# Special case of Complex Vector Space diagonalizable condition

I am trying to look into a special case of the diagonalizable conditions mentioned here and here to get a better understanding of the proof for the general condition. I was wondering if my approach is correct, as there is no hint mentioned in these posts linked, without Jordan Normal Form.

Suppose $$V$$ is a finite-dimensional complex vector space and $$T \in \mathcal{L}(V).$$ If $$T$$ has one distinct eigenvalue and $$V = null(T - \psi I) \oplus range(T- \psi I),$$ $$\forall \psi \in \mathbb{C}$$, then $$T$$ is diagonalizable.

Let $$\lambda$$ denote the distinct eigenvalue of $$T$$. Assume, for contradiction, $$T$$ were not diagonalizable, and thus $$range(T-\lambda I) \neq \{0\}$$

Let $$u_1, ..., u_m$$ constitue a basis for $$null(T-\lambda I)$$ and are thus all eigenvectors with eigenvalue $$\lambda$$, but are not a basis for $$V,$$ since $$range(T-\lambda I) \neq \{0\}$$

Extend $$u_1,...u_m$$ to a basis for $$V,$$ $$u_1,...,u_m,v_1,...v_n$$, where $$v_1,...v_n$$ are not eigenvectors of $$V.$$ Otherwise a contradiction would arise that either $$range(T- \lambda I)$$ did just contain $$0$$ or that $$\lambda$$ wasn't the only distinct eigenvalue of $$T$$.

Thus we now know, $$(T- \lambda I)|_{span(v_1, ..., v_n)}$$ is invertible, and in fact $$T$$ is invertible over this space, since otherwise $$span(v_1, ...,v_n)$$ would contain eigenvectors with eigenvalue $$0$$. And appending one of the $$u's$$ to this basis, WLOG, $$u_1$$, if $$\lambda \neq 0$$, $$T|_{span(u_1, v_1,...,v_n)}$$ is invertible. And denote: $$U = span(u_1, v_1, ... v_n).$$

Since $$T$$ contains an eigenvalue, $$\lambda$$, over $$U$$ (eigenvector $$u_1$$), there exists a basis $$u_1, v'_1, ... v'_n$$, over which $$T|_U$$ has an upper-triangular matrix: $$\mathcal{M}(T|_U)$$.

If $$T|_U$$ is invertible then all the diagonal elements of $$\mathcal{M}(T|_U)$$ are non-zero, and thus there exists other $$\psi$$ such that $$(\mathcal{M}(T|_U) - \psi I)$$ is not invertible. Since $$\lambda$$ is the only distinct eigenvalue of $$T$$, $$\lambda = \psi$$, and $$v'_1,...,v'_n$$ are eigenvectors of $$T$$ and a contradiction arises concluding that $$range(T - \lambda I)$$ does equal $$\{0\}$$.

To handle the case that $$\lambda = 0$$, then $$T$$ is not invertible over $$U$$, and thus $$\mathcal{M}(T|_U)$$ contains a zero diagonal element. Considering $$u_1, v'_1, ..., v'_n$$ as the ordered basis on which $$\mathcal{M}(T|_U)$$ is diagonal, then $$u_1$$ has eigenvalue $$0$$; so, since $$\mathcal{M}(T|_U)$$ is upper triangular:

Either:

1) $$v'_1$$ is an eigenvector, in which case it must have an eigenvalue, by construction, that is not $$0$$, a contradiction. 2) For $$\nu_1,\nu_2 \in \mathbb{C}, v'_1= \nu_1 * v'_1 + \nu_2 * u_1$$. If $$\nu_1 \neq 0$$, there exists an eigenvalue that is non zero, a contradiction. If $$\nu_1 = 0$$, then $$u_1 \in range(T - \lambda I)$$, a contradiction.

Since we have failed to show $$V$$ does not have a basis of eigenvectors, $$range(T - \lambda I) = \{0\}$$ and $$V = null(T - \lambda I)$$ and $$T$$ is diagonalizable. $$\Box$$

Please give me hints as to how I can correct this, if it is wrong. Or let me know if there are any gaps in my understanding. Or even if there is a way to shorten this, if it is correct. I tried multiple other approaches that I was able to catch an error on.

Suppose $$T: V \to V$$ is a linear map, and that $$\lambda \in \mathbb{C}$$ is any scalar. Now, let $$K_\lambda = \ker(T - \lambda)$$ and $$I_\lambda = \operatorname{Im}(T - \lambda)$$.
Since $$K_\lambda$$ is precisely the $$\lambda$$-eigenspace of $$T$$, we have $$T(K_\lambda) \subseteq K_\lambda$$, and since $$T(T - \lambda)v = (T - \lambda)Tv$$, we also have that $$T(I_\lambda) \subseteq I_\lambda$$. Hence $$T$$ restricts to a linear map $$T|_{I_\lambda}: I_\lambda \to I_\lambda$$. Note that $$K_\lambda \cap I_\lambda$$ is precisely the $$\lambda$$-eigenspace of the operator $$T|_{I_\lambda}$$.
Now, let's add in the rest of the assumptions in the question. Suppose that $$\lambda$$ is the only eigenvalue of $$T$$, and that $$V = K_\lambda \oplus I_\lambda$$, and suppose towards a contradiction that $$I_\lambda \neq 0$$. Since $$I_\lambda \neq 0$$ and $$\mathbb{C}$$ is algebraically closed, $$T|_{I_\lambda}$$ has some eigenvector. Since $$\lambda$$ is the only eigenvalue of $$T$$, $$\lambda$$ must also be the only eigenvalue of $$T|_{I_\lambda}$$, and so $$T|_{I_\lambda}$$ must have a nonzero $$\lambda$$-eigenspace. However, the $$\lambda$$-eigenspace of $$T|_{I_\lambda}$$ is $$K_\lambda \cap I_\lambda = 0$$.
• Ah ok, so I guess my proof became long since I formed the subspace $U$, as defined above (which lead to the whole upper-triangular argument), when I could of just said $T|_span(v_1, ..., v_n)$ has an eigenvector. – dylan7 May 30 at 12:51