Proof of Theorem DER

First, the statement of the theorem coincides with Definition DM when $i=1$, so throughout, we need only consider $i>1$.

Given the recursive definition of the determinant, it should be no surprise that we will use induction for this proof (Proof Technique I). When $n=1$, there is nothing to prove since there is but one row. When $n=2$, we just examine expansion about the second row, \begin{align*} (-1)^{2+1}\matrixentry{A}{21}&\detname{\submatrix{A}{2}{1}}+ (-1)^{2+2}\matrixentry{A}{22}\detname{\submatrix{A}{2}{2}}\\ &=-\matrixentry{A}{21}\matrixentry{A}{12}+\matrixentry{A}{22}\matrixentry{A}{11} &&\knowl{./knowls/definition.DM.knowl}{\text{Definition DM}}\\ &=\matrixentry{A}{11}\matrixentry{A}{22}-\matrixentry{A}{12}\matrixentry{A}{21}\\ &= \detname{A}&&\knowl{./knowls/theorem.DMST.knowl}{\text{Theorem DMST}}\\ \end{align*}

So the theorem is true for matrices of size $n=1$ and $n=2$. Now assume the result is true for all matrices of size $n-1$ as we derive an expression for expansion about row $i$ for a matrix of size $n$. We will abuse our notation for a submatrix slightly, so $\submatrix{A}{i_1,i_2}{j_1,j_2}$ will denote the matrix formed by removing rows $i_1$ and $i_2$, along with removing columns $j_1$ and $j_2$. Also, as we take a determinant of a submatrix, we will need to “jump up” the index of summation partway through as we “skip over” a missing column. To do this smoothly we will set \begin{equation*} \epsilon_{\ell j}= \begin{cases} 0 & \ell<j\\ 1 & \ell>j \end{cases} \end{equation*}

Now, \begin{align*} &\detname{A}\\ &\quad= \sum_{j=1}^{n}(-1)^{1+j}\matrixentry{A}{1j}\detname{\submatrix{A}{1}{j}} &&\knowl{./knowls/definition.DM.knowl}{\text{Definition DM}}\\ &\quad= \sum_{j=1}^{n}(-1)^{1+j}\matrixentry{A}{1j} \sum_{\substack{1\leq\ell\leq n\\\ell\neq j}} (-1)^{i-1+\ell-\epsilon_{\ell j}}\matrixentry{A}{i\ell}\detname{\submatrix{A}{1,i}{j,\ell}} &&\text{Induction}\\ &\quad= \sum_{j=1}^{n}\sum_{\substack{1\leq\ell\leq n\\\ell\neq j}} (-1)^{j+i+\ell-\epsilon_{\ell j}} \matrixentry{A}{1j}\matrixentry{A}{i\ell}\detname{\submatrix{A}{1,i}{j,\ell}} &&\knowl{./knowls/property.DCN.knowl}{\text{Property DCN}}\\ &\quad= \sum_{\ell=1}^{n}\sum_{\substack{1\leq j\leq n\\j\neq\ell}} (-1)^{j+i+\ell-\epsilon_{\ell j}} \matrixentry{A}{1j}\matrixentry{A}{i\ell}\detname{\submatrix{A}{1,i}{j,\ell}} &&\knowl{./knowls/property.CACN.knowl}{\text{Property CACN}}\\ &\quad= \sum_{\ell=1}^{n}(-1)^{i+\ell}\matrixentry{A}{i\ell} \sum_{\substack{1\leq j\leq n\\j\neq\ell}} (-1)^{j-\epsilon_{\ell j}} \matrixentry{A}{1j}\detname{\submatrix{A}{1,i}{j,\ell}} &&\knowl{./knowls/property.DCN.knowl}{\text{Property DCN}}\\ &\quad= \sum_{\ell=1}^{n}(-1)^{i+\ell}\matrixentry{A}{i\ell} \sum_{\substack{1\leq j\leq n\\j\neq\ell}} (-1)^{\epsilon_{\ell j}+j} \matrixentry{A}{1j}\detname{\submatrix{A}{i,1}{\ell,j}} &&\text{$2\epsilon_{\ell j}$ is even}\\ &\quad= \sum_{\ell=1}^{n}(-1)^{i+\ell}\matrixentry{A}{i\ell}\detname{\submatrix{A}{i}{\ell}} &&\knowl{./knowls/definition.DM.knowl}{\text{Definition DM}} \end{align*}