If $A$ is orthogonal. how do I show that $\det(A-2I)\not=0$.
I tried writing $A-2I=A-2AA^T=A(I-2A^T)=A(A^TA-2A^T)=AA^T(A-2I)$ but it seems that I am just doing loops after loops.
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2$\begingroup$ If $\det(A-2I)=0$, what can you say about the eigenvalues of A? $\endgroup$– user84413May 24, 2016 at 14:54
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$\begingroup$ $\lambda=2$ is an eigenvalue. $\endgroup$– AdityaMay 24, 2016 at 14:56
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$\begingroup$ That is right, so now see if you can show that A cannot have 2 as an eigenvalue. $\endgroup$– user84413May 24, 2016 at 15:03
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If $2$ is an eigenvalue, then $Ax = 2x$ for some $x\in \Bbb R^n \setminus \{0\}$. Then, $\|x\|^2 = (x,x) = (A^TAx, x) = (Ax, Ax) = (2x,2x) = 4(x,x) = 4\|x\|^2$. Contradiction.
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$\begingroup$ Also prove that any eigenvalue $\lambda$ of $A$ must salsify $|\lambda| = 1$ $\endgroup$– user258700May 24, 2016 at 15:06
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$\begingroup$ Why $(A^TAx, x) = (Ax, Ax)$? $\endgroup$– user261263May 24, 2016 at 15:07
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$\begingroup$ @EugenCovaci Of course I am assuming that the setting takes place in $\Bbb R^n$. We have for any matrix$A$ and vector $x$, $(Ax,x) = (x,A^Tx)$ $\endgroup$– user258700May 24, 2016 at 15:09