I've been studying Linear Algebra Done Right by Axler. I've noticed some matrices pop up over and over again with an interesting property: if $V$ is a finite-dimensional vector space and $T \in \mathcal{L}(V)$ is this particular kind of operator, then $$ V = \operatorname{null} T \oplus \operatorname{range} T. $$ If an inner product is lying around, this is always true of normal operators, for which $$ \operatorname{null} T = (\operatorname{range} T)^\perp, $$ and for any operator $T$ in a vector space it is always true of the operator $T^{\operatorname{dim} V}$. It's also true of invertible operators and projection operators! But having this property does not imply that an operator is normal, or is invertible, or is the $\operatorname{dim} V$-th power of another operator.

Is there a name for the set of operators with the property $V = \operatorname{null}T \oplus \operatorname{range}T$?


2 Answers 2


In case it wasn't clear: Suppose $T$ satisfies the condition given in the question.

If $0$ is an eigenvalue of $T$, its algebraic multiplicity must equal its geometric multiplicity, i.e. when put into Jordan normal form, the Jordan blocks for eigenvalue $0$ are each $1 \times 1$.

Sanity checking that your examples satisfy this condition:

  • Normal matrices are diagonalizable, so the Jordan blocks each have size $1 \times 1$.
  • Similarly, orthogonal projections are symmetric, and are therefore diagonalizable.
  • Raising an $m \times m$ zero-eigenvalue Jordan block to the $m$th power (or higher) yields the zero matrix. So the zero-eigenvalue Jordan blocks of $T^{\dim V}$ each are $1 \times 1$.
  • Invertible operators have no zero eigenvalues so the condition holds vacuously.
  • 1
    $\begingroup$ The statement above that if 0 is an eigenvalue of $T$, then its algebraic multiplicity equals its algebraic multiplicity is false. $\endgroup$ May 26, 2022 at 2:34
  • $\begingroup$ @SheldonAxler Could you clarify why it is false? I am mainly trying to say that the Jordan blocks for the eigenvalue $0$ must all be $1 \times 1$, and the other answer also agrees with this. In this case, the geometric multiplicity (number of Jordan blocks) equals the algebraic multiplicity (number of zeros on the diagonal). $\endgroup$
    – angryavian
    May 31, 2022 at 3:30
  • $\begingroup$ @MorganRodgers Thanks, I've updated my answer to make this clearer. $\endgroup$
    – angryavian
    Jun 4, 2022 at 6:02

I doubt there is a special name for operators $T:V\to V$ with $V=\ker T\oplus\operatorname{im} T$. When $V$ is finite dimensional, one can characterize these operators as those $T$ with the property that $T^2(v)=0\Rightarrow T(v)=0$ ($T^2=T\circ T$). Another way to say this is that the Jordan form decomposition for $T$ has one block, which is diagonal, or no block, for the eigenvalue $0$. Edit: I stated the Jordan form alternative incorrectly. I should have said every Jordan block for the eigenvalue $0$ is $1\times 1$.

To see that this characterization is valid, we show that for any vector space $V$, the condition $T^2(v)=0\Rightarrow T(v)=0$ is equivalent to the condition that $\ker T\cap\operatorname{im} T=0$. Assume first that $T^2(v)=0\Rightarrow T(v)=0$ and $x\in \ker T\cap\operatorname{im} T$. Then $x=T(v)$ for some $v\in V$ and so $T^2(v)=T(x)=0$, whence $x=T(v)=0$. Conversely, if $\ker T\cap\operatorname{im} T=0$ and $T^2(v)=0$, then $T(v)\in \ker T\cap\operatorname{im} T$, whence $T(v)=0$. The equivalence with $V=\ker T\oplus\operatorname{im} T$ in the finite dimensional case now follows from the dimension formula.

I don't see any better characterization of such operators and that makes me think there isn't a special name for such operators.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .