📂Matrix Algebra

The algebraic multiplicity of an eigenvalue is equal to the dimension of the generalized eigenspace.

Theorem¹

Let the distinct eigenvalues $\lambda_{1}, \lambda_{2}, \dots, \lambda_{k}$ of the matrix $A \in M_{n \times n}(\mathbb{C})$ have algebraic multiplicities $m_{1}, m_{2}, \dots, m_{k}$, respectively. And let $\beta_{i}$ be an ordered basis $W_{\lambda_{i}}$ of the generalized eigenspaces $W_{\lambda_{i}}$. Then the following hold.

(a) For $i \ne j$, $\beta_{i} \cap \beta_{j} = \varnothing$

(b) For every $\mathbf{v} \in \mathbb{C}^{n}$, there exists $\mathbf{v}_{i} \in W_{\lambda_{i}}$ ($1 \le i \le k$) satisfying the following.

$$ \mathbf{v} = \mathbf{v}_{1} + \mathbf{v}_{2} + \cdots + \mathbf{v}_{k} $$

In other words, $\mathbb{C}^{n}$ is expressed as the direct sum of the generalized eigenspaces.

$$ \mathbb{C}^{n} = W_{\lambda_{1}} \oplus W_{\lambda_{2}} \oplus \cdots \oplus W_{\lambda_{k}} $$

(c) $\beta = \beta_{1} \cup \beta_{2} \cup \cdots \cup \beta_{k}$ is an ordered basis of $\mathbb{C}^{n}$.

(d) $\dim (W_{\lambda_{i}}) = m_{i}$

Proof

(a)

For $i \ne j$, suppose $\mathbf{v} \in \beta_{i} \cap \beta_{j} \subset W_{\lambda_{i}} \cap W_{\lambda_{j}}$. That is, $\mathbf{v} \ne \mathbf{0}$. Since $W_{\lambda_{j}}$ is $(A - \lambda_{i}I)$–invariant and the map $(A - \lambda_{i} I)|_{W_{\lambda_{j}}}$ has kernel $\left\{ \mathbf{0} \right\}$ (so by generalized eigenspace properties), for any $p$ we have $(A - \lambda_{i} I)^{p} \mathbf{v} \ne \mathbf{0}$. This contradicts the fact $\mathbf{v} \in W_{\lambda_{i}}$, so the assumption is false. Hence

$$ \beta_{i} \cap \beta_{j} = \varnothing $$

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(b)

We prove by mathematical induction.

Base case: $k = 1$

Let $k=1$. Then the characteristic polynomial (특성다항식) of $A$ is $(x - \lambda_{1})^{m_{1}}$. By the Cayley–Hamilton theorem we have $(A - \lambda_{1} I)^{m_{1}} = O$. Hence $V = W_{\lambda_{1}}$.

Inductive step: if true for $k-1$, then true for $k$

Assume the statement holds for $k-1$. Suppose $A$ has $k$ distinct eigenvalues $\lambda_{1}$, $\lambda_{2}$, $\dots$, $\lambda_{k}$, and that the algebraic multiplicity of $\lambda_{k}$ is $m$. Let $K = R(A - \lambda_{k}I)^{m}$. Here $R$ denotes the image (치역). Since $A$ commutes with $(A - \lambda_{k}I)$, the subspace $K$ is $A$–invariant. Now consider the induced map of $(A - \lambda_{k}I)^{m}$ on $W_{\lambda_{i}}$ $(i \lt k)$. This induced map is one-to-one (일대일) and, being a linear map between finite-dimensional spaces, is also onto (전사). Therefore, for $i \lt k$ we have $W_{\lambda_{i}} \subset K$, and $\lambda_{i}$ is an eigenvalue of $A|_{K}$. If $\mathbf{v}$ is an eigenvector of $A$ corresponding to $\lambda_{i}$, then

$$ A|_{K} \mathbf{v} = A \mathbf{v} = \lambda_{i} \mathbf{v} $$

However, $\lambda_{k}$ is not an eigenvalue of $A|_{K}$. For $\mathbf{v} \in K$ let $A \mathbf{v} = \lambda_{k} \mathbf{v}$. Then $\mathbf{v} = (A - \lambda_{k}I)^{m} \mathbf{u}$ and the following holds.

