Prove that the determinant of a square matrix can be computed by expansion about any row or column, in which determinants are defined recursively.

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 $n \times n$ 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 two important theorems, from which a lot of facts about determinants can be easily deduced.

Theorem 1. Suppose that $A$ is an $n \times n$ matrix and that $j$ is a positive integer less than or equal to $n$. Then $$ \det {(A)} = \sum_{i = 1}^{n} {(-1)^{i+j} [A]_{i,j} \det {(A(i|j))}}. $$

Theorem 2. Suppose that $A$ is an $n \times n$ matrix and that $i$ is a positive integer less than or equal to $n$. Then $$ \det {(A)} = \sum_{j = 1}^{n} {(-1)^{i+j} [A]_{i,j} \det {(A(i|j))}}. $$

How does one prove the two theorems with the aforementioned definition of determinants?

Answer 1

Here is a piece of useful notation. Suppose that $i$, $j$ are two integers. We define $$ \rho(i, j) = \begin{cases} 0, & i < j; \\ 1, & i \geq j. \end{cases} $$

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Proof of Theorem 1. Let the proposition $P(n)$ be as follows:

For any $n \times n$ matrix $A$ and for any positive integer $j$ less than or equal to $n$, $$ \det {(A)} = \sum_{i = 1}^{n} {(-1)^{i+j} [A]_{i,j} \det {(A(i|j))}}. $$

We will prove that $P(n)$ is true for $n = 1$, $2$, $3$, $\dots$ by mathematical induction.

$P(1)$ is true by definition.

It is easy to check that $P(2)$ is true (which I leave as an exercise).

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

Let $A$ be an $m \times m$ matrix. Let $j$ be a positive integer less than or equal to $m$. If $j = 1$, it is just the definition. If $j > 1$, we first expand the determinant of $A$ about the first column, and then apply the hypothesis to every $(m - 1) \times (m - 1)$ submatrix: $$ \begin{aligned} & \det {(A)} \\ = {} & \sum_{i = 1}^{m} { (-1)^{i+1} [A]_{i,1} \det {(A(i|1))} } \\ = {} & \sum_{i = 1}^{m} { (-1)^{i+1} [A]_{i,1} \sum_{\substack{1 \leq \ell \leq m \\ \ell \neq i}} { (-1)^{(\ell - \rho(\ell, i)) + (j - 1)} [A]_{\ell,j} \det {(A({i,\ell}|{1,j}))} } } \\ = {} & \sum_{i = 1}^{m} { \sum_{\substack{1 \leq \ell \leq m \\ \ell \neq i}} { (-1)^{(\ell - \rho(\ell, i)) + (j - 1)} (-1)^{i+1} [A]_{i,1} [A]_{\ell,j} \det {(A({i,\ell}|{1,j}))} } } \\ = {} & \sum_{\ell = 1}^{m} { \sum_{\substack{1 \leq i \leq m \\ i \neq \ell}} { (-1)^{(\ell - \rho(\ell, i)) + (j - 1)} (-1)^{i+1} [A]_{i,1} [A]_{\ell,j} \det {(A({i,\ell}|{1,j}))} } } \\ = {} & \sum_{\ell = 1}^{m} { \sum_{\substack{1 \leq i \leq m \\ i \neq \ell}} { (-1)^{\ell} (-1)^{\rho(\ell, i)} (-1)^{i} (-1)^{j} [A]_{i,1} [A]_{\ell,j} \det {(A({i,\ell}|{1,j}))} } } \\ = {} & \sum_{\ell = 1}^{m} { \sum_{\substack{1 \leq i \leq m \\ i \neq \ell}} { (-1)^{\ell + j} (-1)^{i - \rho(i, \ell) + 1} [A]_{i,1} [A]_{\ell,j} \det {(A({i,\ell}|{1,j}))} } } \\ = {} & \sum_{\ell = 1}^{m} {(-1)^{\ell + j} [A]_{\ell,j} \sum_{\substack{1 \leq i \leq m \\ i \neq \ell}} { (-1)^{i - \rho(i, \ell) + 1} [A]_{i,1} \det {(A({i,\ell}|{1,j}))} } } \\ = {} & \sum_{\ell = 1}^{m} {(-1)^{\ell + j} [A]_{\ell,j} \det {(A(\ell|j))} }. \end{aligned} $$ Here are two useful tips:

• $[A]_{\ell,j}$ is the $(\ell - \rho(\ell, i), j-1)$-entry of $A(i|1)$;
• $[A]_{i,1}$ is the $(i - \rho(i, \ell), 1)$-entry of $A(\ell|j)$.

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

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Proof of Theorem 2. Let the proposition $P(n)$ be as follows:

For any $n \times n$ matrix $A$ and for any positive integer $i$ less than or equal to $n$, $$ \det {(A)} = \sum_{j = 1}^{n} {(-1)^{i+j} [A]_{i,j} \det {(A(i|j))}}. $$

We will prove that $P(n)$ is true for $n = 1$, $2$, $3$, $\dots$ by mathematical induction.

$P(1)$ is true by definition.

It is easy to check that $P(2)$ is true (which I leave as an exercise).

