilities for $phi$ s.t.$phi^2=phi$. All of them have $beginpmatrix1&0\0&0endpmatrix$, $beginpmatrix0&0\0&1endpmatrix$, $beginpmatrix0&0\1&1endpmatrix$, etc in their diagonal as blocks and rest elements are $O$ blocks.