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?
Best Answer
Suppose $\lambda_1\neq \lambda_2$.
Then the $\{v_1,v_2\}$ is linearly independent.
Assume $v_1+v_2$ is an eigen vector corresponding to an eigen value $\lambda$.
Then $A(v_1+v_2) =\lambda(v_1+v_2)$
Implies $ Av_1+Av_2=\lambda v_1+\lambda v_2$
i.e $\lambda_1v_1+\lambda_2v_2=\lambda v_1+\lambda v_2$
Implies $(\lambda_1-\lambda) v_1+(\lambda_2-\lambda) v_2=0$
Now by linear independence of $\{v_1, v_2\}$ , we have $\lambda_1=\lambda=\lambda_2$
which contradict our assumption that $\lambda_1\neq \lambda_2$