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I'm reading Hoffman and Kunze linear algebra book and in the end of the page 204, they said:

The theorem 5 is the following:

I didn't understand why we can use the theorem 5 to prove the Cayley-Hamilton theorem, i.e., if $p_c$ is the characteristic polynomial, then $p_c(T)=0$.

We can easily see that

$$p_c(x)=(x-c_1)^{d_1}\ldots(x-c_k)^{d_k}$$

$$p_m(x)=(x-c_1)^{l_1}\ldots(x-c_k)^{l_k}$$

Where $p_m$ is the minimal polynomial.

However I'm having troubles to understand why $l_i\le d_i$.

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    $\begingroup$ $l_i\le d_i$, because $p_m(x)$ is the minimal polynomial. $\endgroup$ – Dietrich Burde Jul 9 '16 at 7:41
  • $\begingroup$ @DietrichBurde but we have to prove before why $p_c(T)=0$ $\endgroup$ – user42912 Jul 9 '16 at 7:43
  • $\begingroup$ Because it holds for triangular matrices, and by Theorem $5$ then also for all complex matrices (because they are triangulable). $\endgroup$ – Dietrich Burde Jul 9 '16 at 7:46
  • $\begingroup$ @DietrichBurde Why does $p_c(A)=0$ for triangular matrices $A$? I don't remember the authors prove this in their book. $\endgroup$ – user42912 Jul 9 '16 at 7:49
  • $\begingroup$ I think a direct calculation works for triangular matrices. I did the 2 by 2 case quickly in my head, the general case is probably similar. $\endgroup$ – Malkoun Jul 9 '16 at 8:22
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I will answer your question for triangular matrices (see the comments). For simplicity of notation, let us just do the 3 by 3 case. Assume you have a linear operator $L$, which maps: $e_1$ to $d_1 e_1$, $e_2$ to $d_2 e_2 + a e_1$ and $e_3$ to $d_3 e_3 + b e_1 + c e_2$.

Let us calculate $p_c(L)=(L-d_1)(L-d_2)(L-d_3)$. The $3$ factors commute, so the order does not matter. $e_1$ is mapped by $L-d_1$ to $0$, so it is mapped by $p_c(L)$ to $0$. $e_2$ is mapped by $L-d_2$ to a constant times $e_1$, which is then mapped to $0$ by $L-d_1$. So $e_2$ is mapped by $p_c(L)$ to $0$. Similarly for $e_3$.

I apologize for my notation, which is kind of ugly, but anyway, you get the general idea.

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