If $A$ is a matrix with real entries, prove that $$\det(AA^T+I)\ge 1.$$
I tried using the eigenvalues. One thing came into my mind: maybe $AA^T$ is positive definite (I don't know whether this is true or not). However, I prefer a solution that does not use properties of positive definite matrix.
So I have 2 questions here:
Is the statement "$AA^T$ is positive definite" true?
Could you help me with a solution that does not use properties of positive definite matrix?
Thanks a lot.
$\require{begingroup} \begingroup$Note$\let\geq\geqslant\newcommand\norm[1]{\|#1\|}$ that $AA^T$ is symmetric (hence diagonalisable) and thus has real nonnegative eigenvalues. Indeed, if $v$ is a (non-zero) eigenvector corresponding to $\lambda$, then $$\lambda\norm v^2=\lambda v^Tv=v^TAA^Tv=\norm{A^Tv}^2\geq0\implies\lambda\geq0.$$
$AA^T+I$ is also symmetric and thus has real eigenvalues.
Furthermore, $\lambda$ is an eigenvalue of $AA^T$ iff $\lambda+1$ is an eigenvalue of $AA^T+I$:
$$AA^Tv=\lambda v\iff(AA^T+I)\cdot v=(\lambda+1)\cdot v.$$
Let $\lambda_1,\ldots,\lambda_n$ be the eigenvalues of $AA^T$ (with multiplicity, so they need not be different). Then $$\det(AA^T+I)=(\lambda_1+1)(\lambda_2+1)\cdots(\lambda_n+1)\geq1.\endgroup$$
Your idea that $AA^T$ is positive definite is a good idea. More precisely, you can show that $AA^T$ is positive semidefinite. This is easiest to show using definitions, because $M$ is positive semidefinite if for all $x$, you have
$$x^TMx \geq 0,$$ and in your case, this inequality is fairly simple to show.
Suppose $A$ has $n$ rows. Let $\mathfrak B$ be the set of all $n\times n$ submatrices of $\begin{bmatrix}A&I_n\end{bmatrix}$. By Cauchy-Binet formula, $$ \det(AA^T+I_n) =\det\left(\begin{bmatrix}A&I_n\end{bmatrix} \begin{bmatrix}A^T\\ I_n\end{bmatrix}\right) =\sum_{B\in\mathfrak B}\det(BB^T) =\sum_{B\in\mathfrak B}\det(B)^2 \ge\det(I_n)^2=1, $$ where the inequality above is due to the fact that $I_n\in\mathfrak B$.
This can also be obtained from the following general formula for the determinant of a sum of two $n\times n$ matrices: $$\det(M+N)=\sum_{p+q=n}\mathop{\mathrm{tr}}(M^{(p)}\mathbin{\square}N^{(q)})\tag{1}$$ Here $M^{(p)}$ and $N^{(q)}$ denote the $p$-th and $q$-th compound powers of $M$ and $N$ respectively, and $\square$ denotes the box product (see here for background).
Applying (1) to $AA^T$ and $I$, we obtain $$\det(AA^T+I)=\sum_{p=0}^n\mathop{\mathrm{tr}}[(AA^T)^{(p)}\mathbin{\square}I^{(n-p)}]\tag{2}$$ It's possible to show that $\mathop{\mathrm{tr}}[(AA^T)^{(p)}\mathbin{\square}I^{(n-p)}]=\mathop{\mathrm{tr}}[(AA^T)^{(p)}]$ (see link above) and $$(AA^T)^{(p)}=A^{(p)}(A^T)^{(p)}=A^{(p)}(A^{(p)})^T$$ so $\mathop{\mathrm{tr}}[(AA^T)^{(p)}]\ge 0$ and from (2) we have $$\det(AA^T+I)=1+\sum_{p=1}^n\mathop{\mathrm{tr}}[(AA^T)^{(p)}]\ge 1\tag{3}$$ Admittedly this is overkill. I'm more fond of the other slick Cauchy-Binet proof that was posted.
Note that $AA^T$ is positive-semi definite. Indeed, $<x,AA^Tx>=<A^Tx,A^Tx>=\|A^Tx\|^2\geq 0.$ Next it is easy to prove.
$AA^T$ is PSD, since it is written as a transpose-product of $A$. Hence $\forall i \in \{1,2,3,...,n\}, \lambda_i(AA^T) \geq 0$.
Note that $\lambda_i(AA^T+I_n) = \lambda_i(AA^T) + 1 \geq 1.$ Hence $\det(AA^T+I_n)=\prod_{i=1}^n (\lambda_i(AA^T) + 1) \geq 1.$
The first equality holds because $AA^T$ is symmetric hence has a full set of orthonormal eigenbasis by spectral theorem, and the same eigenbasis is shared by $I_n$.
