determinant

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The determinant is a number assigned to a system of n\, linear equations and n\, unknowns. By construction, it is zero if two or more of the equations are linearly dependent, and non-zero otherwise.

Definition

Define the determinant of a square n\times n\, matrix A\, to be

\det A \equiv \sum_{ijk...=1}^n \varepsilon_{ijk...} A_{1i} A_{2j} A_{3k} ...\,

where \varepsilon_{ijk...}\, is the Levi-Civita symbol which is equal to 1\, if ijk...\, are an even permutation of (1,2,3...,n)\,, -1\, if they are an odd permutation of (1,2,3...,n)\,, or zero otherwise. For example, if n = 3\,, then \varepsilon_{132} = -1\,. Therefore this can also be written as

\det A \equiv \sum_{\pi} (-1)^{I(\pi)} \prod_{i=1}^n A_{i \pi(i)}\,,

where \pi\, is some permutation of (1,2,3...,n)\,, i.e., an element of the symmetric group S_n\,, and I(\pi)\,, called the inversion number of \pi\,, the number of pairwise exchanges needed on \pi\, to obtain (1,2,3...,n)\,. I.e., (-1)^{I(\pi)}\, is 1\, if \pi\, is an even permutation of (1,2,3...)\, and -1\, otherwise.

Construction

When solving a system of linear equations

\begin{alignat}{7} a_{11} x_1 &&\; + \;&& a_{12} x_2 &&\; + \cdots + \;&& a_{1n} x_n &&\; = \;&&& b_1 \\ a_{21} x_1 &&\; + \;&& a_{22} x_2 &&\; + \cdots + \;&& a_{2n} x_n &&\; = \;&&& b_2 \\ \vdots\;\;\; && && \vdots\;\;\; && && \vdots\;\;\; && &&& \;\vdots \\ a_{m1} x_1 &&\; + \;&& a_{m2} x_2 &&\; + \cdots + \;&& a_{mn} x_n &&\; = \;&&& b_m \\ \end{alignat}

we may inquire whether one or more of these equations are expressible as a linear combination of the others. Let us for a moment consider the homogeneous system

\begin{alignat}{7} a_{11} x_1 &&\; + \;&& a_{12} x_2 &&\; + \cdots + \;&& a_{1n} x_n &&\; = \;&&& 0 \\ a_{21} x_1 &&\; + \;&& a_{22} x_2 &&\; + \cdots + \;&& a_{2n} x_n &&\; = \;&&& 0 \\ \vdots\;\;\; && && \vdots\;\;\; && && \vdots\;\;\; && &&& \,\vdots \\ a_{m1} x_1 &&\; + \;&& a_{m2} x_2 &&\; + \cdots + \;&& a_{mn} x_n &&\; = \;&&& 0 \\ \end{alignat}.

Associate to the i\,th row the row vector

\mathbf{A}_i = (a_{i1}\; a_{i2} \;\cdots \; a_{in})\,.

Exterior product

Our question may be phrased as follows: Do there exist complex numbers c_i\, such that \mathbf{A}_j = \textstyle\sum_{i\neq j} c_i \mathbf{A}_i\, for some j\,? Let us first consider the simpler problem of determining whether two vectors \mathbf{u}\, and \mathbf{v}\, are linearly dependent, i.e., whether \mathbf{u} = c \mathbf{v}\, for some c \in \mathbb{C}\,. We introduce the associative multilinear \wedge\,, known as the exterior product, with the properties that

  • (a \mathbf{u}) \wedge (b \mathbf{v}) = (a b) (\mathbf{u} \wedge \mathbf{v})\,,
  • \mathbf{u} \wedge (\mathbf{v} +\mathbf{w}) = \mathbf{u} \wedge \mathbf{v} + \mathbf{u} \wedge \mathbf{w}\,,
  • (\mathbf{u} + \mathbf{v}) \wedge \mathbf{w} = \mathbf{u} \wedge \mathbf{w} + \mathbf{v} \wedge \mathbf{w}\,,
  • (\mathbf{u} \wedge \mathbf{v}) \wedge \mathbf{w} = \mathbf{u} \wedge (\mathbf{v} \wedge \mathbf{w})\,,
  • \mathbf{u} \wedge \mathbf{u} = 0\,,

which implies \mathbf{u} \wedge \mathbf{v} = -\mathbf{v} \wedge \mathbf{u}\,. Furthermore, the exterior product gives us the following property:

  • \mathbf{u} \wedge \mathbf{v} = 0 \; \iff \mathbf{u} = c \mathbf{v}\,.

