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For reference, Axler's proof was copied in full in this question. In the beginning of that proof, Axler chooses the nonzero vector $v$ without loss of generality. This made me doubt his proof because it is obvious that not all nonzero vectors are necessarily eigenvectors. So I looked for something that would help me to distinguish $v$ from other arbitrary vectors, and I realized that for it to be true that $(v,Tv,\dots,T^nv)$ is linearly dependent, each $T^i$ for $0<i\leq n$ must be nonzero. Thus I am left thinking that Axler's proof must show that such a $v$ can definitely be chosen. Was this step left out of the proof because the existence of such a $v$ trivial in some manner that I am not noticing? Or is my reasoning about the proof incorrect?

Community
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TooWarmInFebruary
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  • If $T^i = 0$, then the vectors $T^i v,, T^{i+1}v,,\dotsc,, T^n v$ are all $0$, then the system is surely linearly dependent too. – Daniel Fischer Mar 02 '14 at 22:11
  • If those are all zero, then the set of $T^jv$ for $0\leq j<i$ might be linearly independent, right? Then adding in those zero vectors shouldn't make it linearly dependent. – TooWarmInFebruary Mar 02 '14 at 22:33
  • Every set of vectors containing a zero is linearly dependent. You then have the nontrivial relation $\vec{0} = 0\cdot v_1 + \dotsc + 0\cdot v_j + 1\cdot \vec{0} + 0\cdot v_{j+2} + \dotsc + 0\cdot v_k$. – Daniel Fischer Mar 02 '14 at 22:37

2 Answers2

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I believe the point of confusion is the assumption

$$0=c(T-\lambda_1 I)\cdots (T- \lambda_m I)v$$

means $(T-\lambda_k I)v = 0$ for some $v$. Given this assumption, it would indeed seem to be the case that $v$ must be an eigenvector for the eigenvalue $\lambda_k$.

However, recall the way multiplication of operators is defined here: The multiplication of the operators $T * T = T^2$ means $(TT)v = T(Tv)$, not $(TT)v = (Tv)(Tv)$, as we can't multiply two vectors in a vector space, only a vector with a scalar. Function multiplication is defined as function composition.

That being said, the main equation is saying we plug $v$ into the operator $T-\lambda_m I$, then take the output and plug it into $T-\lambda_{m-1}I$, etc. until we finally plug the output into $T - \lambda_1 I$ and get $0$. Since we start with a nonzero input and get a zero output, there is some $\lambda_k$ such that $T - \lambda_k I$ takes in a nonzero input and produces a zero output, hence $T - \lambda_k I$ is not invertible and $\lambda_k$ is an eigenvalue.

We cannot definitively conclude anything about the starting vector $v$, since it may "disappear" after the first plugin into $T- \lambda_m I$. For example, say $(T-\lambda_m I)v$ is equal to an eigenvector for the eigenvalue $\lambda_{m-1}$, then when we plug in this eigenvector into the next operator $T - \lambda_{m-1} I$, we get $0$. Note that we can't conclude $v$ is an eigenvector.

BobaFett101
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  • I just want to say that this is a fantastic answer that helped me immensely. Thanks! – EE18 Aug 12 '20 at 16:21
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Well, the proof doesn't claim that $v$ is an eigenvector, so there's no problem if it isn't.

The vectors $(v,Tv,\dots,T^nv)$ are linearly dependent regardless of the values of $v$ and $T$. It doesn't matter whether $v$ is nonzero. It doesn't matter whether $T^i$ is nonzero. It doesn't even matter that $T$ is a linear operator. Any set of $n+1$ vectors is linearly dependent.

Chris Culter
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