for $A$ $n\times n$ real matrix. $B=A^t A$ Let $A$ be a $n\times n$ matrix with real entries. Let $B=A^t A $(where $A^t$ denotes the  transpose of $A$) then which of following is true.


*

*If $B$ is invertible then $A$ is invertible

*If $k$ is eigen value of $B$ then $k\ge 0$

*If $k$ is eigen value of $A$ then $k^2$ is eigen value of $B$

*Let $C=I + B$, then $C$ is invertible.
My attempt: I think $B$ is real symmetric matrix. And since determinant of $B$ is square of $\det(A)$ so option 1 is correct. I need suggestion to see other options. Thanks. Correct options are 1,2,4.

Answer/Hint
You're right for 1. since
$$\det(B)=\det(A^t)\det(A)=(\det(A))^2$$
For 2. let $x$ an eigenvector for $B$ associated to $k$ then
$$k||x||^2=\langle A^tAx,x\rangle=||Ax||^2\implies k\ge0$$
For 3. Take $A$ an orthogonal matrix that's $A^tA=I$ to get a counterexample.
Show the result of 4.  using 2. and the fact that every symmetric real matrix is diagonalizable.
