# 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.

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.

Hint: If $$Ax = \lambda x$$ for a non-zero $$x \in \Bbb F^n$$, then $$A(xy^T) = \lambda(xy^T)$$ for any vector $$y \in \Bbb F^n$$.