Let $T\in\mathcal L(V)$ and $T^{m-1}v\neq 0$ but $T^mv=0$ for some positive integer $m$ and some $v\in V$. Show that $v,Tv,\ldots,T^{m-1}v$ are linearly independent.

I had written a proof but Im not sure if it is correct. And in the case it would be correct I dont know how to write it better and clearly. So I have two questions:

  1. It is the proof below correct?

  2. If so, how I can write it better using the same ideas?

The attempted proof:

1) If $T^{m-1} v$ would be linearly dependent of $T^{m-2} v$ then exists some $\lambda\neq 0$ such that

$$T^{m-1}v=\lambda T^{m-2}v\implies T^mv=\lambda T(T^{m-2}v)=\lambda T^{m-1}v=0\implies \lambda=0$$

Then $T^{m-2}v$ is linearly independent of $T^{m-1}v$.

2) Now observe that

$$\lambda_1v+\lambda_2Tv+\lambda_3T^2v=0\implies T^{m-2}(\lambda_1v+\lambda_2Tv+\lambda_3T^2v)=\lambda_1T^{m-2}v+\lambda_2T^{m-1}v=0$$

so $\lambda_1,\lambda_2=0$ as we had shown previously, so the original equation reduces to

$$\lambda_3T^2v=0\implies \lambda_3=0$$

thus $v,Tv,T^2v$ are linearly independent.

3) Repeating recursively the analysis in 2) for longer lists of vectors of the form $v,Tv,\ldots,T^kv$ for $k< m$ we can show that the list $v,Tv,\ldots,T^{m-1}v$ is linearly independent.

  • $\begingroup$ This is a related post $\endgroup$ – Bijesh K.S Jul 11 '17 at 19:15
  • $\begingroup$ The if and only if in the first line looks problematic as $T$ isn't invertible. $\endgroup$ – DMath Jul 11 '17 at 19:19
  • $\begingroup$ @DMath thank you... I overlook it completely. $\endgroup$ – Masacroso Jul 11 '17 at 19:20
  • 1
    $\begingroup$ I'm afraid that the last part is the bulk of the proof and you just cut it off by saying “repeating the analysis”. Note that for $\{v,Tv,T^2v,T^3v\}$ you need to go down to $T^{m-3}$, which is not as easy as the initial step for $\{T^{m-2}v,T^{m-1}v\}$. $\endgroup$ – egreg Jul 11 '17 at 20:33


$$a_0v+a_1Tv+\ldots+a_{m-1}T^{m-1}v=0 \quad(1)$$

multiply by $T^{m-1}$, then

$$a_0T^{m-1}v=0\to a_0=0$$

so, from $(1)$,


multiply by $T^{m-2}$, then

$$a_1T^{m-1}v=0\to a_1=0$$

can you finish?

  • $\begingroup$ ok but, it is my proof correct? $\endgroup$ – Masacroso Jul 11 '17 at 19:17
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    $\begingroup$ @Masacroso: it seems correct to me. You just need induction to finish it. $\endgroup$ – Arnaldo Jul 11 '17 at 19:19

See below for a simple proof.

Suppose that they are not linear independent; that is for some choice of $\alpha_{k}$,

$T^{m-1}v = \sum_{k=0}^{m-2}\alpha_{k}T^{k}v$.

Then $0 =T^{m}v = T\cdot T^{m-1}v = \sum_{k=0}^{m-2}\alpha_{k}T^{k+1}v$,

so $Tv,...,T^{m-1}v$ are linearly dependent.

But that means that $T^{m-1}v = \sum_{k=1}^{m-2}\beta_{k}T^{k}v$.

Inductively repeat the argument to eventually conclude that for some $\gamma \neq 0$, $T^{m-1}v = \gamma T^{m-2} v$, but then we would have to say that $0 = T T^{m-1} v = \gamma T^{m-1} v \neq 0$, a contradiction.


More simply put: if you had a linear combination $c_0 v + c_1 T v + \ldots c_{m-1} T^{m-1} v = 0$ with $c_j$ not all zero, let $c_i$ be the first nonzero coefficient. Then $$T^i v = - \sum_{j=i+1}^{m-1} (c_j/c_i) T^j v$$ Applying $T^{m-i-1}$ to both sides, $$T^{m-1} v = - \sum_{j=i+1}^{m-1} (c_j/c_i) T^{m+j-i-1} v = 0$$ But all terms on the right are $0$, so $T^{m-1} v = 0$, contradiction!


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