Find out if $A^TA+I$ is a positive definite matrix.

Question

If we have a matrix $A \in \mathbb{R^{mxn}}$ and $I$ as the identity matrix of $n \times n$, with linearly independent columns, can we say that $A^TA+I$ is positive definite? I took an example case where $$A=\begin{pmatrix}1&2\\3&4\\5&6\end{pmatrix}$$ which means that $$A^TA+I= \begin{pmatrix}23&28\\49&64\end{pmatrix}$$ but then I get that $$\begin{vmatrix}23&28\\49&64\end{vmatrix} < 0$$ therefore its not positive definite. Is this correct?

Answer 1

1. $\quad \begin{vmatrix}23&28\\49&64\end{vmatrix} =100 >0.$

2. We have $x^TA^TAx \ge 0$ for all $x \in \mathbb R^n.$ This gives, with $B:=A^TA+I:$

$$x^TBx >0$$

for all $x \in \mathbb R^n$ with $x \ne 0.$ Hence $B$ is positive definite.