Determinants of Matrices

We now focus on actual computation of determinants of matrices. First of all, we recall the definition.

Definition. If $A \in \text{Mat}_n(\mathbb{F})$ then the determinant of $A$ is defined by

\[\det A = \sum_{\sigma \in S_n} \epsilon(\sigma) \left( \prod_{i=1}^n A_{i \sigma(i)} \right)\]

Proposition. Let $D$ be a volume form on $\mathbb{F}^n$ and $A = \begin{pmatrix} A^{(1)} & \cdots & A^{(n)} \end{pmatrix}$. Then

\[D(A^{(1)}, ..., A^{(n)}) = \det A \cdot D(e_1, ..., e_n)\]

or alternatively $D(Ae_1, …, Ae_n) = \det A \cdot D(e_1, …, e_n)$.

Proof.

We compute

\[\begin{align*} D(A^{(1)}, ..., A^{(n)}) &= D \left( \sum_{i_1} A_{i_1 1} e_{i_1}, A^{(2)}, ..., A^{(n)} \right) \\ &= \sum_{i_1} A_{i_1 1} D \left( e_{i_1}, A^{(2)}, ..., A^{(n)} \right) \\ &= \sum_{i_1} \sum_{i_2} A_{i_1 1} A_{i_2 2} D \left( e_{i_1}, e_{i_2}, ..., A^{(n)} \right) \\ &= \sum_{i_1, ..., i_n} \left( \prod_{j=1}^n A_{i_j j} \right) D(e_{i_1}, ..., e_{i_n}) \end{align*}\]

$D(e_{i_1}, …, e_{i_n}) = 0$ unless all $i_j$ are distinct, which means there is $\sigma \in S_n$ such that $i_j = \sigma(j)$. Hence,

\[D(A^{(1)}, ..., A^{(n)}) = \sum_{\sigma \in S_n} \left( \prod_{j=1}^n A_{\sigma(j) j} \right) D(e_{\sigma(1)}, ..., e_{\sigma(n)})\]

and $D(e_{\sigma(1)}, …, e_{\sigma(n)}) = \epsilon(\sigma) D(e_1, …, e_n)$.

We follow with an important result about determinants.

Proposition. Let $A, B \in \text{Mat}_n(\mathbb{F})$. Then

\[\det AB = \det A \det B\]

Proof.

Let $D$ be a non-zero volume form on $\mathbb{F}^n$ and $D_A: \mathbb{F}^n \times \cdots \times \mathbb{F}^n \to \mathbb{F}$ be given by

\[D_A(v_1, ..., v_n) = D(Av_1, ..., Av_n)\]

Then $D_A$ is also a volume form. Thus,

\[D_A(Bv_1, ..., Bv_n) = (\det B) D_A(e_1, ..., e_n) = (\det B) D(Ae_1, ..., Ae_n) = (\det B)(\det A) D(e_1, ..., e_n)\]

and on the other hand

\[D_A(Bv_1, ..., Bv_n) = D(ABe_1, ..., ABe_n) = (\det AB) D(e_1, ..., e_n)\]

and hence $\det AB = \det A \det B$.

Elementary Row/Column Operations

Proposition. Let $A \in \text{Mat}_n(\mathbb{F})$. Then

  • Let $A’$ be a matrix obtained by interchanging $A^{(i)}$ and $A^{(j)}$ for $j \not= i$. Then

    \[D(A') = - D(A)\]

  • Let $A’$ be a matrix obtained by replacing $A^{(i)}$ by $\lambda A^{(i)}$. Then

    \[D(A') = \lambda D(A)\]

  • Let $A’$ be a matrix obtained by replacing $A^{(i)}$ by $A^{(i)} + \lambda A^{(j)}$ for $j \not= i$. Then

    \[D(A') = D(A)\]

Proof.

Follows immediately from axioms and properties.

With that, we can compute the determinants by applying the elementary row/column operations until we get the upper/lower triangular form in which $\det A = \text{tr}\, A’$.

Row/Column Expansions

Definition. $\widehat{A_{ij}}$ denotes the submatrix of $A$ obtained by deleting the $i$th row and $j$th column.

Proposition. [Row/Column Expansions] Let $A \in \text{Mat}_n(\mathbb{F})$. Then

\[\det A = \sum_{j=1}^n (-1)^{i+j} A_{ij} \det \widehat{A_{ij}} = \sum_{i=1}^n (-1)^{i+j} A_{ij} \det \widehat{A_{ij}}\]

Proof.

\[\begin{align*} \det A &= \det(A^{(1)}, ..., \sum_i A_{ij} e_i, ..., A^{(n)}) \\ &= \sum_i A_{ij} \det(A^{(1)}, ..., e_i, ..., A^{(n)}) \\ &= \sum_i A_{ij} (-1)^{i+j} \det A' \end{align*}\]

where

\[A' = \begin{pmatrix} 1 & \ast \\ 0 & \widehat{A_{ij}} \end{pmatrix}\]

which is obtained by interchanging $i-1$ times with the rows above and $j-1$ times with the columns on the left, and introduces the $(-1)^{i+j-2} = (-1)^{i+j}$ term.

For $\sigma \in S_n$, $\prod A_{i \sigma(i)}’ = 0$ unless $\sigma(1) = 1$ so $\det B = \det \widehat{A_{ij}}$ as required.

Singularity and Inverses

Definition. A matrix $A$ is singular if $\det A = 0$.

