# Let $A$ be a real $n × n$ matrix whose columns are orthogonal with respect to the standard scalar product on $R^n$. Find $A^{−1}$.

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!

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 $\Lambda^{-1}A^T = A^{-1}$.
• "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 – user61369 Feb 7 '13 at 14:11
• @user61369: If a $\lambda_k=0$, then the corresponding column is $0$, which is trivially orthogonal to everything. – copper.hat Feb 7 '13 at 17:33