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:


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.


1 Answer 1


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 $K_\lambda \cap I_\lambda = 0$.

  • $\begingroup$ 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. $\endgroup$
    – dylan7
    Commented May 30, 2019 at 12:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .