I had a question about Sheldon Axler's proof of Theorem 5.21 in his book, Linear Algebra Done Right.
I checked this previous post, and had no trouble understanding the substitution part where $T$ is substituted for z.
This picture is from an older edition of the textbook, but it's the same proof.
My question is:
In the second line, Axler appears to choose any arbitrary non-zero vector $ v \in V $, and shows that at least one $(T – \lambda_j I) $ satisfies $ (T – \lambda_j I) v = \vec{0} $.
In other words, this proof appears to argue that for all non-zero vector $ v \in V $, there is at least one $(T – \lambda_j I) $ satisfies $ (T – \lambda_j I) v = \vec{0} $.
But this is clearly not true, as there are plenty of counterexamples even on $ C^2$.
Say, $ T(x,y) = [(3+5i)x, (1+2i)y] $, and choose $ v = ( 1+ i, 1-i) $. Then there is no single eigenvalue for that particular combination.
I know this proof is about the **existence ** of a particular (eigenvalue, eigenvector) combination, and not a proof about an eigenvalue existing for all arbitrary eigenvectors.
But the second line of the proof appears to claim otherwise.
Best Answer
You cannot conclude that $v$ is an eigenvector and indeed the author doesn't state it.
You can recover an eigenvector, though. For simplicity, let $S_k=T-\lambda_kI$ and note that these operators commute with each other. Now consider $$ S_1v,\quad S_2S_1v,\quad \dots,\quad S_{m-1}\dotsm S_2S_1v,\quad S_m\dotsm S_2S_1v=0 $$ Let $r$ be the first index such that $S_r\dotsm S_2S_1v=0$. If $r=1$, you're done. Otherwise $w=S_{r-1}\dotsm S_1v\ne0$ but $S_rw=0$. Hence $w$ is an eigenvalue relative to $\lambda_r$.
With your example, $$ Tv=(-2+8i,3+i),\qquad T^2v=(-46+14i,1+7i) $$ and it turns out that $$ T^2v=(7-11i)v+(4+7i)Tv $$ so the polynomial is $$ x^2-(4+7i)x-(7-11i)=(x-(3+5i))(x-(1+2i)) $$ Now $$ (T-(1+2i)I)v=Tv-(1+2i)v=(-2+8i,3+i)-(-1+3i,3+i)=(-1+5i,0) $$ which is clearly an eigenvector relative to $3+5i$.