% vim: tw=50 % 09/11/2022 11AM \begin{lemma*} $V$ $n$ dimensional over $F = \RR, \CC$, $\alpha \in \mathcal{L}(V)$. Then $\chi_\alpha(t) = (-1)^n t^n + c_{n - 1} t^{n - 1} + \cdots + c_0$, $c_0 = \det A = \det \alpha$, $c_{n - 1} = (-1)^{n - 1} \Trace A$. \end{lemma*} \begin{proof} \begin{itemize} \item $\chi_\alpha(t) = \det(\alpha - t \id)$ \[ \implies \chi_\alpha(0) = \det \alpha = c_0 \] \item Say that $F = \RR$ or $\CC$ (if $F = \RR$, we can think of is as having complex entries as well). We know that $\alpha$ is triangulable over $\CC$, so: \begin{align*} \chi_\alpha(t) &= \det \begin{pmatrix} a_1 - t & \cdots & * \\ \vdots & \ddots & \vdots \\ 0 & \cdots & a_n - t \end{pmatrix} \\ &= \prod_{i = 1}^n )a_i - 1) \\ &= (-1)^n t^n + c_{n - 1} t^{n - 1} + \cdots + c_0 \end{align*} \[ c_{n - 1} = (-1)^{n - 1} \sum{i = 1}^n a_i = \Trace \alpha \] \end{itemize} \end{proof} \subsection{Diagonalisation criterion and minimal polynomial} \begin{notation*}[polynomial of an endomorphism] Pick $p(t)$ polynomial over $F$ \[ p(t) = a_n t^n + \cdots + a_1 t + a_0, \quad a_i \in F \] $A \in \mathcal{M}_n(F)$, for all $n$, $A^n \in \mathcal{M}_n(F)$. We define: \[ p(A) = a_n A^n + \cdots + a_1 A + a_i \id \in \mathcal{M}_n(F) \] If $\alpha \in \mathcal{L}(V)$, we define \[ p(\alpha) = a_n \alpha^n + \cdots + a_1 \alpha + a_0 \id \] where $\alpha = \alpha \circ \cdots \circ \alpha \in \mathcal{L}(V)$. \end{notation*} \noindent $\to$ very useful. \begin{flashcard}[sharp-criterion-of-diagonalisability] \begin{theorem*}[Sharp criterion of diagonalisability] \begin{itemize} \item $V$ vector space over $F$, $\dim_F V < \infty$ \item $\alpha \in \mathcal{L}(V)$ \end{itemize} \cloze{Then $\alpha$ is diagonalisable if and only if there exists a polynomial $p$ which is the product of \emph{distinct linear factors} such that $p(\alpha) = 0$.} \end{theorem*} \end{flashcard} \begin{align*} \alpha \text{ diagonalisable} &\iff \exists (\lambda_1, \dots, \lambda_n) \text{ distinct }, \lambda_i \in F \text{ such that:} \\ &~~~~~ p(t) = \prod_{i = 1}^k (t - \lambda_i) \\ &~~~~~ \text{ and } p(\alpha) = 0 \end{align*} \begin{proof} \begin{enumerate} \item[$\Rightarrow$] Suppose $\alpha$ is diagonalisable, with $\lambda_1, \dots, \lambda_k$ the distinct eigenvalues. Let $p(t) = \prod_{i = 1}^k (t - \lambda_i)$. Let $\mathcal{B}$ be the basis of $V$ made of eigenvectors of $\alpha$ (it is precisely the basis in which $[\alpha]_{\mathcal{B}}$ is diagonal). Then $v \in \mathcal{B}$, then $\alpha(v) = \lambda_i(v)$ for some $i \in \{1, \dots, k\}$, implies $(a - \lambda_i \id)(v) = 0$, implies \[ p(\alpha) = \left[ \prod_{j = 1}^k (\alpha - \lambda_j \id) \right] (v) = 0 \] but the terms in the product commute, i.e. \[ (\alpha - \lambda_j \id)(\alpha - \lambda_k \id) = (\alpha - \lambda_k \id)(\alpha - \lambda_j \id) \] so for all $v \in \mathcal{B}$, $p(\alpha)(v) = 0$, so $p(\alpha)(v) = 0$ for all $v \in V$ (since $\mathcal{B}$ is a basis for $V$). So $p(\alpha) = 0$. \item[$\Leftarrow$] (Kernel lemma, Bezout's theorem for prime polynomials) \begin{itemize} \item Suppose $p(\alpha) = 0$ for: \[ p(t) = \prod_{i = 1}^k (t - \lambda_i) \] $\lambda_i \neq \lambda_j$, $i \neq j$. \item Let $V_{\lambda_i} = \Ker (\alpha - \lambda_i \id)$, we claim: \[ V = \bigoplus_{i = 1}^k V_{\lambda_i} \] Indeed let: \[ q_j(t) = \prod_{\substack{i = 1\\i \neq j}}^k \left( \frac{t - \lambda_i}{\lambda_j - \lambda_i} \right), \quad 1 \le j \le k \] Then \[ q_j(\lambda_i) = \begin{cases} 1 & \text{if $i = j$} \\ 0 & \text{if $i \neq j$} \end{cases} \] Hence let us consider: \[ q(t) = \sum_{j = 1}^k q_j(t) \] Then $\deg q_j \le k - 1$, so $\deg q \le k - 1$. Also $q(\lambda_j) = 1$ for all $1 \le j \le k$. So the polynomial $q(t) - 1$ has degree $\le k - 1$ and at least $k$ roots, so for all $t$, $q(t) = 1$. So for all $t$, \[ q_1(t) + \cdots + q_k(t) = 1 \] \item Let us define the \emph{projector} \[ \i_j q_j(\alpha) \in \mathcal{L}(V) \] Then \begin{align*} \sum_{j = 1}^k \pi_j &= \sum_{j = 1}^k q_j(\alpha) \\ &= \left( \sum_{j = 1}^k q_j \right) (\alpha) \\ &= \id \end{align*} This means for all $v \in V$, \[ v = q(\alpha)(v) = \sum_{j = 1}^k \pi_j(v) = \sum_{j = 1}^k q_j(\alpha)(v) \] Observe: pick $j \in \{1 \dots, k\}$, \[ (\alpha - \lambda_j \id) q_j(\alpha)(v) = \frac{1}{\prod_{i \neq j} (\lambda_i - \lambda_j)}p(\alpha)(v) = 0 \] so \[ \forall j \in \{1, \dots, k\}, \quad (\alpha - \lambda_j \id) \pi_j(v) = 0 \] \[ \implies \forall j \in \{1, \dots, k\} \pi_j(v) \in V_{\lambda_j} \] ($\pi_j$ is a projector on $V_{\lambda_j}$) Now for all $v \in V$, \[ v = \sum_{j = 1}^k \pi_j(v) \] hence \[ V = \sum_{j = 1}^k V_{\lambda_j} \] We need to prove that the sum is \emph{direct}. Indeed, let $v \in V_{\lambda_j} \cap \left( \sum_{i \neq j} V_{\lambda_i} \right)$. \begin{itemize} \item $v \in V_{\lambda_j}$. Then \begin{align*} \pi_j(v) &= q_j(\alpha)(v) \\ &= \prod_{i = 1, i\neq j}^k \frac{(\alpha - \lambda_i \id)}{\lambda_i - \lambda_j} (v) \\ &= \left[ \prod_{i = 1,i\neq j}^k \frac{(\lambda_i - \lambda_j)}{\lambda_i - \lambda_j} \right] (v) \\ &= v \end{align*} so $\pi_j \mid_{V_{\lambda_j}} = \id$. \item By assumption $v \in \sum_{i \neq j} V_{\lambda_i}$. Now, $i_0 \neq j$, $v \in V_{\lambda_{i_0}}$, $\alpha(v) = \lambda_{i_0} v$, \begin{align*} \implies \pi_j(v) &= q_j(\alpha)(v) \\ &= \left[ \prod_{i \neq j} \frac{(\alpha - \lambda_i \id)}{\lambda_i - \lambda_j}\right] (v) \\ &= \left(\prod_{i \neq j} \frac{\lambda_{i_0} - \lambda_j}{\lambda_i - \lambda_j} \right) \\ &= 0 \end{align*} so $\pi_j \mid_{V_{\lambda_i}} = 0$ for $i \neq j$. \end{itemize} \end{itemize} As a conclusion: $v \in V_{\lambda_j} \cap \left( \sum_{i \neq j} V_{\lambda_i} \right)$ \begin{enumerate}[(1)] \item $v \in V_{\lambda_i}$, implies $\pi_j(v) = v$ \item $v \in \sum_{i \neq j} V_{\lambda_i}$ implies $\pi_j(v) = 0$ \end{enumerate} so $v = 0$. We have proved: \[ V = \bigoplus_{j = 1}^k V_{\lambda_j} \] \[ \pi_j \mid_{V_{\lambda_j}} = \id \] \[ \pi_i \mid_{V_{\lambda_j}} = 0 \] for $i \neq j$. \begin{remark*} We have proved the following: if $\lambda_1, \dots, \lambda_k$ are $k$ distinct eigenvalues of $\alpha$, then \[ \sum_{i = 1}^k V_{\lambda_i} = \bigoplus_{i = 1}^k V_{\lambda_i} \] (always true) (and we know the projectors) \end{remark*} This means that the only way diagonalisation fails is if: \[ \bigoplus_{i = 1}^k V_{\lambda_i} = \sum_{j = 1}^k V_{\lambda_j} \neq V \] \end{enumerate} \end{proof} \begin{hiddenflashcard}[proof-of-sharp-criterion-of-diagonalisability] Proof of the sharp criterion of diagonalisability? \\ \cloze{ If $\alpha$ is diagonalisable, then consider $[\alpha]_{\mathcal{B}}$ where $\mathcal{B}$ is a basis in which $\alpha$ is diagonal. Then easily find non-zero polynomial $p$ which is a product of distinct linear factors for which $p(\alpha) = 0$. \myskip For the other direction, $p(\alpha) = 0$ with $p(t) = \prod_{i = 1}^k (t - \lambda_i)$ with $\lambda_i$ distinct. Let $V_{\lambda_i} = \ker(\alpha - \lambda_i \id)$. Define \[ q_j(t) = \prod_{\substack{1 \le i \le k\\i \neq j}} \left( \frac{t - \lambda_i}{\lambda_j- \lambda_i} \right), \qquad \pi_j = q_j(\alpha) \] Note $q_j$ has degree $k - 1$ and $q_j(\lambda_i) = \delta_{ij}$, so $(\sum q_j) - 1$ has $k$ roots but is degree $\le k - 1$ hence identically $0$. So $\sum \pi_j = \id$. Now note $(\alpha - \lambda_j \id) \pi_j (v) = 0$ since $p(\alpha)$ annihilates $v$. So $\pi_j(v) \in V_{\lambda_j}$. \[ v = \ub{\pi_1(v)}_{\in V_{\lambda_1}} + \cdots + \ub{\pi_k(v)}_{\in V_{\lambda_k}} \] so $V = \sum_{i = 1}^k V_{\lambda_k}$ hence there is a generating set of eigenvectors, which can be reduced to a basis. Then in this basis, $\alpha$ is diagonal. } \end{hiddenflashcard} \begin{example*} $A \in \mathcal{M}_n(F)$, $F = \CC$. $A$ has finite order. (there exists $m \in \NN$ such that $A^m = \id$). Then $A$ is diagonalisable (over $\CC$). TODO.. \end{example*}