Let \(M_{n\times n}\) be the real vector space of \(n\times n\) real matrices, with \(n\ge2\), and let \[ \det:M_{n\times n}\to\RR \] be the determinant map.
- Show that \(\det\) is \(C^\infty\).
- Show that the derivative of \(\det\) at \(A\) is the zero linear map if and only if \(\rank A\le n-2\).
Solution
For \(A=(a_{ij})\), write \[ C_{ij}(A) \coloneqq (-1)^{i+j}\det A_{\widehat i,\widehat j} \] for the \((i,j)\)-cofactor of \(A\).
1. The map\[ \det:M_{n\times n}\longrightarrow\RR \]is \(C^\infty\).
In the matrix-entry coordinates, \[ \det(A) = \sum_{\sigma\in S_n} \sgn(\sigma) \prod_{i=1}^n a_{i,\sigma(i)}. \] Thus \(\det\) is a polynomial function on the real vector space \(M_{n\times n}\cong\RR^{n^2}\). Every polynomial function on a finite-dimensional real vector space is \(C^\infty\).
2. For every \(A,H\in M_{n\times n}\),\[ D(\det)_A(H) = \sum_{i,j=1}^n C_{ij}(A)H_{ij}. \]
Since \(\det\) is a polynomial in the entries, its differential is obtained from its coordinate partial derivatives. Laplace expansion along row \(i\) shows that \[ \frac{\partial\det}{\partial a_{ij}}(A)=C_{ij}(A). \] Therefore, for the direction \(H=(H_{ij})\), \[ D(\det)_A(H) = \sum_{i,j=1}^n \frac{\partial\det}{\partial a_{ij}}(A)H_{ij} = \sum_{i,j=1}^n C_{ij}(A)H_{ij}. \]
3. The differential \(D(\det)_A\) is the zero linear map if and only ifevery \((n-1)\times(n-1)\) minor of \(A\) is zero.
By step 2, if every cofactor \(C_{ij}(A)\) is zero, then \(D(\det)_A(H)=0\) for every \(H\). Conversely, if the differential is zero, apply it to the matrix unit \(E_{ij}\) to obtain \[ 0=D(\det)_A(E_{ij})=C_{ij}(A). \] Hence every cofactor vanishes. Cofactors differ only by signs from the \((n-1)\times(n-1)\) minors, so the two conditions are equivalent.
4. Every \((n-1)\times(n-1)\) minor of \(A\) vanishes if and only if\[ \rank A\le n-2. \]
The rank of a matrix over a field is the largest size of a nonzero minor. Thus a matrix has rank at least \(n-1\) exactly when it has a nonzero \((n-1)\times(n-1)\) minor. Negating this condition gives the displayed equivalence.
5. Therefore\[ \boxed{ D(\det)_A=0 \quad\Longleftrightarrow\quad \rank A\le n-2. } \]
Combine steps 3 and 4.
6. Q.E.D.
Step 1 proves part (1), and step 5 proves part (2).