Hint given: Look at the determinants of $A$ and some diagonal matrix $D$.
So far I have:
$A$ is symmetric, therefore $A$ is diagonalizable, thus $A = CDC^{-1}$ for some invertible matrix $A$ and some diagonal matrix $D$.
I know that $\det(A^2+I_n)=\det(D^2+I_n)\neq 0$ and that by definition, when $det \neq 0$ the matrix is invertible.
I don't know if the fact that $A^TA=I_n$ is relevant. I know that the eigenvalues of $I_n$ will be real and positive.
Am I on the right track? I'm not sure how to smash all this together into a proof.
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1$\begingroup$ How do you conclude $$\det (A^2 + I) = \det(CD^2C^{-1} + I) \stackrel{?}{=} \det(D^2 + I)$$ $\endgroup$– AlexRJul 23, 2013 at 18:22
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$\begingroup$ Use the properties of Symmetric matrix. $\endgroup$– Samrat MukhopadhyayJul 23, 2013 at 18:24
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$\begingroup$ $det(A^2+I_n)=det(D^2+I_n)\neq0$ was given as a hint. $\endgroup$– CourtneyJul 23, 2013 at 18:45
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1$\begingroup$ @Courtney: I think you want $C^TC = I$ in your question, not $A^TA = I$, which isn't necessarily true. $\endgroup$– Robert LewisJul 23, 2013 at 18:58
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2 Answers
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Since $A$ is symmetric, $$A=U^T\Lambda U$$ for some unitary matrix $U$ and a diagonal matrix $A$ with eigenvalues of $\Lambda$ in the diagonal. Also, the eigenvalues of $A$ are real. So $$A^2+I_n=U^T\Lambda^2U+I_n=U^T(\Lambda^2+I_n)U$$ is also symmetric with eigenvalues the diagonal elements of $\Lambda^2+I_n$ which are all $\ge 1$. So $\det(A^2+I_n)=\det(\Lambda^2+I_n)>0$ Hence $A^2+I_n$ is invertible.
answered Jul 23, 2013 at 18:31
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1$\begingroup$ Since $A$ is termed symmetric, it is presumably real, since the term Hermitian is usually used in the complex case. Thus $U$ need only be orthogonal, $U^TU = I$, not fully unitary. But the proof looks fine. Picking nits. Cheers. $\endgroup$ Jul 23, 2013 at 18:39
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$\begingroup$ Yes, true @RobertLewis. Actually in mind, I had $A$ Hermitian=D $\endgroup$ Jul 23, 2013 at 18:42
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$\begingroup$ So, I presume the assertion is true even if $A$ is Hermitian, right? $\endgroup$ Jul 23, 2013 at 18:43
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$\begingroup$ A Hermitian matrix is diagonalizable via a unitary matrix $U$, which as such satisfies $U'U = I$ (here I am using the sysmbol ' in place of the usual dagger for Hermitian adjoint, since I cannot find the dagger on my keyboard!). D = $U'AU$ will be real, since A is Hermitian. So the proof flies in this case as well. $\endgroup$ Jul 23, 2013 at 18:53
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An easier hint: $x^T(A^2+I)x=\|Ax\|^2+\|x\|^2$.
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$\begingroup$ @Courtney By the way, your proof looks fine to me, but as Robert Lewis comments, your claim that $A^TA=I$ is not necessarily true, but this claim is irrelevant to your proof anyway. $\endgroup$– user1551Jul 23, 2013 at 21:44