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I don't really understand what it means by having columns that are orthogonal with respect to the standard scalar product. Does that mean:

If $v_1$ and $v_2$ are some of the columns of $A$, then $v_1 . v_2 = 0$ ?

If not, what does it mean? If so, how does that help me in finding $A^{-1}$ ?

Thank you!

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You have $(A e_i)^TA e_j = \delta_{ij} \lambda_k$, or in other words $e_i^T A^TA e_j = \delta_{ij} \lambda_k$, from which it follows that $A^T A = \Lambda$, where $\Lambda$ is a diagonal matrix with entries $\lambda_k$. Assuming that $\lambda_k \neq 0 $ for all $k$, this gives $(\Lambda^{-1}A^T) A = I $, from which it follows that $\Lambda^{-1}A^T = A^{-1}$.

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  • $\begingroup$ "Assuming that $\lambda_k \neq 0 $ for all $k$" - does this not follow from orthogonality? If a $\lambda_k $ was 0 then the columns would not be orthogonal $\endgroup$ – user61369 Feb 7 '13 at 14:11
  • $\begingroup$ @user61369: I have converted your answer to a comment. Answers should be reserved for posts that answer the question. But because you do not have 50 reputation points yet, you can only comment on your own questions and answers. So, you didn't do anything wrong; the "add comment" button will only appear for you once you gain 50 points. Here is an explanation of reputation points. $\endgroup$ – Zev Chonoles Feb 7 '13 at 14:13
  • $\begingroup$ @user61369: If a $\lambda_k=0$, then the corresponding column is $0$, which is trivially orthogonal to everything. $\endgroup$ – copper.hat Feb 7 '13 at 17:33

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