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minimal.pdf
Lemma 4 If $U \stackrel{\beta}{\longrightarrow} V \stackrel{\gamma}{\longrightarrow} W$ are finite-dimensional vector spaces and linear maps, then
$$
\operatorname{dim} \operatorname{Ker}\left(\gamma\circ \beta\right) \leq \operatorname{dim} \operatorname{Ker}(\gamma)+\operatorname{dim} \operatorname{Ker}(\beta) .
$$
Proof. Observe that $\operatorname{Ker}(\gamma \circ \beta)=\beta^{-1}(\operatorname{Ker}(\gamma))$. (By using “$\beta^{-1}$”, I am not suggesting that $\beta$ is invertible; this is the notation for the pre-image or inverse image of a subset under a function, as explained in Numbers and Sets.)
Now, consider the function
$$
\begin{array}{rcl}
\beta^{\prime}: \beta^{-1}(\operatorname{Ker}(\gamma)) & \longrightarrow & \operatorname{Ker}(\gamma), \\
u & \longmapsto& \beta(u) .
\end{array}
$$
Applying the rank-nullity formula we get
$$
\operatorname{dim} \beta^{-1}(\operatorname{Ker}(\gamma))=\operatorname{dim} \operatorname{Im}\left(\beta^{\prime}\right)+\operatorname{dim} \operatorname{Ker}\left(\beta^{\prime}\right),
$$
and adding to this our initial observation and the facts that $\operatorname{Im}\left(\beta^{\prime}\right) \leq \operatorname{Ker}(\gamma)$ and $\operatorname{Ker}\left(\beta^{\prime}\right) \leq \operatorname{Ker}(\beta)$, the lemma is proved. $_\blacksquare$
The hard work is now done. Supposing that all the $s_i$ are 1 , the composite of the maps
$$
V \stackrel{\alpha-\lambda_k I}{\longrightarrow} V \stackrel{\alpha-\lambda_{k-1} I}{\longrightarrow} \cdots \stackrel{\alpha-\lambda_1 I}{\longrightarrow} V
$$
is 0 . So
$$
\begin{aligned}
\operatorname{dim}(V) & =\operatorname{dim} \operatorname{Ker}\left(\left(\alpha-\lambda_1 I\right) \circ \cdots \circ\left(\alpha-\lambda_k I\right)\right) \\
& \leq \operatorname{dim} \operatorname{Ker}\left(\alpha-\lambda_1 I\right)+\cdots+\operatorname{dim} \operatorname{Ker}\left(\alpha-\lambda_k I\right) \\
& =\operatorname{dim}\left(\operatorname{Ker}\left(\alpha-\lambda_1 I\right) \oplus \cdots \oplus \operatorname{Ker}\left(\alpha-\lambda_k I\right)\right)
\end{aligned}
$$
where the inequality comes from the Lemma (and an easy induction), and the second equality is justified by the fact that the sum of the eigenspaces is a direct sum. Hence the sum of the eigenspaces has the same dimension as $V$, i.e. this sum is $V$, and $\alpha$ is diagonalizable. |
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