Let $A$ be a square complex matrix of order n such that exists $k \geq 1 : A^k = A^{k+1}$. Problem about invariant subspaces.

Question

I'm stuck on this problem for two days.

Notation: $SpA$ is the spectrum of $A$, $\mu_A(t)$ is the minimal polynomial of $A$.

Let $A$ be a square complex matrix of order n such that exists $k \geq 1 : A^k = A^{k+1}$. Prove that if there is only one $n-1$ dimensional $A$-invariant subspace $W$ of $\mathbb{C}^n$ then $k \geq n$.

Now: since we consider a complex matrix we know that is triangularizable, which implies that the $A$ minimal polynomial $\mu_A (t)$ is completely factorizable. By $A^k = A^{k+1}$ we deduce that $\mu_A(t) | t^k(t-1)$ and $SpA = \{ 0, 1 \}$ or $SpA = \{ 0 \}$(it could be also $SpA = \{ 1 \}$ but I'm not interested in this case). So if we prove that $\mu_A(t) = t^n(t-1)$ or $\mu_A(t) = t^n$ we got the thesis, but I don't know how to show that.

Can you give me a hint please?

Thanks. English is not my mother tongue, please excuse any errors on my part.

Answer 1

Since $A$ is a complex matrix, it admits a Jordan canonical form; recall that the algebraic multiplicity of an eigenvalue $\lambda$, i.e., the multiplicity of the linear factor $t-\lambda$ in the characteristic polynomial of $A$, is the sum of the sizes of the Jordan blocks corresponding to $\lambda$, whilst the multiplicity of the linear factor $t-\lambda$ in the minimal polynomial of $A$ is the largest size of any Jordan block corresponding to $A$.

Anyhow, the basic idea is that invariant subspaces of $A$ can be constructed from the Jordan blocks of $A$, and in particular, that any set of Jordan blocks of $A$ whose sizes sum to $n$ gives rise to an $(n-1)$-dimensional $A$-invariant subspace. Explicitly, let $J$ be a Jordan matrix and $S$ an invertible matrix such that $J = S^{-1}AS$, so that the columns of $S$ are an ordered basis for $\mathbb{C}^n$ consisting of a disjoint union of Jordan chains, each of which corresponds precisely to a Jordan block of $J$; in particular, the size of each Jordan block is precisely the cardinality of the corresponding Jordan chain and the rank of the generalized eigenvector in the Jordan chain of highest rank, i.e., the generator of that Jordan chain.

1. Choose a set of Jordan blocks of $J$ whose sizes sum to $n$.
2. Choose a distinguished Jordan block, take the Jordan chain corresponding to that block and delete the the generalized eigenvector of highest rank, i.e., the generator of that Jordan chain.
3. For each remaining Jordan block, take the Jordan chain corresponding to that block, without any modifications.
4. Take the union of the sets of generalised eigenvectors chosen/constructed in steps 2. and 3.

You can now check that the span of the set constructed in step 4. is an $(n-1)$-dimensional $A$-invariant subspace.

But now, from this perspective, what can you do if $A$ has at least two Jordan blocks?