Is $A A^\top + I$ invertible Let $A$ be $n×n$ real matrix. Is $A A^\top + I$ invertible?
I know that $A^\text{*} A + I$ is invertible but don't know will it make any change for only transpose.
 A: $AA^T+I$ is a (real) symmetric matrix, so it is diagonalizable. Let $AA^T+I=PDP^{-1}$, where $D$ consists of the eigenvalues of $AA^T+I$, which are real. But eigenvalues of $AA^T+I$ is grater than or equals to $1$ as $AA^T$ is a positive symmetric matrix. Hence $D$ is invertible, and so does $AA^T+I$.
A: YES $(AA^T+I)$ is invertible, given $(A^TA+I)$ is invertible.
To prove this $C = (AA^T+I)$ is invertible -
1) C is square matrix $n * n$ (no issues).
2) We need to comment on the determinant of $C$ $\implies$. 
case 1: If $A$ is invertible $\implies$ $det(AA^T+I) = det(A^{-1}).det(AA^T+I).det(A)$ and by applying $det(AB) = det(A).det(B)$ we can say $det(AA^T+I) = det(A^TA+I)$ $\implies$ our $C$ is invertible.
case 2: More generally - consider we have two matrices X = \begin{matrix}
    I & -B\\
    A & I \\
    \end{matrix}
and matrix Y =  \begin{matrix}
    I & B\\
    0 & I \\
    \end{matrix}
so here if we see $$det(X).det(Y) == det(XY) == det(I+AB)$$ and $$det(Y).det(X) == det(YX) == det(I+BA)$$ $\therefore$ we can say that $$det(X).det(Y)==det(Y).det(X)==det(I+AB)==det(I+BA)$$ for the question asked we can just put $B = A^T$ $\therefore$ it is invertible.
