Let $A$ be a $10\times 10$ matrix and $A^{10,000} = 0$ Show $A^{10} = 0$

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I found the titular question

Let $A$ be a $10\times 10$ matrix and $A^{10,000} = 0$. Show $A^{10} = 0$

On a practice qualifier. I am very rusty on linear algebra and could not remember any specific theories other than maybe finding the characteristic polynomial of $A^{10}$ and reasoning that if it is not equal to the $\lambda^{10}$ then the characteristic polynomial of $A^{10,000}$ is certainly not equal to $\lambda^{10,000}$ but drew a blank when trying to rationalize why this would be so and why it would matter. So maybe that is a dead end. Then I saw this post Let A Be a Symmetric Matrix of Order $A^2 = 0$ which included many unique ways of solving that (possibly similar) question. So what are some possible proofs for this question? Ideally I would start completely reviewing linear-algebra but I am catching up on a lot of subjects and won't have the time to dive deep into it until the summer. In the mean time I'd like to do a little targeting refreshing on practice qualifiers like this one. Thank you for any help!

p.s. I included an abstract algebra tag because that is where I have the strongest background so feel free to use generalized theorems and techniques from abstract algebra along with any linear-algebra proofs you like.

I am almost certain that this sort of question has been asked on this site before, but I'm having trouble finding it. In brief, the proof is as follows: show that $0$ is the only possible eigenvalues of $A$, then find the characteristic polynomial of $A$ and the result follows from the Cayley-Hamilton theorem. — Ben Grossmann, Jan 11 '20 at 17:09
If you're interested in avoiding the Cayley-Hamilton theorem, there's also this proof — Ben Grossmann, Jan 11 '20 at 17:12
You are bouncing around the right circle of ideas. The minimal polynomial of $A$ must divide what? Since the degree of the minimal polynomial is at most the degree of its characteristic polynomial, what can we conclude? — hardmath, Jan 11 '20 at 17:20
I reopened this because this question is not restricted to proofs without using the minimal polynomial, as was the proposed dupe. — Bill Dubuque, Jan 14 '20 at 18:59

score 5 · Answer 1 · answered Jan 11 '20 at 17:24

Hint:

If $A^{10\,000}=0$, the matrix is a root of the monomial $x^{10\,000}$, so its minimal polynomial, being a divisor of this monomial, is a power of $x$. However, we know that the characteristic and minimal polynomials of a matrix have the same irreducible factors and the degree of the characteristic polynomial is the dimension of the matrix.

score 0 · Answer 2 · answered Jan 11 '20 at 18:39

Fix a particular non-zero vector $\vec v$. There is a smallest positive integer $k$ such that $A^k\vec v=\vec 0$ (and $A^{k-1}\vec v\not=\vec 0$). Look at the set of vectors $\{\vec v,A\vec v,A^2\vec v,\dots,A^{k-2}\vec v, A^{k-1}\vec v\}$. I claim that this set must be linearly independent, but I shall postpone the proof.

If the set is linearly independent, the set must contain no more than $10$ vectors because the space you are considering is $10$-dimensional only. This means $k$ is no more than $10$, and this holds true for all vector $\vec v$, meaning $A^{10}\vec v=\vec0$ for all vectors $\vec v$.

Here is a proof that the set is linearly independent:

To prove that the set is linearly independent, let $c_r$ be some scalars such that $$\sum_{r=0}^{k-1}c_rA^r\vec v=\vec0.\label{a}\tag{1}$$ We will be proving each $c_r$ must be $0$. Apply $A^{k-1}$ to both sides of the equation $\ref{a}$ and we obtain the equation $$c_0A^{k-1}\vec v=\vec0,$$ meaning $c_0=0$. Apply $A^{k-2}$ to both sides of the equation $\ref{a}$ and we obtain the equation $$c_0A^{k-2}\vec v+c_1A^{k-1}\vec v=\vec0,$$ meaning $c_1=0$ (we already know $c_0=0$). Keep repeating this process and you would also obtain $c_2=0,c_3=0,\dots,c_{k-1}=0$.

score 0 · Answer 3 · answered Jan 11 '20 at 19:15

0

A nilpotent matrix $A\in M_n(K)$ satisfies $A^n=0$, see this duplicate:

$A^n=0$ for $A$ nilpotent without the use of minimal polynomial

Because of $A^{10000}=0$, we know that $A$ is nilpotent. Thus we have $A^{10}=0$.

answered Jan 11 '20 at 19:15

Dietrich Burde

140,055

Let $A$ be a $10\times 10$ matrix and $A^{10,000} = 0$ Show $A^{10} = 0$

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