# Prove that Either $T$ Is Diagonalizable or $T$ Is Nilpotent.

(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 $rank \ (T) = 1$. Prove that either $T$ is diagonalizable or $T$ is nilpotent, not both.

Here's how I proceeded: Let $dim\ V = n$. By the $rank \ (T) + nullity \ (T) = dim \ V$ theorem, $Nullity \ (T) = n-1$. 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 $Nullspace \ (T)$ . Now, let $A = [T]_B$ = 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?

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 $v^Tu \neq 0$.
• @Omnomnomnom How do you know there could be one Jordan block of size 2 associated with $\lambda =0$? – Sarah Aug 23 '16 at 1:27