Proving $3\times3$ nilpotent matrix satisfies $A^3=0$ Let $A$ be a $3$ $\times$ $3$ matrix. Prove that if there exists $n$ $>$ $3$ such that $A^n$= $0$, then $A^3$ = $0$. Can someone help explain how to prove this? I know that since A is nilpotent, its only eigenvalue ($\lambda$) is $0$. So, I was thinking that I could argue the following: $A^3$$\dot{\vec{v}}$=$\lambda^3$$\dot{\vec{v}}$=$\dot{\vec{0}}$, which implies that $A^3$=$0$. I don't think this proof is fully correct, however, because it doesn't use the fact that $n>3$. I could have made the same argument to suggest that $A^2=0$, which is not necessarily true when a $3$ $\times$ $3$ matrix A is nilpotent.
 A: There are a few ways to do that. If you know about minimal polynomials, then you can use that the minimal polynomial of $A$ has degree at most $3$ and divides $A^n$.
A less advanced way: Note that for each $k$, the range of $A^k$ is a subspace of ${\mathbf R}^3$, and $range(A^3) \subset range(A^2)\subset range(A) \subset {\mathbf R}^3$. The dimensions must correspondingly be nonincreasing as you go down the list. If $range(A^3)$ weren't of dimension zero, then you'd have $range(A^k) = range(A^{k+1}) \neq \emptyset$ for some $k \leq 2$. Then by induction $range(A^l) = range(A^k)$ for all $l \geq k$. This would include $l = n$, a contradiction.
A: Let $v$ be a non-zero vector and consider the sequence
$$
v, Av, A^2v, ..., A^kv, 0, 0, ...
$$
where $k < n$ is the largest integer such that $A^kv \ne 0$. We will see that the non-zero elements of the sequence are linearly independent. Therefore, in a three-dimensional space, at most three, i.e. $v, Av, A^2v$, can be non-zero. Thus, $A^3v = 0$ for every $v$.
To prove linear independence, note first that the last non-zero element of the sequence, i.e. $A^kv$ lies in the kernel of $A$, but the other vectors, i.e. $v, Av, A^2v, ..., A^{k-1}v$ do not. Therefore none of the vectors
$v, Av, A^2v, ..., A^{k-1}v$ can be a multiple of $A^kv$. Similarly, the last two non-zero vectors of the sequence, i.e. $A^{k-1}v, A^kv$ lie in the kernel of $A^2$, but the preceding vectors do not. Therefore none of the preceding vectors is a linear combination of $A^{k-1}v, A^kv$.
Continuing this reasoning we see that $A^iv, ..., A^kv$ lie in the kernel of $A^{k-i+1}$ while none of $v, Av, ..., A^{i-1}v$ do. Therefore, none of $v, Av, ..., A^{i-1}v$ is a linear combination of $A^iv, ..., A^kv$. Therefore, the non-zero vectors in the sequence $v, Av, A^2v, ..., A^kv, ...$ are linearly independent.
Therefore there cannot be more than three of them, so $A^3 = 0$.
A: Since $M_{n\times n}$, the space of $n\times n$ matrices, is an $n^2$ dimensional vector space, there must be a linear dependence between the matrices $A^{0}, A^1, A^2,\ldots, A^{n^2}$ for any $A$.  Therefore, $A$ satisfies some polynomial of degree at most $n^2$.  However, if $p(A)=0$ and $q(A)=0$, then for any other polynomials $a, b$, we have $a(A)p(A)+b(A)q(A)=0$, so by the Bezout lemma, $\gcd(p,q)(A)=0$.  On the one hand, this establishes that $A$ satisfies a unique monic minimal degree polynomial, but on the other hand establishes that if $A$ satisfies that if $p(A)=0$, then the minimal polynomial of $A$ will be some divisor of $p$.
By the Cayley-Hamilton theorem, $A$ satisfies its characteristic polynomial, which shows that the minimal polynomial of $A$ is actually of degree at most $n$.
Putting this all together, in your particular case, the minimal polynomial divides $x^n$ and has degree at most $3$.  In particular, it divides $x^3$.
However, if we do not have the Cayley-Hamilton theorem, then we must make use of something else to get our bound of $3$.  One good way is the following lemma:
If $\ker A^k=\ker A^{k+1}$, then $\ker A^k=\ker A^i$ for all $i>k$.
Once you show this, we note that we have a chain of subspaces
$$\ker A^0\subset\ker A \subset \ker A^{2} \subset \ker A^3 \subset \cdots.$$
Since $A$ is acting on a 3 dimensional vector space, and since we have a proper inclusion only when two subspace have different dimensions, the dimension of the kernel must increase by at least $1$ every time until it stabilizes (at 3, since it is eventually the whole space).  In particular, if $\dim \ker A^n=3$ for some $n$, then $\dim A^3=3$.
A: Since you know that the eigenvalues of a nilpotent matrix are $0$, consider the Jordan normal form.
The Jordan normal form of this matrix will be strictly upper-triangular. Now, you can easily prove that for any strictly upper-triangular matrix of order $n \times n$, you have $A^n = O$, where $O$ is the zero matrix of the same order.
And your question is the case when $n=3$.
A: Pick an arbitrary nonzero vector $v$. Let $f$ be the monic polynomial of the lowest degree such that $f(A)v=0$. Such $f$ must exist because $A^nv=0$. Also, $k:=\deg(f)\le3$ because $v,Av,A^2v,A^3v$, being four vectors in a three-dimensional vector space, must be linearly dependent.
However, $f$ must divide $x^n$, otherwise $g=\operatorname{gcd}(f,x^n)$ will satisfy $g(A)v=0$ (because $g=af+bx^n$ for some polynomials $a$ and $b$) but $\deg(g)<\deg(f)$, which contradicts minimality of $\deg(f)$. Therefore $f(x)=x^k$ for some $k\le 3$. Hence $A^3v=A^{3-k}f(A)v=0$ and we conclude that $A^3=0$ because $v$ is arbitrary.
A: Because $p(A)=0$ where $p(\lambda)=\lambda^3$, then the minimal polynomial $m(\lambda)$ must divide $p$, forcing $m(\lambda)=\lambda^n$ for either $n=0,1,2$ or $3$. The conclusion is that $A$ is nilpotent or $0$.