Proposition. Let $A \in \text{Mat}_n(\mathbb{F})$. Then the following statements are equivalent:

  • $A$ is invertible;

  • $A$ is non-singular;

  • $r(A) = n$.

Proof.

If $A$ is invertible, there exists $A^{-1}$ such that $AA^{-1} = I_n$ therefore $\det A \det A^{-1} = 1$ which implies $\det A \not= 0$ and $A$ is non-singular.

If $r(A) < n$, then columns of $A$ are linearly dependent and $\det A = 0$. Contrapositively, if $A$ is non-singular, $r(A) = n$.

As proved before, for linear map $\alpha: V \to V$, surjectivity implies isomorphism. Therefore, $r(A) = n$ implies $A$ is invertible.

Proposition. $\det A^{-1} = 1 / \det A$.

Proof.

$1 = \det I_n = \det(A A^{-1}) = \det A \det A^{-1}$.

Definition. The adjugate matrix, denoted by $\text{adj}\, A$, is the element of $\text{Mat}_n(\mathbb{F}^n)$ such that

\[(\text{adj}\, A)_{ij} = (-1)^{i+j} \det \widehat{A_{ji}}\]

Proposition. Let $A \in \text{Mat}_n(\mathbb{F})$. Then

\[(\text{adj}\, A)A = A(\text{adj}\, A) = (\det A) I_n\]

Thus, if $\det A \not= 0$, then

\[A^{-1} = {1 \over \det A} \text{adj}\, A\]

Proof.

\[((\text{adj}\, A)A)_{jk} = \sum_i (\text{adj}\, A)_{ji} A_{ik} = \sum_i (-1)^{i + j} \det \widehat{A_{ij}} A_{ik}\]

The sum is in column expansion form. If $k = j$, the sum is equal to $\det A$. If $k \not= j$, the sum is the expansion of a matrix obtained by replacing the $j$th column by $k$th column. Since the matrix has two identical column, the sum is equal to $0$. Therefore, $((\text{adj}\, A)A) = (\det A)I_n$ as required.

In general, computing determinant and finding inverse is more efficient with the use of elementary operations instead of finding the adjugate.

Determinantal Rank

Definition. The determinantal rank of a matrix $A$ is the largest integer $r$ such that there exists $r \times r$ minor of $A$ with non-zero determinant.

Proposition. The determinantal rank of a matrix is equivalent to its rank.

Proof.

Let $A$ be a $n \times n$ matrix of determinantal rank $r$. There exists a $r \times r$ minor $M$ with non-zero determinant, i.e. $r(M) = r$. Therefore, the columns/rows of $M$ are $r$ linearly independent vectors and their extensions to $n \times n$ are also linearly independent, i.e. $r(A) \ge r(M)$.

On the other hand, let $A$ be a $n \times n$ matrix of rank $r$. Then there are $r$ rows of linearly independent vectors which can be used to form a $r \times n$ matrix $B$ in which $r(B) = r$. Since the column rank of $B$ is also $r$, we can form a full-rank $r \times r$ minor $M$ in which $\det M \not= 0$, i.e. $r(M) \ge r(A)$.

Therefore, the column, row and determinantal ranks of a matrix are all equivalent.

Block Triangular Matrices

Proposition. Let $A \in \text{Mat}_k(\mathbb{F})$ and $B \in \text{Mat}_l(\mathbb{F})$ with $k, l \ge 1$. Then

\[\det \begin{pmatrix} A & C \\ 0 & B \end{pmatrix} = \det A \det B\]

Proof.

Define

\[X = \begin{pmatrix} A & C \\ 0 & B \end{pmatrix}\]

which is a $(k + l) \times (k + l)$ matrix. Then

\[\det X = \sum_{\sigma \in S_{k+l}} \epsilon(\sigma) \left( \prod_{i=1}^{k+l} X_{i \sigma(i)} \right)\]

Since $X_{ij} = 0$ when $i > k$ and $j \le k$. We can restrict the sum to those $\sigma$ such that $\sigma(i) > k$ for $i > k$ (so $X_{i \sigma(i)}$ are entries of $B$). Therefore, we can factorize these $\sigma$ as $\sigma = \sigma_1 \sigma_2$ with $\sigma_1 \in S_k$ and $\sigma_2$ is a permutation of $\Set{k+1, …, n}$. Thus,

\[\begin{align*} \det X &= \sum_{\sigma_1} \sum_{\sigma_2} \epsilon(\sigma_1 \sigma_2) \left( \prod_{i=1}^k X_{i \sigma_1(i)} \right) \left( \prod_{j=1}^l X_{j+k \sigma_2(j + k)} \right) \\ &= \left( \sum_{\sigma_1} \epsilon(\sigma_1) \left( \prod_{i=1}^k A_{i \sigma_1(i)} \right) \right) \left( \sum_{\sigma_2} \epsilon(\sigma_2) \left( \prod_{j=1}^l B_{j \sigma_2(j)} \right) \right) \\ &= \det A \det B \end{align*}\]

The core idea behind is that product term in determinant chooses one entry from each row and then sum over all combinations. The structure of the triangular block matrix makes any choice of entry in $C$ will result with a product of $0$ since we have to choose an entry from the $0$ block. Therefore, the only choices that is non-zero are those choosing $k$ entries from $A$ and $l$ entries from $B$.

In general, it is not true that for a matrix $M \in \text{Mat}_{2n}(\mathbb{F})$, $\det M = \det A \det D - \det B \det C$ where $A, B, C, D \in \text{Mat}_n(\mathbb{F})$.

References