Prove the generalized Laplace expansion.

Question

Suppose that $A$ is an $m \times n$ matrix. The $(i, j)$-entry of $A$ is denoted as $[A]_{i,j}$.

Suppose that $A$ is an $m \times n$ matrix. Suppose that $i_1$, $i_2$, $\dots$, $i_s$ are distinct integers less than or equal to $m$, and that $j_1$, $j_2$, $\dots$, $j_t$ are distinct integers less than or equal to $n$. The matrix obtained from $A$ by removing rows $i_1$, $i_2$, $\dots$, $i_s$ and columns $j_1$, $j_2$, $\dots$, $j_t$ is denoted as $A({i_1,\dots,i_s}|{j_1,\dots,j_t})$.

If $A$ is an $n \times n$ matrix, we see $A({1,2,\dots,n}|{1,2,\dots,n})$ as a "zero-by-zero matrix" for the sake of convenience.

Suppose that $A$ is an $m \times n$ matrix. Suppose that $i_1$, $i_2$, $\dots$, $i_s$ are distinct integers less than or equal to $m$, and that $j_1$, $j_2$, $\dots$, $j_t$ are distinct integers less than or equal to $n$. The matrix obtained from $A$ by removing all rows other than rows $i_1$, $i_2$, $\dots$, $i_s$ and all columns other than $j_1$, $j_2$, $\dots$, $j_t$ is denoted as $\displaystyle A\binom{i_1,\dots,i_s}{j_1,\dots,j_t}$.

Example. Let $$ A = \begin{bmatrix} 1 & 2 & 3 & 4 & 5 \\ 6 & 7 & 8 & 9 & 10 \\ 11 & 12 & 13 & 14 & 15 \\ 16 & 17 & 18 & 19 & 20 \\ \end{bmatrix}.$$ Then $$A({1,2}|{3}) = \begin{bmatrix} 11 & 12 & 14 & 15 \\ 16 & 17 & 19 & 20 \\ \end{bmatrix},$$ and $$A\binom{1,2}{3} = \begin{bmatrix} 3 \\ 8 \\ \end{bmatrix}.$$

Suppose that $A$ is an $n \times n$ matrix. The determinant of $A$, $\det {(A)}$ is defined as follows: $$ \det {(A)} = \begin{cases} [A]_{1,1}, & n = 1; \\ \displaystyle \sum_{i = 1}^{n} {(-1)^{i+1} [A]_{i,1} \det {(A(i|1))}}, & n \geq 2. \end{cases} $$

It should be noted that we can simply write $$ \det {(A)} = \sum_{i = 1}^{n} {(-1)^{i+1} [A]_{i,1} \det {(A(i|1))}} $$ if we define the determinant of a "zero-by-zero matrix" to be $1$.

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Here are the generalized Laplace expansions, one with respect to columns and the other with respect to rows.

Theorem 1. Suppose that $A$ is an $n \times n$ matrix, that $j_1$, $j_2$, $\dots$, $j_k$ are distinct positive integers less than or equal to $n$, and that $j_1 < j_2 < \dots < j_k$. Then $$ \begin{aligned} \det {(A)} = \sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n} {\det {\left( A\binom{i_1, i_2, \dots, i_k} {j_1, j_2, \dots, j_k} \right)}} (-1)^{i_1 + i_2 + \dots + i_k + j_1 + j_2 + \dots + j_k} \det {(A({i_1,i_2,\dots,i_k}|{j_1,j_2,\dots,j_k}))}. \end{aligned} $$

Theorem 2. Suppose that $A$ is an $n \times n$ matrix, that $i_1$, $i_2$, $\dots$, $i_k$ are distinct positive integers less than or equal to $n$, and that $i_1 < i_2 < \dots < i_k$. Then $$ \begin{aligned} \det {(A)} = \sum_{1 \leq j_1 < j_2 < \dots < j_k \leq n} {\det {\left( A\binom{i_1, i_2, \dots, i_k} {j_1, j_2, \dots, j_k} \right)}} (-1)^{i_1 + i_2 + \dots + i_k + j_1 + j_2 + \dots + j_k} \det {(A({i_1,i_2,\dots,i_k}|{j_1,j_2,\dots,j_k}))}. \end{aligned} $$

How does one prove the two theorems?

Answer 1

It will be easier if it is first proved that the determinant of a square matrix can be computed by expansion about any row or column; see this for the details.

I will use the $\rho$-notation which is defined in the above answer.

I will prove only the first theorem; the proof of the second one is very similar to that of the first one, which I leave as an exercise.

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Let the proposition $P(k)$ be as follows:

For any $n \times n$ matrix (in which $n \geq k$) and for any $k$ distinct integers $j_1$, $j_2$, $\dots$, $j_k$ less than or equal to $n$ (in which $j_1 < j_2 < \dots < j_k$), $$ \begin{aligned} \det {(A)} = \sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n} {\det {\left( A\binom{i_1, i_2, \dots, i_k} {j_1, j_2, \dots, j_k} \right)}} (-1)^{i_1 + i_2 + \dots + i_k + j_1 + j_2 + \dots + j_k} \det {(A({i_1,i_2,\dots,i_k}|{j_1,j_2,\dots,j_k}))}. \end{aligned} $$

We will prove that $P(k)$ is true for $k = 1$, $2$, $3$, $\dots$.

$P(1)$ is true, because the determinant of a square matrix can be computed by expansion about any column.

Suppose that $P(k-1)$ is true. Our goal is to prove that $P(k)$ is true.

