Let $n\in \mathbb{Z}\_{>0}$. Let $A\in \mathcal{M}\_{n,n}(\mathbb{F})$ and $v\_1,v\_2$ be eigenvectors of $A$ for the eigenvalues $\lambda \_1$ and $\lambda \_2$. Assume that $v\_1+v\_2$ is also an eigenvector. Show that $\lambda \_1=\lambda \_2$.

Assume there are $n$ eigenvalues $\lambda$.

$\lambda \_1$ has eigenvector $v\_1$.

$\lambda \_2$ has eigenvector $v\_2$.

...

$\lambda \_n$ has eigenvector $v\_n$.

Let $\lambda \_3$ be the eigenvalue to $v\_3=v\_1+v\_2$.

The eigenvectors build a base and are therefore linearly independent.

But:

$$\alpha\_1v\_1+\alpha\_2v\_2+\alpha\_3v\_3+...+\alpha\_nv\_n=0$$
$$\alpha\_1v\_1+\alpha\_2v\_2+\alpha\_3\cdot(v\_1+v\_2)+...+\alpha\_nv\_n=0$$

With $\alpha\_1=\alpha\_2=1$ and $\alpha\_3=-1$ and $\alpha\_i=0$,$i\in\mathbb{N\_{>2}}$

Therefore the eigenvectors aren't linearly independent.

If the eigenvector $v\_1+v\_2$ belongs to eigenvalue $\lambda\_1$ or $\lambda\_2$
The eigenspace would look (somewhat) like this: $E\_A(\lambda\_1)=(r\cdot v\_1+n\cdot(v\_1+v\_2),r,n\in \mathbb{R})$
or $E\_A(\lambda\_2)=(r\cdot v\_2+n\cdot(v\_1+v\_2),r,n\in \mathbb{R})$

With algeabric multiplicity of 1 but geometric mulitplicity of 2 this would be wrong. Therefore the algebraic multiplicity of the eigenspace corresponding to $\lambda\_1$ has to be 2. If $\lambda\_1\neq\lambda\_2$ the eigenvectors can't form a base as there would be one eigenvector too much, therefore $\lambda\_1=\lambda\_2$

Is this proof correct or did I go wrong somewhere?