Is $A A^\top + I$ invertible

Question

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.

Answer 1

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.

Answer 2

$AA^T+I$ 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 $AA^T+I$.