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From Alan J Laub, Matrix Analysis For Scientists and Engineers, 2004, p 79

Theorem 9.15. Let $A \in \mathbb{C}^{n\times n} $ have distinct eigenvalues $\lambda_1, \ldots, \lambda_n$ and let the corresponding right eigenvectors form a matrix $X = [x_1, \ldots ,x_n]$. Similarly, let $Y = [y_1, \ldots ,y_n]$ be the matrix of corresponding left eigenvectors. Furthermore, suppose the left and right eigenvectors have been normalized so that $y_i^Hx_i=1, i \in \{1,\ldots,n\}$. Finally, let $\Lambda = \text{diag}(\lambda_1,\ldots,\lambda_n) \in \mathbb{R}^{n\times n}$ ...

The theorem goes on, but I am wondering about the line where he lets $\Lambda = \text{diag}(\lambda_1,\ldots,\lambda_n) \in \mathbb{R}^{n\times n}$. In general, matrices over the complex numbers with distinct eigenvalues need not have real eigenvalues. How then can we assume that $\Lambda \in \mathbb{R}^{n\times n}$?

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  • $\begingroup$ Could very well be a typo, impossible to say without the complete statement of the theorem. $\endgroup$ – daw Oct 19 '18 at 6:03
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In a theorem, we can impose conditions.

We can use the theorem when the condition holds. The conclusion need or need not hold when the assumption is not true.

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  • $\begingroup$ So you think the eigenvalues are real by assumption? Definitely possible, but it just strikes me as an odd placement for such an assumption in the statement of the theorem. If he were assuming real eigenvalues, why not say so when defining them initially? "Let A in C^nxn have distinct real eigenvalues..." $\endgroup$ – Chad Oct 19 '18 at 4:32
  • $\begingroup$ I do not have access to the book, you might like to read the proof and see if we need the assumption. $\endgroup$ – Siong Thye Goh Oct 19 '18 at 4:33

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