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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.

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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.

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

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