(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 $A = [T]_B$ = enter image description here 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 $v^Tu \neq 0$.

  • $\begingroup$ Thank you for the encouraging remark, and the suggested approaches! Question: what's the "SVD"? $\endgroup$ – Abdul3333 Dec 10 '14 at 19:17
  • 1
    $\begingroup$ @Abdul3333 Singolar Value Decomposition. $\endgroup$ – Bman72 Jan 5 '15 at 16:17
  • $\begingroup$ @Omnomnomnom How do you know there could be one Jordan block of size 2 associated with $\lambda =0$? $\endgroup$ – Sarah Aug 23 '16 at 1:27
  • $\begingroup$ @Sarah I don't quite understand what you mean to ask. Perhaps you should post your own question asking for clarification on this point. $\endgroup$ – Omnomnomnom Aug 23 '16 at 4:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.