I'm a little stuck on this problem. I know that since $A$ is symmetric, $A=A^{T}$. I'm also pretty sure that $AA^{T}$ is invertible. Therefore $A^2$ would be invertible. I'm not really sure how to account for the identity matrix (In).
Thanks in advance!
Best Answer
For an arbitrary real valued matrix $A$, $A^TA$ is positive-semidefinite, so all its eigenvalues are nonnegative. It follows that all the eigenvalues of $A^TA+I$ are no less than $1$, so $A^TA+I$ is invertible(and positive definite). In particular, when $A$ is symmetric, $A^2+I$ is invertible(and positive definite).
A more direct way is to show that for any nonzero column vector $v$, $(A^TA+I)v\ne 0$. To see this, note that for every column vector $w$, $w^Tw\ge 0$ and the equality holds if and only if $w=0$. Then $$v^T(A^TA+I)v=(Av)^T(Av)+v^Tv>0.\tag{1}$$ $(1)$ implies that $(A^TA+I)v\ne 0$, which completes the proof. In fact, $(1)$ also shows that $A^TA+I$ is positive definite.