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I searched the http://math.stackexchange.com but most of the answer for this question use eigenvalues etc., but I need more elementary ways to show that why $\det(A A^T) \geq 0$ for any real matrix, and $A^T$ is the transpose of $A$. I could easily prove that when $m = n$. I need to show that when $m < n$ or $m > n$.

I guess also when $m > n$, then $\det(A A^T) = 0$. Is it right? If yes, how?

Thanks!

Jendrik Stelzner
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asad
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    What does your proof for $m=n$ look like? Note that $\det(AA^T) = 0$ iff $A^T$ has a nontrivial kernel (easy exercise). – Dominik Nov 21 '16 at 15:08
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    @Dominik, if $m=n$, we have square matrices and so the determinant is defined for both $A$ and $A^T$, and by a theorem in many of linear algebra textbooks we have $det A=det A^T$, therefore, $detAA^T=detA.det A^T=(det A)^2\geq0$. – asad Nov 21 '16 at 15:11
  • See e.g. last answer to http://math.stackexchange.com/questions/726351/is-determinant-of-matrix-multiplied-its-transpose-always-positive. $AA^\top$ is a Gram matrix and as such positive semi-definite. Thus, it has non-negative eigenvalues and a non-negative determinant. – Dr_Be Nov 21 '16 at 15:17
  • http://math.stackexchange.com/q/158219/9464 –  Nov 21 '16 at 15:18
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    Hint: Cauchy-Binet theorem. – darij grinberg Nov 21 '16 at 15:53
  • @asad how does the inequality show up though ? – Michael Walker Mar 17 '22 at 21:25

2 Answers2

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Without eigenvalues:

If $A^T$ has rank deficit, so does $AA^T$. Thus, in that case we have $\det(AA^T) = 0$. Now, assume $m \le n$ and rank of $A$ equals $m$. Let $$ A^T = Q \begin{bmatrix} R \\ 0 \end{bmatrix} $$ be a QR decomposition of $A^T$ (for example given by Gram–Schmidt), that is, $Q$ is square and orthogonal and $R$ is invertible and upper triangular. Then, we have $$ \det(AA^T) = \det(R^T R) = \det(R)^2 > 0. $$

Jendrik Stelzner
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user251257
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Why without using eigenvalues? For $A\in \mathbb{C}^{n\times m}$, $AA^H$ is a Hermitian $n\times n$ matrix. Thus, its eigenvectors form an orthonormal basis for $\mathbb{C}^n$ and their corresponding eigenvalues are real. Let $(\lambda,x)$ be an eigen-pair of $AA^H$, then $$AA^Hx=\lambda x \Longrightarrow 0\leq\|Ax\|_2^2=\lambda$$ where last equation by taking inner product with $x$. Since $\lambda$ was arbitrary eigenvalue, det$(A^HA)\geq 0$.

Ruzayqat
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