Prove the formula for det(I + xA)

Question

How do I show that the determinant and cofactors$(B_{kl})$ of matrix $B$ are given as follows? $$ B=I+xA=\left(\begin{array}{cccc} 1+a_{11}x & a_{12}x & \cdots & a_{1n}x \\\\ a_{21}x & 1+a_{22}x & \cdots & a_{2n}x \\\\ \vdots & \vdots & \ddots & \vdots \\\\ a_{n1}x & a_{n2}x & \cdots & 1+a_{nn}x \end{array}\right) $$ $$ |B|=1+x\sum_{j}a_{jj}+\sum_{m=2}^{n}\frac{x^{m}}{m!}\sum_{j_{1},\ldots,j_{m}=1}^{n}\left|\begin{array}{cccc} a_{j_{1}j_{1}} & \cdots & a_{j_{1}j_{m}} \\\\ \vdots & \ddots & \vdots \\\\ a_{j_{m}j_{1}} & \cdots & a_{j_{m}j_{m}} \end{array}\right| $$ $$ B_{kl(k\neq l)}=-a_{lk}x-\sum_{m=2}^{n-1}\frac{x^{m}}{(m-1)!}\sum_{j_{1},\ldots,j_{m-1}=1}^{n}\left|\begin{array}{cccc} a_{lk} & a_{lj_{1}} & \cdots & a_{lj_{m-1}} \\\\ a_{j_{1}k} & a_{j_{1}j_{1}} & \cdots & a_{j_{1}j_{m-1}} \\\\ \vdots & \vdots & \ddots & \vdots \\\\ a_{j_{m-1}k} & a_{j_{m-1}j_{1}} & \cdots & a_{j_{m-1}j_{m-1}} \end{array}\right| $$

After a long time of working out my own solution, I was able to post an Answer below. Please let me know if there is a better way.

Answer 1

We have $$|1+Ax|= \sum_{\sigma\in S_n} \epsilon_\sigma\prod_{i=1}^n (a_{i\sigma_i}x+\delta_{i\sigma_i}) $$ $$ =\sum_{k=0}^n x^k \sum_{j_1<\cdots<j_k\in1,\cdots,n} \quad \sum_{\sigma\in {\rm Stab}\left(\{j_i,\cdots,j_k)\}^C\right) }\epsilon_{\sigma}\prod_{i=1}^k a_{j_i\sigma_{j_i}} $$ $$ =\sum_{k=0}^n x^k \sum_{j_1<\cdots<j_k\in1,\cdots,n} \quad \left|\begin{array}{cccc} a_{j_{1}j_{1}} & \cdots & a_{j_{1}j_{k}} \\\\ \vdots & \ddots & \vdots \\\\ a_{j_{k}j_{1}} & \cdots & a_{j_{k}j_{k}} \end{array}\right| $$

Here the first and final equality are just the definition of determinant and minor. $S_n$ denotes the set of permutations of the indices. $\epsilon_\sigma$ is the sign of the permutation $\sigma$. $\delta_{ij}$ is $1$ if $i=j$ and $0$ otherwise. Finally ${\rm Stab}\left(\{j_i,\cdots,j_k)\}^C\right)$ is the set of permutations of the indices which only move (some of) $j_1,\cdots,j_k$. Note the sign of such a permutation is the same as the sign of its restriction to $j_1,\cdots,j_k$.

The key step was the middle equality, where we collated all the terms where $x$ had exponent $k$.

Answer 2

Using the box product and letting $M^{(k)}$ denote the $k$-th compound power (i.e. exterior power) of a square matrix $M$, we have $$(xA+I)^{(k)}=\sum_{p=0}^k x^pA^{(p)}\mathbin{\square} I^{(k-p)}\tag{1}$$ Taking the trace on both sides of (1) with $k=n$ and using the fact that $\mathop{\mathrm{tr}}(A^{(p)}\mathbin{\square}I^{(n-p)})=\mathop{\mathrm{tr}}(A^{(p)})$, we obtain $$\det(xA+I)=\sum_{p=0}^n x^p\mathop{\mathrm{tr}}(A^{(p)})=1+x\mathop{\mathrm{tr}}(A)+\cdots+x^n\det(A)\tag{2}$$ Note here $\mathop{\mathrm{tr}}(A^{(p)})$ is just the sum of the $p\times p$ principal minor determinants of $A$.

Taking $k=n-1$ in (1) and then applying the Poincaré dual map, you can obtain the formula for the cofactors. For more details see my videos here and here.

Answer 3

My approach is to use the Maclaurin expansion and compute the derivative of the determinant. $$ |B|=\sum_{m=0}^{n}\frac{x^{m}}{m!}\frac{\mathrm{d}^{m}|B|}{\mathrm{d}x^{m}}(x=0) $$ The problem then becomes to show the following formula for the derivative. $$ \frac{\mathrm{d}^{m}|B|}{\mathrm{d}x^{m}}(x=0)=\sum_{j_{1},\ldots,j_{m}=1}^{n}\left|\begin{array}{ccc} a_{j_{1}j_{1}} & \cdots & a_{j_{1}j_{m}} \\\\ \vdots & \ddots & \vdots \\\\ a_{j_{m}j_{1}} & \cdots & a_{j_{m}j_{m}} \end{array}\right| $$ It can be verified as follows.

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$$ \frac{\mathrm{d}|B|}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}\sum_{\sigma\in S_{n}}\text{sgn}(\sigma)b_{1\sigma(1)}b_{2\sigma(2)}\cdots b_{n\sigma(n)} $$ $$ =\sum_{j=1}^{n}\sum_{\sigma\in S_{n}}\text{sgn}(\sigma)b_{1\sigma(1)}\cdots b_{(j-1)\sigma(j-1)}a_{j\sigma(j)}b_{(j+1)\sigma(j+1)}\cdots b_{n\sigma(n)} $$ $$ =\sum_{j=1}^{n}\left|\begin{array}{ccc} \vdots & & \vdots \\\\ b_{(j-1)1} & \cdots & b_{(j-1)n} \\\\ a_{j1} & \cdots & a_{jn} \\\\ b_{(j+1)1} & \cdots & b_{(j+1)n} \\\\ \vdots & & \vdots \end{array}\right|\xrightarrow{x\to0}\sum_{j=1}^{n}a_{jj} $$ $$ \frac{\mathrm{d}^{m}|B|}{\mathrm{d}x^{m}}=\sum_{j_{1},\ldots,j_{m}}^{\text{No duplicates}}\left|\begin{array}{ccc} \vdots & & \vdots \\\\ b_{(j_{2}-1)1} & \cdots & b_{(j_{2}-1)n} \\\\ a_{j_{2}1} & \cdots & a_{j_{2}n} \\\\ b_{(j_{2}+1)1} & \cdots & b_{(j_{2}+1)n} \\\\ \vdots & & \vdots \\\\ b_{(j_{1}-1)1} & \cdots & b_{(j_{1}-1)n} \\\\ a_{j_{1}1} & \cdots & a_{j_{1}n} \\\\ b_{(j_{1}+1)1} & \cdots & b_{(j_{1}+1)n} \\\\ \vdots & & \vdots \\\\ b_{(j_{m}-1)1} & \cdots & b_{(j_{m}-1)n} \\\\ a_{j_{m}1} & \cdots & a_{j_{m}n} \\\\ b_{(j_{m}+1)1} & \cdots & b_{(j_{m}+1)n} \\\\ \vdots & & \vdots \\\\ \end{array}\right| $$ $$ =\sum_{j_{1},\ldots,j_{m}}^{\text{No duplicates}}\sum_{\sigma\in S_{n}}\text{sgn}(\sigma)b_{1\sigma(1)}\cdots b_{(j_{2}-1)\sigma(j_{2}-1)}a_{j_{2}\sigma(j_{2})}b_{(j_{2}+1)\sigma(j_{2}+1)}\cdots b_{(j_{1}-1)\sigma(j_{1}-1)}a_{j_{1}\sigma(j_{1})}b_{(j_{1}+1)\sigma(j_{1}+1)}\cdots b_{(j_{m}-1)\sigma(j_{m}-1)}a_{j_{m}\sigma(j_{m})}b_{(j_{m}+1)\sigma(j_{m}+1)}\cdots b_{n\sigma(n)} $$ $$ \xrightarrow{x\to0}\sum_{j_{1},\ldots,j_{m}}^{\text{No duplicates}}\sum_{\sigma\in S_{m}}\text{sgn}(\sigma)a_{j_{1}\sigma(j_{1})}\cdots a_{j_{m}\sigma(j_{m})}=\sum_{j_{1},\ldots,j_{m}}\left|\begin{array}{ccc} a_{j_{1}j_{1}} & \cdots & a_{j_{1}j_{m}} \\\\ \vdots & \ddots & \vdots \\\\ a_{j_{m}j_{1}} & \cdots & a_{j_{m}j_{m}} \end{array}\right| $$ $$ \left(=m!\sum_{j_{1}<\ldots<j_{m}}\left|\begin{array}{ccc} a_{j_{1}j_{1}} & \cdots & a_{j_{1}j_{m}} \\\\ \vdots & \ddots & \vdots \\\\ a_{j_{m}j_{1}} & \cdots & a_{j_{m}j_{m}} \end{array}\right|\right) $$

