# Using determinants, show that for all $n \times n$ symmetric matrices $A$, the matrix $A^2 + I_n$ is invertible.

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.

• How do you conclude $$\det (A^2 + I) = \det(CD^2C^{-1} + I) \stackrel{?}{=} \det(D^2 + I)$$ – AlexR Jul 23 '13 at 18:22
• Use the properties of Symmetric matrix. – Samrat Mukhopadhyay Jul 23 '13 at 18:24
• $det(A^2+I_n)=det(D^2+I_n)\neq0$ was given as a hint. – Courtney Jul 23 '13 at 18:45
• @Courtney: I think you want $C^TC = I$ in your question, not $A^TA = I$, which isn't necessarily true. – Robert Lewis Jul 23 '13 at 18:58

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.
• 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. – Robert Lewis Jul 23 '13 at 18:39
• Yes, true @RobertLewis. Actually in mind, I had $A$ Hermitian=D – Samrat Mukhopadhyay Jul 23 '13 at 18:42
• So, I presume the assertion is true even if $A$ is Hermitian, right? – Samrat Mukhopadhyay Jul 23 '13 at 18:43
• 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. – Robert Lewis Jul 23 '13 at 18:53
An easier hint: $x^T(A^2+I)x=\|Ax\|^2+\|x\|^2$.
• @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. – user1551 Jul 23 '13 at 21:44