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Let $A$ be an $n\times n$ matrix over some field $K$. Then its characteristic polynomial $\mathrm{char}_A(X)\in K[X]$ is monic of degree $n$ and annihilated by $A$ (Cayley-Hamilton). It follows that

Corollary. $A^n, A^{n-1},\dots,\mathrm{id}_n$ are linearly dependent.

I wonder if this Corollary can be obtained without making use of the determinant (which is required to define $\mathrm{char}_A(X)$)?

Otherwise, can we show that there is some $m<n^2$ for which $A^m, A^{m-1},\dots,\mathrm{id}_n$ are linearly dependent? Note that if we take $m=n^2$, then the statement is obvious, because the $n\times n$ matrices form a vector space of dimension $n^2$, and $A^m, A^{m-1},\dots,\mathrm{id}_n$ are $n^2+1$ matrices.

Zuy
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    Possible duplicate of https://math.stackexchange.com/q/1310609/589 – lhf Oct 02 '23 at 20:52
  • I see, misread. The book you want is Axler, Linear Algebra Done Right. There were earlier (and more difficult) determinantphobic books by Werner Greub – Will Jagy Oct 02 '23 at 21:24
  • I have the third edition. Chapter 5 is Invariant Subspaces. Page 136 he proves at most $n$ eigenvalues. Page 145 he shows there is at least one eigenvalue (perhaps complex). That's theorem 5.21; it is about a cyclic vector, really: given operator $T$ and nonzero vector $v,$ the set $v, Tv, T^2 v,..., T^n v$ is linearly dependent. Determinant is on page 307 – Will Jagy Oct 02 '23 at 21:39
  • perhaps determinaphobic – Will Jagy Oct 02 '23 at 22:24
  • @WillJagy Greub is definitely not determinant phobic, he has beautiful coverage of the determinant and uses it to define the characteristic polynomial. – blargoner Oct 03 '23 at 00:42
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    If the field is algebraically closed, there is a well-known proof. First, upper-triangularise $A$ as $P^{-1}TP$. Then for each $k$, $T-t_{kk}I$ is an upper triangular matrix whose $k$-th diagonal element is zero. By mathematical induction, for each $k=1,2,\ldots,n$, the first $k$ columns of $(T-t_{11}I)\cdots(T-t_{22}I)\cdots(T-t_{kk}I)$ are zero. Therefore $\prod_{k=1}^n(T-t_{kk}I)=0$. Expand the product into a monic polynomial, we see that $T^n,T^{n-1},\ldots,T,I$ are linearly dependent. Therefore, so are $A^n,A^{n-1},\ldots,A,I$. – user1551 Oct 03 '23 at 06:22
  • For an algebraically closed field: 1st prove nilpotent $A$ satisfies $A^n = \mathbf 0$ (by manipulation of rank inequalites). For general $A$ w/ eigenvalue $\lambda$ write $V = \text{image }(A-\lambda I)^n \oplus \ker (A-\lambda I)^n$ where the latter has dimension $r$, $A$ and $(A-\lambda I)^n$ commute so $A$ respects these invariant subspaces If $r=n$ then we have the result by nilpotence and otherwise $1\leq r \leq n-1$ so use strong induction hypothesis that some non-zero polynomial of degree $\leq $ dimension kills $A$'s restriction to each invariant subspace. – user8675309 Oct 03 '23 at 15:58

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