Eigenvalues of matrix $A^TA+I$ are real and greater than 1? In this paper, the author states that the eigenvalues of the matrix $A^TA + I$ are real and greater than 1, since $A^TA$ is symmetric positive definite.
But why?
 A: Here's a useful fact: if $ x $ is an eigenvector of $ M $ with eigenvalue $\lambda$, then $ x $ is an eigenvector of $ M + c I $ with eigenvalue $\lambda + c $.
Proof: $ (M + c I) x = Mx + cx = (\lambda + c) x. $
(Similarly, if $ x $ is an eigenvector of $ M + c I $ with eigenvalue $\lambda $, then $ x $ is an eigenvector of $ M $ with eigenvalue $\lambda -c $.)
The matrix $ A^T A $ is positive definite, so its eigenvalues are real and strictly positive.
It follows that the eigenvalues of $ A^T A + I $ are strictly greater than $1$.
A: Well, since $A^T A$ is positive definite, it is diagonalisable with positive real eigenvalues. That is, if $A$ is $n \times n$, then for some $n \times n$ invertible matrix $P$, we have:
$$PA^TAP^{-1} = \left(\begin{matrix} \lambda_1 & 0 & \ldots & 0 \\ 0 & \lambda_2 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & \lambda_n \end{matrix}\right),$$
where $\lambda_1, \lambda_2, \ldots, \lambda_n > 0$ are the positive real eigenvalues of $A^T A$. But then,
\begin{align*}
P(A^TA + I)P^{-1} &= PA^TAP^{-1} + PIP^{-1} = PA^TAP^{-1} + I \\
&= \left(\begin{matrix} \lambda_1 + 1 & 0 & \ldots & 0 \\ 0 & \lambda_2 + 1 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & \lambda_n + 1 \end{matrix}\right)
\end{align*}
Therefore, the eigenvalues of $A^TA + I$ are just $\lambda_1 + 1, \lambda_2 + 1, \ldots, \lambda_n + 1 > 1$.
A: There is no need to take normal forms, you just have to check that we have positive definiteness by its definition. Let $M=A^T A+I$. Obviously $M=M^T$ and we have:
$$\forall v\neq 0,\qquad \langle v,Mv\rangle = \|v\|^2 +\|A v\|^2\color{red}{>}0 $$
as well as:
$$ \inf_{v:\|v\|=1}\langle v,Mv\rangle \geq 1.$$
