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Question:

Suppose we already have m linearly independent eigenvectors $\mu_1, \mu_2, ... \mu_m$. Say $A\mu_i = \lambda\mu_i$ for $i = 1, 2, ..., m$ where $\lambda_1, \lambda_2, ..., \lambda_m$ are not necessarily distinct. For a new eigenvalue $\mu(\mu \ne \lambda_i \;~for~\; i = 1, 2, ..., m)$ of A, let $\{ v_1, v_2, ..., v_p\}$ be a basis for the eigenspace $E_{\mu}$.

Prove that $\{ \mu_1, \mu_2 ..., \mu_m, v_1, v_2, ..., v_p\}$ is linearly independent. (Hint: consider the vector equation

$a_1\mu_1 + a_2\mu_2 + ... + a_m\mu_m + b_1v_1 + b_2v_2 + ... + b_pv_p = 0$ --- (1)

By using the property of eigenvectors, show that

$a_1(\lambda_1-\mu)\mu_1 + a_2(\lambda_2-\mu)\mu_2 + ... + a_m(\lambda_m-\mu)\mu_m = 0$ ---(2)

I wonder how proving equation(2) in Hint help proving equation(1) in which $a_1, a_2, ..., a_m, b_1, b_2, ..., b_p$ are all 0?

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  • $\begingroup$ Perhaps you should not use $\mu$ to denote both eigenvectors and eigenvalues (just a suggestion). $\endgroup$ Apr 26 '14 at 7:34
  • $\begingroup$ I just copy the Question in book. In the book, it is using that notation, so I didn't change anything. should I change some notation to re-formulize this question? $\endgroup$ Apr 26 '14 at 9:47
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If you can prove equation (2), then you basically prove that $a_1=\cdots=a_m=0$.

Why?

Since $\{\mu_1,\ldots,\mu_m\}$ is linearly independent we must have $a_i(\lambda_i-\mu)=0$ for each $1\leq i \leq m$, and since $\lambda_i$ is distinct from $\mu$, we have $\lambda_i-\mu \neq 0$...it follows that $a_i=0$. This reduces equation (1) to $b_1v_1 + b_2v_2 + ... + b_pv_p = 0$ ... you can take it from here.

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