Geometric Multiplicity of Eigenvalues Let $A\in\mathbb{F}^{n×n}$ and define $L: \mathbb{F}^{n×n}\rightarrow \mathbb{F}^{n×n}$ by $L(X)=AX$. If $\lambda$ is an eigenvalue of $A$ with geometric multiplicity 1, show that the geometric multiplicity of $\lambda$ as an eigenvalue of $L$ is at least $n$.
I have already proven that $\sigma (A)=\sigma (L)$. This connects with the problem above.
Any hint will do and is a great help. Thanks.
 A: Hint: If $Ax = \lambda x$ for a non-zero $x \in \Bbb F^n$, then $A(xy^T) = \lambda(xy^T)$ for any vector $y \in \Bbb F^n$.
A: I don't think too much spectral theory is needed to solve this problem, to wit:
If $\lambda \in \Bbb F$ is an eigenvalue of $A$ of geometric multiplicity $1$, then the eigenspace corresponding to $\lambda$ is a $1$-dimensional subspace of $\Bbb F^n$; thus there is a vector
$0 \ne \vec x \in \Bbb F^n \tag 1$
with
$A\vec x = \lambda \vec x; \tag 2$
furthermore, any non-zero vector
$0 \ne \vec y \in \Bbb F^n \tag 3$
satisfying
$A \vec y = \lambda \vec y \tag 4$
must be collinear with $\vec x$:
$\exists 0 \ne \alpha \in \Bbb F, \; \vec y = \alpha \vec x. \tag 5$
Next, consider the $n$ matrices
$X_i \in \Bbb F^{n \times n}, \; 1 \le i \le n, \tag 6$
where the $i$-th column of $X_i$ is $\vec x$, and all other columns are $0$; that is, the $X_i$ take the form
$X_i = [0 \; 0 \; \ldots \; \vec x \; \ldots \; 0 \; 0]. \tag 7$
Now any
$Y \in \Bbb F^{n \times n} \tag 8$
may be written
$Y = [\vec y_1 \; \vec y_2 \; \ldots \; \vec y_n], \tag 9$
with each
$\vec y_i \in \Bbb F^n; \tag{10}$
the action of $L$ on $Y$ is thus given by
$LY = [A\vec y_1 \; A\vec y_2 \ldots \; A\vec y_n]; \tag{11}$
it follows that
$LX_i = [0 \; 0 \; \ldots \; A\vec x \; \ldots \; 0 \; 0] = [0 \; 0 \; \ldots \; \lambda \vec x \; \ldots \; 0 \; 0] = \lambda [0 \; 0 \; \ldots \; \vec x \; \ldots \; 0 \; 0] = \lambda X_i; \tag {12}$
that is, each $X_i$ is a $\lambda$-eigenvector of $L$ in $\Bbb F^{n \times n}$; furthermore, the $X_i$ are linearly independent over $\Bbb F$, for given any
$a_i \in \Bbb F, \; 1 \le i \le n, \tag{13}$
we have
$\displaystyle \sum_1^n a_iX_i = [a_1 \vec x \; a_2\vec x \; \ldots \; a_n \vec x] \ne 0 \tag{14}$
provided at least one $a_i \ne 0$.
We have thus demonstrated the existence of $n$ linearly independent $\lambda$-eigenvectors of $L$ in $F^{n \times n}$, that is, that the geometric multiplicity of $\lambda$ as an eigenvalue of $L:\Bbb F^n \to \Bbb F^n$ is at least $n$. $OE\Delta$.
Nota Bene: Based upon what we have done above, we can, with only a little extra work, show that in fact the geometric multiplicity of $\lambda$ as an eigenvalue of $L$ is in fact precisely $n$; for with $Y$ as in (8)-(11), 
$LY = \lambda Y \Longrightarrow A \vec y_i = \lambda \vec y_i, \; 1 \le i \le n; \tag{15}$
then as noted above in (3)-(5) we have
$\vec y_i = \alpha_i \vec x, \; 1 \le i \le n, \tag{16}$
and so
$Y = \displaystyle \sum_1^n \alpha_i X_i; \tag{17}$
that is, every eigenvector $Y$ of $L$ lies in $\text{span}\{ X_i, \; 1 \le i \le n \}$; this proves the dimension of the $\lambda$-eigenspace of $L$ is precisely $n$, and hence the geometric multiplicity of $\lambda$ is exactly $n$ as well.
End of Note.
