Is a Jordan block not further block diagonalizable?

Question

We can always find a Jordan canonical form of a matrix A. It is a block diagonal matrix. Is it true that each block cannot be reduced to a matrix with more blocks in diagonal? In other words, for a given linear operator T, is it true that "each T-invariant subspace corresponding to a Jordan block cannot be decomposed into a direct sum of T-invariant proper subspaces of it" ?

Answer 1

is it true that "each T-invariant subspace corresponding to a Jordan block cannot be decomposed into a direct sum of T-invariant proper subspaces of it" ?

Yes. In fact this holds more generally for the decomposition of a finite-dimensional vector space (over an arbitrary field) under the action of a linear endomorphism into cyclic primary subspaces: see these notes on invariant subspaces, especially Theorem 5.2. A Jordan block is precisely a cyclic primary subspace for a polynomial of the form $(t-\alpha)^d$. But in general a subspace which is cyclic -- i.e., is spanned by the $T$-orbit of a single vector -- and primary -- i.e., the minimal polynomial is a power of a prime polynomial $p(t)^d$ is indecomposable: it cannot be written as a direct sum of nonzero $T$-invariant subspaces. One way to see this is to observe that since the minimal polynomial of $V_1 \oplus V_2$ is the lcm of the minimal polynomials of $V_1$ and $V_2$, if $V$ has minimal polynomial $p(t)^d$, then at least one of $V_1$ and $V_2$ must have minimal polynomial $p(t)^d$, contradicting uniqueness of primary decmposition.

Alternately, and maybe preferably, Theorem 3.2 of loc. cit. gives a complete description of all the invariant subspaces of any cyclic space $V$. From this description one sees that if (and only if!) the minimal polynomial of $V$ is a prime power, these invariant subspaces form a chain (i.e., are simply ordered under inclusion), and this certainly implies indecomposability.