(Linear Algebra – Hoffman, Kunze, 2nd Ed., Sec 6.8, Q6)
Let $V$ be a finite-dimensional vector space over the field $\mathbb{F}$, and let $T$ be a linear operator on $V$ such that $\textrm{rank} (T) = 1$. Prove that either $T$ is diagonalizable or $T$ is nilpotent, not both.
Here's how I proceeded:
Let $\textrm{dim} V = n$. By the $\textrm{rank}(T) + \textrm{null}(T) = \textrm{dim} V$ theorem, $\textrm{null} (T) = n-1$. Now let $B = \{a_1, a_2,… , a_{n-1}, a_n\}$ be a basis for $V$, with $B' = \{a_2,…,a_n\}$ a basis for $\ker(T)$ .
Now, let $$A = [T]_B= \left(\begin{array}[cccc]
(c_1 & 0 & \cdots & 0\\
c_2 & 0 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
c_n & 0 & \cdots & 0 \end{array}\right).$$
Where $T(a_1) = c_1(a_1) + c_2(a_2) + … + c_n(a_n)$
Now I argue as follows: if $c_1 = 0$, then $A$ (and $T$) is nilpotent with $A^2 = 0$. Since the minimal polynomial is not a linear factor, $A$ (and $T$) is not diagonilazable.
On the other hand, if $c_1\neq 0$, then $A$ (and $T$) is diagonalizable, since the minimal polynomial = $x(x-c_1)$.
Is my solution correct?
Best Answer
Yes, your solution is completely correct. In a complete answer, you should show some computation verifying that $A^2 = 0$ when $c_1 = 0$.
Another approach is as follows: since the nullity of $T$ is $n-1$, we know that the Jordan form of $T$ has $n-1$ blocks associated with $\lambda = 0$. Either all blocks are size $1$ (giving us a diagonalizable matrix), or $1$ block is size $2$ (giving us a nilpotent matrix).
Yet another approach: $T$ must have the SVD $$ T = \sigma_1 uv^T $$ for column vectors $u,v$. Consider the cases $v^Tu = 0$ and $v^Tu \neq 0$.