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I need help proving $p(A)=0$ without Cayley-Hamilton theorem when $A$ is upper triangular, as part of the proof of the Cayley-Hamilton theorem

The result makes a lot of sense but I can't prove it properly

If $A \in \Bbb C^{n \times n }$ is upper triangular then its characteristic polynomial is $p= (x-c_1)^{r _1}...(x-c_k)^{r_k}$ then $p(A)$ will be the product of $k$ upper triangular matrices with interspersed zeros on the diagonal...

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  • $\begingroup$ Write it out in detail, all symbols, for $n=2$ and $n=3.$ With small $n$ you can see exactly how $p(x)$ factors and how to interpret $(x-c_j)^{r_j}$ when $x=A$ $\endgroup$ – Will Jagy Aug 20 '14 at 18:00
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If $c_1,\ldots c_n$ are the entries on the diagonal of an upper triangular matrix $A$, then the characteristic polynomial is $p(x) = (x-c_1) \cdots (x-c_n)$.

$A-c_iI$ is upper triangular with the $i$th diagonal entry being zero. See what happens when you then compute $p(A)=(A-c_1 I) \cdots (A-c_nI)$.

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  • $\begingroup$ I'll have a product of matrices each with a zero in a different place on the diagonal... but why is the product zero? I see that's the case for $3 \times 3$ and $2 \times 2$ matrices but I don't see the general case $\endgroup$ – Shomar Aug 20 '14 at 21:26
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    $\begingroup$ @Shomar I think you can show that $A-c_1I$ has the first column all zero, $(A-c_1 I)(A-c_2 I)$ has the first two columns all zero, and so on. $\endgroup$ – angryavian Aug 20 '14 at 22:42
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    $\begingroup$ The idea in this answer is quite beautiful and worthy of many more upvotes. Notice that this gives a full proof of Cayley-Hamilton, by passing to an algebraic closure and using Schur decomposition (or for an overkill, Jordan canonical form). @angryavian: I suggest that you update your answer, by fleshing out your last comment. $\endgroup$ – zcn Aug 22 '14 at 3:18
  • $\begingroup$ It's clear that $(A-c_1 I)(A-c_2 I)$ has the first two columns all zero. By induction, let's say $(A-c_1 I) \cdots (A-c_kI)=C$ has the first $k$ columns all zero. What happens with the $k+1$ column of $C(A-c_{k+1} I)=B$? If we take the $i$th element from that column $[B]_{i \ k+1}=\sum_{j=1}^n [C]_{ij} [A-a_{k+1}I]_{j \ k+1} $ But if $j \le k$ then $[C]_{ij} = 0$ . If $j=k+1$ then $[A-a_{k+1}I]_{k+1 \ k+1}=0$ and if $j \gt k+1$ then $[A-a_{k+1}I]_{j \ k+1}=0$ because $A-a_{k+1}I$ is upper triangular. Then the $k+1$ column is all zeros and $p(A)=(A-c_1 I) \cdots (A-c_nI)=0$. Is this correct? $\endgroup$ – Shomar Aug 23 '14 at 15:23

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