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...

  • $\begingroup$ 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

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 $p(A)=(A-c_1 I) \cdots (A-c_nI)$.

| cite | improve this answer | |
  • $\begingroup$ 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
  • 2
    $\begingroup$ @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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.