# eigenvalue of matrix $A$ given that $A$ multiplied by its transpose minus identity is negative definite

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.

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