Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 2 / 10F / Solution

First $N(A)$ is finite: by Cauchy-Schwarz, $\Vert Ax{\Vert }_2\le ({\sum }_{ij}{a}_{ij}^2{)}^{1/2}\Vert x{\Vert }_2$. It is nonnegative and absolutely homogeneous. If $N(A)=0$, every $A{e}_j=0$, so every column is zero and $A=0$; the converse is immediate. The triangle inequality for the Euclidean norm gives $N(A+B)\le N(A)+N(B)$ after taking suprema. Thus $N$ is a norm.

For any vector $x$, rescaling gives $\Vert Ax{\Vert }_2\le N(A)\Vert x{\Vert }_2$. Applying this twice givesFurthermore $|a_{ij}|\le \Vert A{e}_j{\Vert }_2\le N(A)$, proving the requested strict entry bound when $N(A)<\epsilon$.

For a polynomial in matrix entries, expand each monomial at $A+H$. The terms involving exactly one entry of $H$ give a linear map $Df(A)[H]$, with coefficients polynomial in the entries of $A$. Every remaining term has at least two entries of $H$ and is $O(N(H{)}^2)$ for $N(H)\le 1$, because $|h_{ij}|\le N(H)$. This proves differentiability. Its derivative is continuous as an operator-valued function: each coefficient is continuous, and for $N(H)\le 1$ every $h_{ij}$ is bounded by one, so the operator-norm difference is bounded by the sum of the coefficient differences. Coordinate continuity follows from the same entry bound. Hence every such polynomial function is continuously differentiable.

In the Leibniz expansion of $\det (I+H)$, the identity permutation supplies $1+{\sum }_i{h}_{ii}$ plus terms of degree at least two. Any nonidentity permutation moves at least two indices, so each of its terms has at least two $H$ entries. ConsequentlyIf $A$ is invertible, multiplicativity of the determinant and the preceding expansion giveThe adjugate identity gives $\mathrm{adj}A=\det (A)A^{-1}$. Thus $d^{\prime }(A)[H]=\mathrm{tr}((\mathrm{adj}A)H)$ for invertible $A$.

For singular $A$, take invertible $A_r\to A$, using the permitted density result. Both $d^{\prime }$ and the adjugate entries are continuous polynomial expressions, so taking limits for every fixed $H$ proves the derivative of the determinant formula on all matrices:Here the standard adjugate is the transpose of the cofactor array, as required by $(\mathrm{adj}A)A=(\det A)I$. In components this derivative is ${\sum }_{ij}{\mathrm{cof}}_{ij}(A)h_{ij}$, also valid at singular matrices.