📂Linear Algebra

Diagonalizability of Linear Transformations, Multiplicity of Eigenvalues, and the Relationship with Eigenspaces

Theorem¹

Let $T : V \to V$ be a linear transformation on a finite-dimensional vector space $V$. Suppose the characteristic polynomial of $T$ splits and $\lambda_{1}, \lambda_{2}, \dots, \lambda_{k}$ are distinct eigenvalues of $T$. Then,

1. $T$ is diagonalizable if and only if, for all $i$, the multiplicity of $\lambda_{i}$ and the dimension $\dim(E_{\lambda_{i}})$ of the eigenspace are the same.
   $$T \text{ is diagobalizable. } \iff \text{multiplicity of } \lambda_{i} = \dim(E_{\lambda_{i}}),\quad \forall i$$

2. If $T$ is diagonalizable, and if $\beta_{i}$ is an ordered basis of $E_{\lambda_{i}}$, then $\beta = \beta_{1} \cup \cdots \cup \beta_{k}$ is an ordered basis of $V$ containing the eigenvectors of $T$.

3. $T$ is diagonalizable if and only if $V$ is the direct sum of the eigenspaces of $T$. $$T \text{ is diagobalizable. } \iff V = E_{\lambda_{1}} \oplus \cdots \oplus E_{\lambda_{k}}$$

Proof

1.

Let $m_{i}$ be the multiplicity of $\lambda_{i}$. Suppose $d_{i} = \dim(E_{\lambda_{i}})$, $n = \dim(V)$.

• $(\Longrightarrow)$

  Assume $T$ is diagonalizable. This is equivalent to saying that there exists a basis of $V$ consisting of the eigenvectors of $T$. Let $\beta$ be such a basis of $V$, consisting of the eigenvectors of $T$. For each $i$, let $\beta_{i} = \beta \cap E_{\lambda_{i}}$. In other words, $\beta_{i}$ is a set of eigenvectors corresponding to the eigenvalue $\lambda_{i}$ that belong to $\beta$. Also, suppose $n_{i} = \left| \beta_{i} \right|$. Then, since $\beta_{i}$ is a linearly independent subset of $E_{\lambda_{i}}$, $n_{i} \le d_{i}$ holds. Moreover, since the algebraic multiplicity of an eigenvalue is greater or equal to its geometric multiplicity, $d_{i} \le m_{i}$ holds. Therefore, $n = \left| \beta \right|$ implies, by adding $n_{i}$ for all $i$, we get $n$, and by definition of multiplicity, adding $m_{i}$ for all $i$ also results in $n$. Therefore, $$n = \sum\limits_{i=1}^{k}n_{i} \le \sum\limits_{i=1}^{k}d_{i} \le \sum\limits_{i=1}^{k}m_{i} = n$$ $$\implies \sum\limits_{i=1}^{k}(m_{i} - d_{i}) = 0$$ But since $m_{i} - d_{i} \ge 0$, for all $i$, $m_{i} = d_{i}$ holds.

• $(\Longleftarrow)$

  Suppose that for all $i$, $m_{i} = d_{i}$ holds. Let $\beta_{i}$ be an ordered basis of $E_{\lambda_{i}}$. And let $\beta = \beta_{1} \cup \cdots \cup \beta_{k}$. Then, because the union of linearly independent sets from different eigenspaces is also linearly independent, $\beta$ is linearly independent. Also, by assumption, $\sum\limits_{i=1}^{k} d_{i} = \sum\limits_{i=1}^{k} m_{i} = n$ holds, so $\beta$ contains $n$ linearly independent eigenvectors. Therefore, $\beta$ is an ordered basis of $V$, and $T$ is diagonalizable.

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2.

The proof of $(\Longleftarrow)$ in 1. already covered this.

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3.

Let $\lambda_{1}, \dots, \lambda_{k}$ be distinct eigenvalues of $T$.

• $(\Longrightarrow)$

  Assume $T$ is diagonalizable. Let $\gamma_{i}$ be an ordered basis of the eigenspace $E_{\lambda_{i}}$. Then, by 2., $\gamma_{1} \cup \cdots \cup \gamma_{k}$ is an ordered basis of $V$.

  Properties of direct sums

  Let $W_{1}, W_{2}, \dots, W_{k}$ be subspaces of a finite-dimensional vector space $V$. The following propositions are equivalent:

  1. $V = W_{1} \oplus W_{2} \oplus \cdots \oplus W_{k}$
  2. If $\gamma_{i}$ is an ordered basis of $W_{i}$, then $\gamma_{1} \cup \cdots \cup \gamma_{k}$ is an ordered basis of $V$.
  3. There exists an ordered basis $\gamma_{i}$ of $W_{i}$ such that $\gamma_{1} \cup \cdots \cup \gamma_{k}$ is an ordered basis of $V$. Thus, by the theorem above, $V = E_{\lambda_{1}} \oplus \cdots \oplus E_{\lambda_{k}}$ holds.

• $(\Longleftarrow)$

  Assume $V = E_{\lambda_{1}} \oplus \cdots \oplus E_{\lambda_{k}}$. Let $\gamma_{i}$ be an ordered basis of $E_{\lambda_{i}}$. Then, by the properties of direct sums, $\gamma_{1} \cup \cdots \cup \gamma_{k}$ is an ordered basis of $V$. Since this basis consists of eigenvectors, $T$ is diagonalizable.

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