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To prove that for any $n \times n$ matrix there are $n$ linearly independent eigen vectors if all the eigen values are distinct, I see that in a book it starts with the following matrix

$$ \textbf{Z}=(\alpha_2\textbf{I-A})\dots(\alpha_n\textbf{I-A})$$

where $\alpha_i; i=1,\dots,n$ are the eigen values. I did not understand the reason behind this. It looks like a trick.

Actually, the quantities $\textbf{Z$v_k$=0};k=2,3,\dots,n$ (which means $v_k;k=2,3,\dots,n$ are in the kernel space of $Z$) where $v_k$s are eigen vectors whereas $\textbf{Z$v_1$}$ is a scalar multiple of $v_1$ (which means $v_1$ is an eigen vector of $Z$). So, the matrix $\textbf{Z}$ has been chosen like this. I am trying to understand the meaning of these things.

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  • $\begingroup$ what is the complete flow of proof in that book? $\endgroup$ – al-Hwarizmi Aug 25 '13 at 7:44
  • $\begingroup$ Are you sure the indices in the product start at $2$ (so there is not $\alpha_1$ in the formula)? $\endgroup$ – Marc van Leeuwen Aug 25 '13 at 9:13
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For every eigenvalue $\alpha$, the matrix $\alpha I - A$ contains the respective eigenvectors of $A$ in its kernel:

$$ Av = \alpha v \qquad \Longleftrightarrow \qquad (\alpha I - A) v = 0 \ . $$

So, I guess your book could proceed like this: let $v_1, \dots , v_n$ be eigenvectors of $\alpha_1, \dots , \alpha_n$ eigenvalues. Assume there is a linear combination such that

$$ \lambda_1 v_1 + \dots + \lambda_n v_n = 0 \ . $$

Applying matrix $Z$ to both sides of this equality, we get

$$ \lambda_1 (\alpha_2 - \alpha_1) \dots (\alpha_n - \alpha_1) v_1 = 0 \ . $$

Since $v_1 \neq 0$, by definition of eigenvector, and all the eigenvalues $\alpha_i$ are different, it follows that

$$ \lambda_1 = 0 \ . $$

Applying induction, or changing matrix $Z$ suitably, you would also get that all coeffitients $\lambda_i$ are necessarily zero. Hence vectors $v_1, \dots , v_n$ are linearly independent.

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  • $\begingroup$ Roig: yes, this is the solution. I was trying to understand the intuition behind this trick of choosing $Z$ using the fact that it is some kind of linear map. $\endgroup$ – RIchard Williams Aug 25 '13 at 9:39

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