Is it true that $\ker(T) = \ker(T^2) = ...$ if $T$ is a diagonalizable operator?

Question

We covered ways that linear operators $T:V \rightarrow V$ could fail to be diagonalizable. In particular we talked about the operator $T: F^2 \rightarrow F^2$, where $\beta = \{e_1,e_2\}$ is the standard basis for $F^2$ with $T(e_1) = 0$ and $T(e_2) = e_1$. If you take the characteristic polyomial of $T$ then you get $$ p(t) = t^2 $$ and hence the only eigenvalue is $\lambda = 0$, and because $E_0 \neq F^2$ it's straightforward to see that the operator fails to be diagonalizable. So far this is all fair, but then we went on to note that if we consider $T^2 = T \circ T$, that $$ \ker(T) \subset \ker(T^2), $$ because $T^2$ is the zero operator on $F^2$ and hence the kernel is all of $F^2$. Apparently this is "typical behaviour" of non-diagonalizable operators. Another question online here showed that if $T$ is diagonalizable that $\ker(T) = \ker(T^2)$ for an arbitrary diagonalizable operator, but is it also true that $$ \ker(T^n) = \ker(T^{+1}), \; \forall n. $$ Is this a sufficient way to characterize diagonalizable operators? If this is true how should I go about proving it; would the approach to take be an induction on $n$? Thanks in advance for the clarifications!

Answer 1

The statement is not true. As an example, the matrix $$ A = \pmatrix{0&0&0\\0&1&1\\0&0&1} $$ is non-diagonalizable, but satisfies $\ker(A^n) = \ker(A^{n+1})$ for all integers $n \geq 1$.

Here is a sufficient condition to characterize diagonalizable over an algebraically closed field $F$. Such an operator is diagonalizable iff for all $\lambda \in F$, $\ker((A - \lambda I)^n) = \ker((A - \lambda I)^{n+1})$ for all integers $n \geq 1$ (where $I$ denotes the identity operator).

Answer 2

If $T$ is diagonalizable, let $B=\{v_1,\ldots,v_n\}$ be a basis consisting of eigenvectoes of $T$. Then, for each $n\in\Bbb N$, $\ker(T^n)$ is the space spanned by those $v_k$'s whose corresponding eigenvalue is $0$. In particular, $\ker(T^n)$ is the same for all $n$'s.

However$$\begin{array}{rccc}T\colon&F^2&\longrightarrow&F^2\\&(x,y)&\mapsto&(x+y,y)\end{array}$$is not diagonalizable, but nevertheless $(\forall n\in\Bbb N):\ker(T^n)=\{0\}$.

Answer 3

Requiring $\ker(T^n)=\ker(T^{n+1})$ for all integers $n\geq1$ (which simplifies to just requiring $\ker(T)=\ker(T^2)$) is not enough for an operator to be diagonalizable. For instance, take your example $T$ and add $I$. The resulting linear map $T+I$ is invertible, so all $\ker(T^n)$ are trivial, yet the map is still just as undiagonalizable.

You will note that this new map has $1$ as its only eigenvalue, yet the corresponding eigenspace is only of dimension $1$. And this is a characteristic property of non-diagonalizable linear transformations:

A linear transformation is non-diagonalizable iff there is at least one eigenvalue whose multiplicity as a root of the characteristic polynomial is unequal to the dimension of the corresponding eigenspace.

Or, said in jargon, iff there is at least one eigenvalue where the algebraic multiplicity is not equal to the geometric multiplicity.

Going back to kernels, if we have a non-diagonalizable $T$ and such an eigenvalue $\lambda$, then we have $$ \ker(T-\lambda I)\subset \ker((T-\lambda I)^2) $$ So on the other hand, if $$ \ker(T-\lambda I)=\ker((T-\lambda I)^2) $$ for all scalars $\lambda$ (or more specifically for all eigenvalues $\lambda$) then $T$ is diagonalizable. (One caveat is that, for instance, real matrices will still require you to check this for all complex eigenvalues.)