Condition for $\det(A^{T}A)=0$

Question

Is it always true that $\det(A^{T}A)=0$, $\hspace{0.5mm}$ for $A$ an $n \times m$ matrix with $n<m$?

From some notes I am reading on Regression analysis, and from some trials, it would appear this is true.

It is not a result I have seen, surprisingly.

Can anyone provide a proof?

Thanks.

Answer 1

From the way you wrote it, the product is size $m.$ However, the maximum rank is $n$ which is smaller. The matrix $A^T A$ being square and of non-maximal rank, it has determinant $0.$

Answer 2

hint

$\mathrm{Rank}(AB)\leq \min (\mathrm{Rank} A, \mathrm{Rank}B)$ Notice that $A^{\mathrm{T}}A$ is $m\times m$

Answer 3

$A^TA$ is an $m\times m$ matrix and has determinant 0 unless its rank is $m$.

However, the rank is the dimension of the image of $\mathbb R^m$ under the linear transformation defined by the matrix ... and this image is a subset of the image of $\mathbb R^n$ under the transformation defined by $A^T$. Since a linear transformation can never increase the dimension of the space it applies to, the rank is at most $n$, which is, by assumption, less than $m$.

Answer 4

A possible short-cut to prove this:

Let $$ \tilde{A}:=\begin{pmatrix}A\\0\end{pmatrix} $$ such that $A\in\mathbb{R}^{m\times m}$, then immediately you have $A^TA=\tilde{A}^T\tilde{A}$, but the latter clearly has determinant 0.

For a general result, consult: http://en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula