Linear transformations and eigen values

Question

Let $V$ be a finite dimensional vector-space over $\mathbb{K}$ and $\phi:V\to V$ 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 $v\in V$ and $v\neq 0$ with $ \phi(v)=\lambda v$ and $ \lambda \in \mathbb{K}$

Since $\phi \circ \phi = \phi$, we have $(\phi \circ \phi)(v) = \lambda\lambda v\iff \phi(v) = \lambda^2v$

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

2.) With $\ker\phi$ 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?

Answer 1

Hints:

1. If $\phi(v) = \lambda v$ then $$\lambda v =\phi(v) = (\phi\circ\phi)(v) = \phi(\lambda v) = \lambda^2 v$$ 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 $\phi : \mathbb{R}^2 \to \mathbb{R}^2$, $\phi(x,y) = (x,0)$.

Answer 2

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 $\iff$ $\varphi =\alpha \cdot \text{id}_X$ 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 $m(x)=x(x-1)$.

Answer 3

1. $\lambda v = \phi(v) = \phi^{2}(v) = \lambda^{2}v$. 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 $V = \mathbb{R}^{2}$, $\phi = \left( \begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix} \right)$. One can easily prove that $\phi$ satisfies $\phi^{2} = \phi$ and that the eigenvalues of $\phi$ are $1$ and $0$.

Answer 4

If $\lambda$ denote the eigenvalue of $\phi$ then $\lambda^2=\lambda\implies\lambda(\lambda-1)=0$ 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 $\phi-I=O\implies \phi=I$

• $4$. 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 $\begin{pmatrix}1&0\\0&0\end{pmatrix}$, $\begin{pmatrix}0&0\\0&1\end{pmatrix}$, $\begin{pmatrix}0&0\\1&1\end{pmatrix}$, etc in their diagonal as blocks and rest elements are $O$ blocks.

Answer 5

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 $\vec v\in V$ of $\phi,$ say with eigenvalue $\lambda.$ Then by definition, we have that $\vec v\ne\vec 0,$ and $\phi\left(\vec v\right)=\lambda\vec v.$ Since $\phi$ is a linear transformation, then we have $$\phi\left(\lambda\vec v\right)=\lambda\phi\left(\vec v\right)=\lambda^2\vec v,$$ but on the other hand, $$\phi\left(\lambda\vec v\right)=\phi\bigl(\phi\left(\vec v\right)\bigr)=(\phi\circ\phi)\left(\vec v\right)=\phi\left(\vec v\right)=\lambda\vec v.$$ Thus, $$\lambda^2\vec v=\lambda\vec v,$$ so $$\left(\lambda^2-\lambda\right)\vec v=\vec 0,$$ and since $\vec v\ne\vec 0$ 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 $\lambda=1.$