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.

  • 2
    $\begingroup$ 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 $\lambda$. $\endgroup$ Nov 30, 2019 at 2:28
  • $\begingroup$ @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 $\endgroup$
    – MathMan
    Nov 30, 2019 at 7:13

2 Answers 2


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 $i<k$. 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.


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