Let $A$ be an $n \times n$ matrix, with $n \geq k$. Let $j_1$, $j_2$, $\dots$, $j_k$ be any $k$ distinct integers less than or equal to $n$, with $j_1 < j_2 < \dots < j_k$. Then $$ \begin{aligned} & \det {(A)} \\ = {} & \sum_{d_1 = 1}^{n} {[A]_{d_1,j_1} (-1)^{d_1 + j_1} \det {(A(d_1|j_1))}} \\ = {} & \sum_{d_1 = 1}^{n} {[A]_{d_1,j_1} (-1)^{d_1 + j_1}} \sum_{\substack{ 1 \leq d_2 < \dots < d_k \leq n \\ d_1 \neq d_2,\dots,d_k }} {\det {\left( A\binom{d_2, \dots, d_k}{j_2, \dots, j_k} \right)}} (-1)^{d_2 - \rho(d_2, d_1) + \dots + d_k - \rho(d_k, d_1) + j_2 + \dots + j_k - (k - 1)} \det {(A({d_1,d_2,\dots,d_k}|{j_1,j_2,\dots,j_k}))} \\ = {} & \sum_{\substack{ 1 \leq d_1 \leq n \\ 1 \leq d_2 < \dots < d_k \leq n \\ d_1 \neq d_2,\dots,d_k }} {[A]_{d_1,j_1} (-1)^{d_1 + j_1}} {\det {\left( A\binom{d_2, \dots, d_k}{j_2, \dots, j_k} \right)}} (-1)^{d_2 - \rho(d_2, d_1) + \dots + d_k - \rho(d_k, d_1) + j_2 + \dots + j_k - (k - 1)} \det {(A({d_1,d_2,\dots,d_k}|{j_1,j_2,\dots,j_k}))} \\ = {} & \sum_{\substack{ 1 \leq d_1 \leq n \\ 1 \leq d_2 < \dots < d_k \leq n \\ d_1 \neq d_2,\dots,d_k }} (-1)^{ (d_1 + \dots + d_k) + (j_1 + \dots + j_k) + (\rho(d_1, d_2) + \dots + \rho(d_1, d_k)) - 2(k-1) } [A]_{d_1,j_1} {\det {\left( A\binom{d_2, \dots, d_k}{j_2, \dots, j_k} \right)}} \det {(A({d_1,d_2,\dots,d_k}|{j_1,j_2,\dots,j_k}))} \\ = {} & \sum_{\substack{ 1 \leq d_1 \leq n \\ 1 \leq d_2 < \dots < d_k \leq n \\ d_1 \neq d_2,\dots,d_k }} (-1)^{ (d_1 + \dots + d_k) + (j_1 + \dots + j_k) + (\rho(d_1, d_2) + \dots + \rho(d_1, d_k)) } [A]_{d_1,j_1} {\det {\left( A\binom{d_2, \dots, d_k}{j_2, \dots, j_k} \right)}} \det {(A({d_1,d_2,\dots,d_k}|{j_1,j_2,\dots,j_k}))} \\ = {} & \sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n} \sum_{\substack{ 1 \leq s \leq k\\ d_1 = i_s \\ d_r = i_{r-1}, \ 2 \leq r \leq s \\ d_r = i_{r}, \ r > s \\ }} (-1)^{ (d_1 + \dots + d_k) + (j_1 + \dots + j_k) + (\rho(d_1, d_2) + \dots + \rho(d_1, d_k)) } [A]_{d_1,j_1} {\det {\left( A\binom{d_2, \dots, d_k}{j_2, \dots, j_k} \right)}} \det {(A({d_1,d_2,\dots,d_k}|{j_1,j_2,\dots,j_k}))} \\ = {} & \sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n} \sum_{1 \leq s \leq k} (-1)^{ (i_1 + \dots + i_k) + (j_1 + \dots + j_k) + (s-1) } [A]_{i_s,j_1} {\det {\left( A\binom{i_1,\dots,i_{s-1},i_{s+1},\dots,i_k}{j_2, \dots, j_k} \right)}} \det {(A({i_1,i_2,\dots,i_k}|{j_1,j_2,\dots,j_k}))} \\ = {} & \sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n} \left( (-1)^{s-1} \sum_{1 \leq s \leq k} [A]_{i_s,j_1} {\det {\left( A\binom{i_1,\dots,i_{s-1},i_{s+1},\dots,i_k}{j_2, \dots, j_k} \right)}} \right) (-1)^{ (i_1 + \dots + i_k) + (j_1 + \dots + j_k) } \det {(A({i_1,i_2,\dots,i_k}|{j_1,j_2,\dots,j_k}))} \\ = {} & \sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n} \left( (-1)^{s+1} \sum_{1 \leq s \leq k} [A]_{i_s,j_1} {\det {\left( A\binom{i_1,\dots,i_{s-1},i_{s+1},\dots,i_k}{j_2, \dots, j_k} \right)}} \right) (-1)^{ (i_1 + \dots + i_k) + (j_1 + \dots + j_k) } \det {(A({i_1,i_2,\dots,i_k}|{j_1,j_2,\dots,j_k}))} \\ = {} & \sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n} { \det {\left( A\binom{i_1, i_2, \dots, i_k} {j_1, j_2, \dots, j_k} \right)} } (-1)^{ (i_1 + \dots + i_k) + (j_1 + \dots + j_k) } \det {(A({i_1,i_2,\dots,i_k}|{j_1,j_2,\dots,j_k}))}. \end{aligned} $$

We have used the hypothesis in the second equality.

Hence by mathematical induction, $P(k)$ is true for $k = 1$, $2$, $3$, $\dots$.

Note that the proof uses nothing more than the usual Laplace expansion.