# matrices with nullspace the complement of the range

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 $$V = \operatorname{null}T \oplus \operatorname{range}T$$?

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.
• The statement above that if 0 is an eigenvalue of $T$, then its algebraic multiplicity equals its algebraic multiplicity is false. May 26, 2022 at 2:34
• @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). May 31, 2022 at 3:30
• @MorganRodgers Thanks, I've updated my answer to make this clearer. 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.