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For the cofactors, I first calculated for $B_{21}$. $$ B_{21}=-\left|\begin{array}{cccc} b_{12} & b_{13} & \cdots & b_{1n} \\\\ b_{32} & b_{33} & \cdots & b_{3n} \\\\ \vdots & \vdots & \ddots & \vdots \\\\ b_{n2} & b_{n3} & \cdots & b_{nn} \end{array}\right|=\sum_{m=0}^{n-1}\frac{x^{m}}{m!}\frac{\mathrm{d}^{m}B_{21}}{\mathrm{d}x^{m}}(x=0) $$ $$ B_{21}(x=0)=0,\quad \frac{\mathrm{d}B_{21}}{\mathrm{d}x}(x=0)=-\left|\begin{array}{cccc} a_{12} & a_{13} & \cdots & a_{1n} \\\\ 0 & 1 & \cdots & 0 \\\\ \vdots & \vdots & \ddots & \vdots \\\\ \end{array}\right|-\sum\left|\begin{array}{ccc} 0 & \cdots & 0 \\\\ \vdots & & \vdots \\\\ \end{array}\right|=-a_{12} $$ $$ \frac{\mathrm{d}^{2}B_{21}}{\mathrm{d}x^{2}}=-\frac{\mathrm{d}}{\mathrm{d}x}\left|\begin{array}{cccc} a_{12} & a_{13} & \cdots & a_{1n} \\\\ b_{32} & b_{33} & \cdots & b_{3n} \\\\ \vdots & \vdots & \ddots & \vdots \\\\ \end{array}\right|-\frac{\mathrm{d}}{\mathrm{d}x}\left|\begin{array}{cccc} b_{12} & b_{13} & \cdots & b_{1n} \\\\ \vdots & \vdots & \ddots & \vdots \\\\ a_{j2} & a_{j3} & \ddots & a_{jn} \\\\ \vdots & \vdots & \ddots & \vdots \\\\ \end{array}\right| $$ $$ =-\sum_{j}\left|\begin{array}{cccc} a_{12} & a_{13} & \cdots & a_{1n} \\\\ b_{32} & b_{33} & \cdots & b_{3n} \\\\ \vdots & \vdots & \ddots & \vdots \\\\ a_{j2} & a_{j3} & \ddots & a_{jn} \\\\ \vdots & \vdots & \ddots & \vdots \\\\ \end{array}\right|-\sum_{j}\left|\begin{array}{cccc} a_{12} & a_{13} & \cdots & a_{1n} \\\\ b_{32} & b_{33} & \cdots & b_{3n} \\\\ \vdots & \vdots & \ddots & \vdots \\\\ a_{j2} & a_{j3} & \ddots & a_{jn} \\\\ \vdots & \vdots & \ddots & \vdots \\\\ \end{array}\right| $$ $$ =-2\sum_{j}\sum_{\sigma\in S_{n}}\text{sgn}(\sigma)a_{1\sigma(1)}a_{j\sigma(j)}b_{2\sigma(2)}\cdots $$ $$ \xrightarrow{x\to0}-2\sum_{j}\sum_{\sigma\in S_{2}}\text{sgn}(\sigma)a_{1\sigma(1)}a_{j\sigma(j)}=-2\left|\begin{array}{ccc} a_{12} & a_{j2} \\\\ a_{1j} & a_{jj} \\\\ \end{array}\right| $$ Thus, since $B_{kl}$ loses the diagonal components of the $l$th column and $k$th row and $x\to0$, the $k$th column (not row) and $l$th row (not column) become zero, we need to differentiate the $l$th row somewhere in the middle of differentiating $m$ times to make the whole $l$th row $\boldsymbol{a}_{l}$. When $m$ rows containing the $l$th row are selected for differentiation, there are $m$ times to differentiate the $l$th row, so $$ B_{kl}(x=0)=0,\quad \frac{\mathrm{d}^{m}B_{kl}}{\mathrm{d}x^{m}}(x=0)=(-1)^{k+l}m\sum_{j_{1},\ldots,j_{m-1}}\left|\begin{array}{cccc} a_{lk} & a_{lj_{1}} & \cdots & a_{lj_{m-1}} \\\\ a_{j_{1}k} & a_{j_{1}j_{1}} & \cdots & a_{j_{1}j_{m-1}} \\\\ \vdots & \vdots & \ddots & \vdots \\\\ a_{j_{m-1}k} & a_{j_{m-1}j_{1}} & \cdots & a_{j_{m-1}j_{m-1}} \end{array}\right| $$

Please let me know if there is a better way.