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enter image description here $$$$The mentioned theorems are: Theorem 9.2 Eigenvectors corresponding to distinct eigenvalues are linearly independent. Theorem 9.3 A linear mapping $ f:V->V$ is diagonalisable if and only if V has basis consisting of eigenvectors of $f$. $$$$ $1.$ Is there a typo and $v_i\in B_i$ should be replaced by $v_i\in E_ {\lambda_ i}$? $$$$ $2.$ If I view the line $v_i=\mu_{i1}e_{i1}+...+\mu_{id_i}e_{id_i} \in B_i$ through matrices then the matrix represented by the the enries $\mu _{ij}$ can be transpose, whence there is a matrix multiplication formula hence $v_i$ is also an $nxn$ matrix. Is that the right way to see this? $$$$Thanks in advance for any help you are able to provide.

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  1. There is no typo. Every vector in $B_i$ is in $E_{\lambda_i}$ since $B_i$ is a basis for $E_{\lambda_i}$. You could replace $B_i$ by $E_{\lambda_i}$ in the sentence that you are asking about, and it would still read perfectly. However, the author then goes on to talk about the linear independence of $\cup B_i$. Linearly independence is a property of sets of vectors (the $B_i$) and not of vector subspaces (the $E_{\lambda_i}$).
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  • $\begingroup$ But then it suppose to be $v_i\in SpanB_i$ because $0$ can't be a basis, isn't it right? $\endgroup$ – proofy May 3 '13 at 15:53
  • $\begingroup$ The way I understand it is that each $v_i$ is an arbitrary basis vector from each $B_i$. However, having read on, the author writes $\mu_1e_1 + \cdots + \mu_de_d \in B$. This is impossible. If $B$ is a basis and the $\mu_k$ are freely chosen numbers then $B$ is an uncountably infinite vector space. I think the author may mean to say $\text{span}(B)$ when he writes $B$. $\endgroup$ – Fly by Night May 3 '13 at 16:00
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  1. I see this as a typo: It should be $v_i\in E_{\lambda_i}$ in this row and also in the second row $v_i=\sum \mu_{ij} e_{ij}\in E_{\lambda_i}$. In my opinion this way the proof makes more sense.
  2. I don't see how you would view $v_i$ as an $n\times n$ matrix. I would rather say that the vector $(\mu_{ij})_{j=1,\dots,d_i}$ is the coordinate vector of $v_i$ with respect to the basis $B_i$ of $E_{\lambda_i}$. Filling in zeroes, you could also see it as the coordinate vector of $v_i$ with respect to the basis $B$ of the span of all eigenspaces.
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