2

Proof from Axler

I have a few questions about the last part. How does applying the 8.11 make the rest of the terms 0?

and for the next equality, it looks like we move (T - $\lambda_1$I)$^k$ to operate on $v_1$ first. How can we do that without knowing all terms commute with each other?

Harry
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  • I don't understand your first question. However, $(T-\lambda_1)^k$ can pass to the right of $(T-\lambda_1)^k (T-\lambda_2)^n \cdots (T-\lambda_m)^n$ because $\lambda_iI$ is just a diagonal matrix with the entry $\lambda_i$ along the diagonal entries, and $T$ commutes with sums and powers of $\lambda_iI$ and itself $T$. – Mee Seong Im Jul 07 '17 at 23:02
  • What does $G(\lambda,T)$ denote? – Bernard Jul 07 '17 at 23:03
  • it means the generalized eigenspace with $\lambda$ as eigenvalue – Harry Jul 07 '17 at 23:07
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    The page shown in the question above comes from my book Linear Algebra Done Right, third edition (http://www.linear.axler.net). The online errata for the book shows that there is a typo in the displayed equation three lines above 8.15: the second "$w$" should be "$v_1$". – Sheldon Axler Jul 08 '17 at 00:50

2 Answers2

3

They do commute: $$(T-\lambda_i I)(T - \lambda_j I) = T^2 - (\lambda_i + \lambda_j) T + \lambda_i \lambda_j I = (T-\lambda_j I) (T - \lambda_i I).$$ This answers your second question.

This also answers your first question, because for $i > 1$, you can commute the terms in the operator to obtain $$ (T-\lambda_1)^k (T-\lambda_2 I)^n \cdots (T-\lambda_m I)^n= (\text{other operators}) (T - \lambda_i I)^n v_i = 0.$$

angryavian
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0

We have $$ (T- xI)(T-yI) = T^2-(x+y)T+xyI = (T-yI)(T-xI).$$

This can be generalized to show that all the factors commute with one another.

spaceisdarkgreen
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