$$ (A - \lambda_{k}I) \mathbf{v} = (A - \lambda_{k}I) (A - \lambda_{k}I)^{m} \mathbf{u} = (A - \lambda_{k}I)^{m+1} \mathbf{u} = \mathbf{0} $$

Hence $\mathbf{u} \in K$ and $\mathbf{v}=$ $(A - \lambda_{k}I)^{m} \mathbf{u} = \mathbf{0}$ holds. Therefore $\lambda_{k}$ is not an eigenvalue of $A|_{K}$. Because every eigenvalue of $A|_{K}$ is also an eigenvalue of $A$, the map $A|_{K}$ has $k-1$ distinct eigenvalues $\lambda_{1}, \lambda_{2}, \dots, \lambda_{k-1}$.

Now let $\mathbf{x} \in V$. Then $(A - \lambda_{k}I)^{m} \mathbf{x} \in K$, and by the inductive hypothesis, for any element $\mathbf{w}_{i}$ of the generalized eigenspaces $W_{\lambda_{i}}^{\prime}$ corresponding to the eigenvalues $\lambda_{i}$ of $A|_{K}$, the following holds.

$$ (A - \lambda_{k}I)^{m} \mathbf{x} = \mathbf{w}_{1} + \mathbf{w}_{2} + \cdots + \mathbf{w}_{k-1} $$

The map $(A - \lambda_{k}I)^{m}$ is a surjection (전사함수) with target $W_{\lambda_{i}}^{\prime}$, so there exists some $\mathbf{v}_{i} \in W_{\lambda_{i}}$ such that $\mathbf{w}_{i} = (A - \lambda_{k}I)^{m} \mathbf{v}_{i}$ $(i \lt k)$ holds. Hence we get

$$ (A - \lambda_{k}I)^{m} \mathbf{x} = (A - \lambda_{k}I)^{m} \mathbf{v}_{1} + (A - \lambda_{k}I)^{m} \mathbf{v}_{2} + \cdots + (A - \lambda_{k}I)^{m} \mathbf{v}_{k-1} $$

$$ (A - \lambda_{k}I)^{m} \left( \mathbf{x} - \mathbf{v}_{1} + \mathbf{v}_{2} + \cdots + \mathbf{v}_{k-1} \right) = \mathbf{0} $$

Thus $\mathbf{x} - \mathbf{v}_{1} + \mathbf{v}_{2} + \cdots + \mathbf{v}_{k-1} \in W_{\lambda_{k}}$, and if $\mathbf{v}_{k} = \mathbf{x} - \mathbf{v}_{1} + \mathbf{v}_{2} + \cdots + \mathbf{v}_{k-1}$, then the following holds.

$$ \mathbf{x} = \mathbf{v}_{1} + \mathbf{v}_{2} + \cdots + \mathbf{v}_{k} $$

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(c), (d)

Let $\mathbf{x} \in \mathbb{C}^{n}$. By (b) we have $\mathbf{x} = \mathbf{v}_{1} + \cdots \mathbf{v}_{k}$ $(\mathbf{v}_{i} \in W_{\lambda_{i}})$, and since $\beta_{i}$ is an ordered basis of $W_{\lambda_{i}}$, each $\mathbf{v}_{i}$ can be expressed as a linear combination of the elements of $\beta_{i}$. Therefore $\beta$ spans (generates) $\mathbb{C}^{n}$. If $q = | \beta |$, then $n \le q$. Since $\dim (W_{\lambda_{i}}) \le m_{i}$ (by generalized eigenspace properties), the following inequality holds.

$$ q = \sum_{i}^{k} \dim (W_{\lambda _{i}}) \le \sum_{i}^{k} m_{i} = n $$

But since we already had $n \le q$, it follows that $n = q$. Because $|\beta| = n$ and $\beta$ spans $\mathbb{C}^{n}$, $\beta$ is an ordered basis of $\mathbb{C}^{n}$.

From the above results we also have $\sum\limits_{i}^{k} \dim (W_{\lambda_{i}}) = \sum\limits_{i}^{k} m_{i}$. Hence, because $\dim (W_{\lambda_{i}}) \le m_{i}$, we obtain $\dim (W_{\lambda_{i}}) = m_{i}$.

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