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

If $$\textsf{V}$$ is a finite-dimensional complex vector space, $$\textsf{T}\in\mathcal{L}(\textsf{V})^1$$, $$\lambda$$ is arbitrary in $$\mathbb{C}$$ and if $$\textsf{V} = \text{null}(\textsf{T}-\lambda\textsf{I}) \oplus \text{range}(\textsf{T}-\lambda\textsf{I})$$ then prove that $$\textsf{T}$$ is diagonalizable.

Attempt : I am solving Axler's 3 edition book in Exercise $$5c.$$ The book hasn't introduced the Jordan normal form or the generalized eigenvectors. Could someone please give a direction to move ahead.

Thanks a lot for your help.

$$^1$$ $$\mathcal{L}(\textsf{V})$$ is the set of all linear maps from $$\textsf{V}$$ to itself.

• To be clear, the hypotheses for this exercise are that $V$ is a finite-dimensional complex vector space, $T$ is a linear map from $V$ to $V$, and $V = \text{null}(T - \lambda I) \oplus \text{range}(T - \lambda I)$ for every complex number $\lambda$. Commented Nov 30, 2019 at 2:28
• @SheldonAxler Yes that's affirmative. I have an intuition that we need to show that $V = \bigoplus null (T- \lambda I)$. I am not very sure how to move ahead though Commented Nov 30, 2019 at 7:13

Let us enumerate the eigenvalues $$\lambda_1,\cdots \lambda_k$$. As eigenspaces corresponding to different eigenvalues have trivial intersection, we have that $$\text{null}(T-\lambda_{i+1} I) \subset \text{range} (T-\lambda_i I)$$ for each $$i. By induction and the condition given by the problem, this gives us $$V=\bigoplus_{i=1}^k \text{null} (T-\lambda_i I) \,\oplus \,\bigcap_{i=1}^k \text{range}(T-\lambda_i I)$$. Thus, it remains to show that the intersection of ranges is trivial. Observe that $$W:=\,\bigcap_{i=1}^k \text{range}(T-\lambda_i I)$$ is an invariant subspace of T. If $$\dim W>0$$, then $$T|_W$$ has an eigenvector, which is not possible since all the eigenspaces of $$T$$ are accounted for in $$\bigoplus_{i=1}^k \text{null} (T-\lambda_i I)$$, so we must have $$W=\{0\}$$ as required. It follows that $$V$$ decomposes into a direct sum of eigenspaces of $$T$$, so $$T$$ is diagonalizable.
The condition is automatically satisfied if $$\lambda$$ is not an eigenvalue. This suggests that one should apply the condition to eigenvalues. I am not sure what is the simplest way to solve the problem but equipped with Cayley-Hamilton's theorem with rank/nullity argument, it is not difficult to show that $$V=\bigoplus_i {\rm null}(T-\lambda_iI)^{r_i},$$ assuming that the characteristic polynomial for $$T$$ is $$\prod_i(x-\lambda_i)^{r_i}.$$ Then one just needs to prove for each $$i$$ that $${\rm null}(T-\lambda_iI)={\rm null}(T-\lambda_iI)^{r_i}$$ using the given condition. Then by dimension count, one gets a basis consisting of eigenvectors, hence $$T$$ is diagonalizable.