3

Let $A \in M_{m \times n}({\Bbb F})$, prove $$ \det\left(I_m + AA^t\right) = \det\left(I_n + A^tA\right) $$

I don't need the full answer, maybe a hint. I've tried using Sylvester's identity but I can't solve either way. I tried to replicate the proof from here, but I wondered if there were other proofs, that is why I asked for a hint, this is not homework.

Rodrigo de Azevedo
  • 1
ipreferpi
  • 476

2 Answers2

3

Here's a sketch. Assume without loss of generality that $m < n$. Writing $A = A' \oplus 0$ (over some basis) with $A'\in M_{m\times m}$, reduce to the case of $m = n$. If $A$ is invertible, then $$\det (1 + AA^t) = \det A(A^{-1} + A^t) = \det(A^t) \det ((A^{-1})^t + A)^t = \det(1 + A^t A).$$ But the given equation is a polynomial in $A_{ij}$, and the space of invertible $A$ is dense.

anomaly
  • 25,364
  • 5
  • 44
  • 85
  • I didn't understand the last line, why worry to say such $A$ is dense? – ipreferpi May 09 '19 at 23:54
  • 1
    We assumed that $A$ is invertible midway through the proof sketch. The displayed equation doesn't make sense otherwise; but the result holds for general $A$ if it holds for invertible $A$ by continuity. – anomaly May 09 '19 at 23:55
3

For square matrices, take determinant on both sides of $$\begin{pmatrix}I&0\\A^T&I\end{pmatrix}\begin{pmatrix}I+AA^T&A\\0&I\end{pmatrix}\begin{pmatrix}I&0\\-A^T&I\end{pmatrix}=\begin{pmatrix}I&A\\0&I+A^TA\end{pmatrix}$$ For non-square matrices you can pad it with zeros to a square matrix and use the formula above.

logarithm
  • 401
  • Good solution, but it might be worth to mention that the block triangular determinants are not entirely trivial (except when the matrix is indeed triangular). See https://math.stackexchange.com/questions/75293/determinant-of-a-block-lower-triangular-matrix – Jean-Claude Arbaut May 19 '19 at 17:31