Question: Let $A$ be $n\times n$ and nilpotent such that $A^{n-1}\neq 0$. Show that $A$ has exactly one Jordan Block.
My Attempt: Since $A$ is nilpotent, $A^k=0$ some $1\leq k\leq n$. So, the minimal polynomial, $m(x)$, of $A$ is a factor of $x^k$, and so $m(x)=x^a$ where $1\leq a\leq k\leq n$. But, since $A^{n-1}\neq 0$, we have that $A^n=0$, and so $a=k=n$ thus the size of the largest Jordan block, the degree of the minimal polynomial, is $n$. Since our matrix is $n\times n$, we can only fit one Jordan Block.
First, is what I have written correct? To show the "exactly" part, is the idea to get a sort of "minimum" length bound of $n$ by using the degree of the min poly, and then a "maximum" length bound of $n$, the size of the matrix?
Thank you for any help!
If you choose $x$ such that $A^{n-1}x\ne 0$, then $\{ x,Ax,A^2x,\cdots,A^{n-1}x\}$ is a linearly independent set of $n$ vectors, which matches the dimension of the space. So it is a basis of the underlying vector space $X$. And $A$ is a single Jordan block of length $n$ with respect to this basis.