Upper-Triangularizability

For a finite-dimensional vector space $\mathbf{V}$, an operator $T \in \mathscr{L}(\mathbf{V})$ is upper-triangularizable if there exists a basis $\beta = v_1 , \dots, v_n$ of $\mathbf{V}$ for which $[T]_\beta$ is upper-triangular.

$$ [T]_\beta = \begin{pmatrix} \lambda _1 & \dots & \star \\ \vdots & \ddots & \vdots \\ 0 & \dots & \lambda _m \end{pmatrix} $$

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Proposition. If $[T]\beta$ is upper-triangular for a basis $\beta = v_1, \dots, v_n$, the linear map $T - \lambda_k I$ maps vectors in $\text{span}(v_1, \dots, v_k)$ to vectors in $\text{span}(v_1, \dots, v{k-1})$.

$$ T-\lambda_k I : \text{span} (v_1, \dots, v_k) \rightarrow \text{span}(v_1, \dots, v_{k-1}) $$

• Proof </aside>

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Lemma. If $[T]_\beta$ is upper-triangular with diagonal entries $\lambda_1 , \dots, \lambda_n$ then,

$$ q(z)=\prod _{i=1} ^n (z-\lambda_i) \text{ annihilates $T$} $$

• Proof </aside>

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Lemma. If $[T]_\beta$ is upper-triangular with diagonal entries $\lambda _1 , \dots , \lambda _n$ then,

$$ \sigma(T) = \{\lambda_1, \dots, \lambda_n\} \text{ or, equivalently, } \lambda_k \in \sigma (T) $$

• Proof </aside>

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Theorem. The following statements are equivalent and characterize the upper-triangularizability of $T$ given a basis $\beta = v_1, \dots, v_n$ of $\mathbf{V}$.

1. $[T]_\beta$ is upper-triangular.
2. $\text{span}(v_1,\dots, v_k)$ is $T$-invariant.
3. $Tv_k \in \text{span}(v_1, \dots, v_k)$. </aside>

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Theorem. An operator $T \in \mathscr{L}(\mathbf{V})$ is upper-triangularizable iff $\mu _T$ splits into linear factors containing, not necessarily distinct, roots $\lambda_1, \dots, \lambda _m$.

$$ \text{$T$ is upper-triangularizable} \iff \mu _T(z) = \prod _{\ell=1} ^m (z - \lambda _\ell) \hspace{15pt} m \leq n $$

• Proof </aside>

Diagonalizability

For a finite-dimensional vector space $\mathbf{V}$, an operator $T \in \mathscr{L}(\mathbf{V})$ is diagonalizable if there exists a basis $\beta = v_1 , \dots, v_n$ of $\mathbf{V}$ for which $[T]_\beta$ is diagonal.

$$ [T]_\beta = \begin{pmatrix} \lambda _1 & \dots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \dots & \lambda_m \end{pmatrix} $$

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Theorem. The following statements are equivalent and characterize the diagonalizability of $T$ given its spectrum $\sigma(T) = \{\lambda_1, \dots, \lambda_m\}$.

1. $T$ is diagonalizable.
2. There exists a basis of $\mathbf{V}$ consisting only of eigenvectors of $T$.
3. $\mathbf{V} = \bigoplus _{i=1}^n E(\lambda_i , T)$
4. $\dim \mathbf{V} = \sum _{i=1}^n \dim E(\lambda _i, T)$.

• Proof </aside>

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Theorem. An operator $T \in \mathscr{L}(\mathbf{V})$ is diagonalizable iff $\mu_T$ splits into linear factors containing distinct roots $\lambda_1, \dots, \lambda_m$.

$$ \text{$T$ is diagonalizable} \iff \mu _T(z) = \prod {\ell=1}^m(z-\lambda\ell) \hspace{15pt} m\leq n $$

• Proof </aside>