# If $V = \text{null}(T-\lambda I) \oplus \text{range}(T-\lambda I)$, then $T$ is diagonalizable?

$$V$$ is a finite-dimensional complex vector space and $$T \in L(V)$$ ($$L(V)$$ is the set of all linear maps from $$V$$ to itself), and $$\lambda$$ is arbitrary in $$\mathbb{C}$$.

I know $$T$$ is diagonalizable if it has $$\text{dim}(V)$$ distinct eigenvalues, or if $$V$$ has a basis consisting of eigenvectors of $$T$$. I was thinking it might also have something to do with eigenspaces (there are a few theorems with diagonalizability and eigenspaces in my text), since those specifically handle $$\text{null}(T-\lambda I)$$, but then that would leave out the range.

I have the definition of an eigenspace, direct sum, diagonalizability, and some conditions equivalent to diagonalizability, i.e. a matrix is diagonalizable if V consists of eigenvectors of T, there exist 1-dimensional subspaces of V that are invariant under T, where V is the direct sum of these 1D subspaces, V is the direct sum of eigenspaces of T corresponding to each eigenvalue, and the dimension of V is the same as the dimension of the sum of these eigenspaces. Also, the book is "Linear Algebra Done Right" by Sheldon Axler.

I also have that $$T$$ is diagonalizable if it contains $$dim(V)$$ distinct eigenvalues.

• Comments are not for extended discussion; this conversation has been moved to chat. Mar 7, 2019 at 18:58

[This is basically an inelegant version of Arturo Magidin's great comment.]

I will use the "hammer" that [the matrix of] a finite-dimensional linear operator $$T$$ has a Jordan normal form. Let $$A$$ be the matrix of $$T$$ in the basis such that $$A$$ is in Jordan normal form.

If $$A$$ has a Jordan block with eigenvalue $$\lambda$$ that has size larger than one, for example $$\begin{bmatrix} \lambda & 1 \\ & \lambda & 1 \\ && \lambda \end{bmatrix}.$$ [This is just one block; there may be other Jordan blocks in the matrix.] In $$A - \lambda I$$ this block becomes $$\begin{bmatrix} 0& 1 \\ & 0& 1 \\ && 0\end{bmatrix}.$$ From here you can see that $$(1, 0, 0)$$ is in both the nullspace and the range of this block; thus you can then find something in both the nullspace and range of the full matrix $$A - \lambda I$$. This contradicts the given assumption regarding the direct sum. Thus all Jordan blocks must be $$1 \times 1$$, i.e. $$A$$ is diagonalizable.

• Sorry, I just saw OP posted a comment not knowing about generalized eigenvalues, so this answer might be inappropriate... I might delete this answer later. Mar 4, 2019 at 6:45
• Could you please post a solution without using the jordan form as well please Nov 29, 2019 at 21:19

By induction on $$n:=\dim V$$. The case $$n=0$$ (or $$1$$) is clear. If $$n>0$$, let $$\lambda_0$$ be an eigenvalue of $$T$$. Then $$V=E(\lambda_0,T)\oplus\operatorname{range}(T-\lambda_0)$$ by assumption. Let $$W=\operatorname{range}(T-\lambda_0)$$ and $$S=T\rvert_W$$. It is clear that $$W$$ is $$T$$-invariant, so $$S:W\to W$$. We claim that $$\ker(S-\lambda)\oplus\operatorname{range}(S-\lambda)=W$$ for all $$\lambda$$. Then we may apply the induction hypothesis as $$\dim W and get that $$W$$ is the direct sum of the eigenspaces of $$S$$. Hence $$V=E(\lambda_0,T)\oplus W$$ is the direct sum of eigenspaces of $$T$$ and thus $$T$$ is diagonalizable.

To prove $$\ker(S-\lambda)\oplus\operatorname{range}(S-\lambda)=W$$, it suffices to prove that $$\ker(S-\lambda)\cap\operatorname{range}(S-\lambda)=0$$ (by dimension reasons, see also exercise 3). But this is clear as $$\ker(S-\lambda)\cap\operatorname{range}(S-\lambda)\subseteq \ker(T-\lambda)\cap\operatorname{range}(T-\lambda)=0.$$

The following proof by contradiction uses results 5.27(b), 5.62, and Exercise 4 in Section 5D in Axler, Sheldon (2024). Linear Algebra Done Right, Undergraduate Text in Mathematics (4th ed.), Springer Publishing, https://doi.org/10.1007/978-3-031-41026-0

Proof. Suppose $$V=\text{null}(T-\lambda I) \oplus \text{range}(T-\lambda I)$$ for all $$\lambda \in \mathbb{C}$$. By 5.27(b), the minimal polynomial of $$T$$ equals $$(z-\lambda_{1})^{n_{1}}\dots(z-\lambda_{m})^{n_{m}}$$ for some distinct $$\lambda_{1},\dots,\lambda_{m} \in \mathbb{C}$$, where $$n_{1},\dots,n_{m}$$ is a list of positive integers. Suppose $$n_{1}>1$$, without loss of generality. Then $$\prod_{k=1}^{m}(T-\lambda_{k}I)^{n_{k}}v=(T-\lambda_{1}I)(T-\lambda_{1}I)^{n_{1}-1}\prod_{k= 2}^{m}(T-\lambda_{k}I)^{n_{k}}v=0$$ for all $$v \in V$$. This implies $$(T-\lambda_{1}I)^{n_{1}-1}\prod_{k= 2}^{m}(T-\lambda_{k}I)^{n_{k}}v=0$$ because $$(T-\lambda_{1}I)^{n_{1}-1}\prod_{k= 2}^{m}(T-\lambda_{k}I)^{n_{k}}v \in \text{null}(T-\lambda_{1} I) \cap \text{range}(T-\lambda_{1} I)=\left\{0\right\}$$ by Exercise 4 in Section 5D, which is a contradiction. Therefore $$T$$ is diagonalizable by 5.62. QED.