# Linear transformations and eigen values

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?

• For 4), you need to provide an explicit example. (For example, a linear transformation given by a particular matrix.) – Minus One-Twelfth Jul 4 at 12:22
• Is $V$ finite dimensional ? – Surb Jul 4 at 12:25
• It is not given if $V$ is finite dimensional or not, but I guess so. (The creator of the task might forgot to add that.) Let's just suppose, it's finite dimensional. (I've added it to my question.) – Doesbaddel Jul 4 at 12:26
• Such transformations are known as projections. The map $\phi$ projects along its eigenspace $E_0$ onto the eigenspace $E_1$. If you want an orthogonal projection, insist on $\phi^* = \phi$ as well; this makes the eigenspaces orthogonal! – Theo Bendit Jul 4 at 12:37

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 $$\phi : \mathbb{R}^2 \to \mathbb{R}^2$$, $$\phi(x,y) = (x,0)$$.

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 $$m(x)=x(x-1)$$.

• Oh ok, did'nt knew the characteristic polynomial has degree $n$ if the vector-space has dimension $n$. – Doesbaddel Jul 4 at 12:48
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 $$0$$.

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 $$\lambda=1.$$

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.