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

Let $A$ be an $m \times m$ matrix. Let $i$ be a positive integer less than or equal to $m$. We first expand the determinant of $A$ about the first column, and then apply the hypothesis to every $(m - 1) \times (m - 1)$ submatrix except $A(i|1)$ (in which we write $f = (-1)^{i+1} [A]_{i,1} \det {(A(i|1))}$ for brevity): $$ \begin{aligned} & \det {(A)} \\ = {} & f + \sum_{\substack{1 \leq k \leq m \\k \neq i}} {(-1)^{k+1} [A]_{k,1} \det {(A(k|1))}} \\ = {} & f + \sum_{\substack{1 \leq k \leq m \\k \neq i}} {(-1)^{k+1} [A]_{k,1} \sum_{j = 2}^{m} {(-1)^{i - \rho(i,k) + j - 1} [A]_{i,j} \det {(A(k,i|1,j))}}} \\ = {} & f + \sum_{\substack{1 \leq k \leq m \\k \neq i}} {\sum_{j = 2}^{m} {(-1)^{k+1} [A]_{k,1} (-1)^{i - \rho(i,k) + j - 1} [A]_{i,j} \det {(A(k,i|1,j))}}} \\ = {} & f + \sum_{j = 2}^{m} {\sum_{\substack{1 \leq k \leq m \\k \neq i}} {(-1)^{k+1} [A]_{k,1} (-1)^{i - \rho(i,k) + j - 1} [A]_{i,j} \det {(A(k,i|1,j))}}} \\ = {} & f + \sum_{j = 2}^{m} {\sum_{\substack{1 \leq k \leq m \\k \neq i}} {(-1)^{i+j} [A]_{i,j} (-1)^{k - \rho(k,i) + 1} [A]_{k,1} \det {(A(k,i|1,j))}}} \\ = {} & f + \sum_{j = 2}^{m} {(-1)^{i+j} [A]_{i,j} \sum_{\substack{1 \leq k \leq m \\k \neq i}} {(-1)^{k - \rho(k,i) + 1} [A]_{k,1} \det {(A(k,i|1,j))}}} \\ = {} & f + \sum_{j = 2}^{m} {(-1)^{i+j} [A]_{i,j} \det {(A(i|j))}}. \end{aligned} $$ Here are two useful tips (in which $k \neq i$ and $j \neq 1$):

• $[A]_{i,j}$ is the $(i - \rho(i, k), j-1)$-entry of $A(k|1)$;
• $[A]_{k,1}$ is the $(k - \rho(k, i), 1)$-entry of $A(i|j)$.

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

Answer 2

To prove that the determinant of a square matrix can be computed by expansion about any row, one can first prove that the determinant of a square matrix can be computed by expansion about the first row. Then one can just prove that the determinant of a square matrix can be computed by expansion about any row, as one can prove that the determinant of a square matrix can be computed by expansion about any column. (I feel that I have repeated so many words; I would like some advice on reduce repetition....)

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

For any $n \times n$ matrix $A$, $$ \det {(A)} = \sum_{j = 1}^{n} {(-1)^{1+j} [A]_{1,j} \det {(A(1|j))}}. $$

We will prove that $P(n)$ is true for $n = 1$, $2$, $3$, $\dots$ by mathematical induction.

$P(1)$ is true by definition.

It is easy to check that $P(2)$ is true (which I leave as an exercise).

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

Let $A$ be an $m \times m$ matrix. We first expand the determinant of $A$ about the first column, and then apply the hypothesis to every $(m - 1) \times (m - 1)$ submatrix except $A(1|1)$ (in which we write $f = (-1)^{1+1} [A]_{1,1} \det {(A(1|1))}$ for brevity): $$ \begin{aligned} & \det {(A)} \\ = {} & f + \sum_{i = 2}^{m} { (-1)^{i+1} [A]_{i,1} \det {(A(i|1))} } \\ = {} & f + \sum_{i = 2}^{m} { (-1)^{i+1} [A]_{i,1} \sum_{j = 2}^{m} { (-1)^{1+j-1} [A]_{1,j} \det {(A(i,1|1,j))} } } \\ = {} & f + \sum_{i = 2}^{m} { \sum_{j = 2}^{m} { (-1)^{i+1} [A]_{i,1} (-1)^{1+j-1} [A]_{1,j} \det {(A(i,1|1,j))} } } \\ = {} & f + \sum_{j = 2}^{m} { \sum_{i = 2}^{m} { (-1)^{i+1} [A]_{i,1} (-1)^{1+j-1} [A]_{1,j} \det {(A(i,1|1,j))} } } \\ = {} & f + \sum_{j = 2}^{m} { \sum_{i = 2}^{m} { (-1)^{1+j} [A]_{1,j} (-1)^{i-1+1} [A]_{i,1} \det {(A(i,1|1,j))} } } \\ = {} & f + \sum_{j = 2}^{m} { (-1)^{1+j} [A]_{1,j} \sum_{i = 2}^{m} { (-1)^{i-1+1} [A]_{i,1} \det {(A(i,1|1,j))} } } \\ = {} & f + \sum_{j = 2}^{m} { (-1)^{1+j} [A]_{1,j} \det {(A(1|j))} }. \end{aligned} $$ Here are two useful tips (in which $k \neq 1$ and $j \neq 1$):

• $[A]_{1,j}$ is the $(1, j-1)$-entry of $A(k|1)$;
• $[A]_{k,1}$ is the $(k-1, 1)$-entry of $A(1|j)$.

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