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$V$ 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.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Aloizio Macedo
    Mar 7, 2019 at 18:58

3 Answers 3

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[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.

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  • $\begingroup$ 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. $\endgroup$
    – angryavian
    Mar 4, 2019 at 6:45
  • $\begingroup$ Could you please post a solution without using the jordan form as well please $\endgroup$
    – MathMan
    Nov 29, 2019 at 21:19
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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 $\dim W<n$ 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 $$\ker(S-\lambda)\cap\operatorname{range}(S-\lambda)\subseteq \ker(T-\lambda)\cap\operatorname{range}(T-\lambda)=0.$$

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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.

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