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I have the following problem: I would like to show that

$A^TA < I$ implies that the eigenvalues of $A$ are less than $1$.

$A$ is a symmetric matrix and $I$ is the identity.

This claim is also made here: Identity minus a matrix times its transpose positive semidefinite But no justification is given.

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If $Av=\lambda v$ with $\|v\|=1$, $$ |\lambda|^2=\langle Av,Av\rangle=\langle A^TAv,v\rangle\leq\langle v,v\rangle=1. $$ You don't need $A$ symmetric for this to hold.

The claim in the answer you mention is a bit stronger, although the computation is the same. It says that $$ \|Ax\|^2=\langle Ax,Ax\rangle=\langle A^TAx,x\rangle\leq\langle x,x\rangle=\|x\|^2. $$ So $\|Ax\|\leq\|x\|$ for all $x$.

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Hint:

Let $A=UDU^T$ where $U$ is orthogonal and $D$ is a diagonal matrix, then we can write $UU^T=I$

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