Suppose \mathbf{w} = a \mathbf{u} + b \mathbf{v}\,. Then \mathbf{u} \wedge \mathbf{v} \wedge \mathbf{w} = 0\,. In general, \mathbf{u} \wedge \mathbf{v} \wedge \mathbf{w} \wedge ...\, vanishes if any of the factors in the product are linearly dependent on the others. In practice, we may express \mathbf{u},\mathbf{v},\mathbf{w}...\, in terms of a basis \{ \mathbf{e}_1, \mathbf{e}_2,...\mathbf{e}_n\}\, of V^n\, as \mathbf{u} = \textstyle \sum_{i = 1}^n A_i \mathbf{e}_i\,, etc. The exterior product is then completely determined in terms of the basis. For example,

\mathbf{u} \wedge \mathbf{v} \wedge \mathbf{w} = \sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n A_i B_j C_k (\mathbf{e}_i \wedge \mathbf{e}_j \wedge \mathbf{e}_k)\,.

Non-vanishing contributions come from the terms in which i,j,k\, are distinct.

Determinant

Suppose that our product contains exactly n\, factors:

\mathbf{A}_1 \wedge \mathbf{A}_2 \wedge ... \wedge \mathbf{A}_n\,,

where \mathbf{A}_i \in V^n\,. Then

\mathbf{A}_1 \wedge \mathbf{A}_2 \wedge ... \wedge \mathbf{A}_n = \sum_{i,j,k...=1}^n A_{1i} A_{2j} A_{3k}... \mathbf{e}_i \wedge \mathbf{e}_j \wedge \mathbf{e}_k ...\,.

All non-vanishing terms are then proportional to \mathbf{e}_1 \wedge \mathbf{e}_2 \wedge ... \wedge \mathbf{e}_n\,, and we may write

\mathbf{e}_i \wedge \mathbf{e}_j \wedge \mathbf{e}_k ... = \varepsilon_{ijk..} \mathbf{e}_1 \wedge \mathbf{e}_2 \wedge ... \wedge \mathbf{e}_n\,,

where \varepsilon_{ijk...}\, is the Levi-Civita symbol which is equal to 1\, if ijk...\, are an even permutation of (1,2,3...,n)\,, -1\, if they are an odd permutation of (1,2,3...,n)\,, or zero otherwise. For example, if n = 3\,, then \varepsilon_{132} = -1\,. Then

\mathbf{A}_1 \wedge \mathbf{A}_2 \wedge ... \wedge \mathbf{A}_n = \left(\sum_{i,j,k...=1}^n \varepsilon_{ijk...} A_{1i} A_{2j} A_{3k}... \right)\mathbf{e}_1 \wedge \mathbf{e}_2 \wedge \mathbf{e}_3 ...\wedge \mathbf{e}_n\,.

The quantity in brackets is a number known as the determinant of the square matrix with elements A_{ij}\,, written as

\det A \equiv \sum_{ijk...=1}^n \varepsilon_{ijk...} A_{1i} A_{2j} A_{3k} ...\,,

so we may write

\mathbf{A}_1 \wedge \mathbf{A}_2 \wedge ... \wedge \mathbf{A}_n = \det A \; \mathbf{e}_1 \wedge \mathbf{e}_2 \wedge \mathbf{e}_3 ...\wedge \mathbf{e}_n\,.

It follows that if \det A = 0\, then at least one of the rows of A\, are linearly dependent. If A \neq 0\, then they are linearly independent.

Calculation

Most elementarily, the determinant can be calculated, practical only for small n\,, using the Laplace expansion.

Properties

  • \ln \det A = \operatorname{tr} \ln A\,.
  • If |\epsilon| \ll 1\,,
\det (I+\epsilon A) \approx 1 + \epsilon\,\operatorname{tr} A + \frac{\epsilon^2}{2}(\operatorname{tr}(A)^2-\operatorname{tr}(A^2)) \, (see characteristic polynomial)

Derivative of a determinant

The derivative of a determinant of a matrix is given by Jacobi's formula:

d|A|\ = \operatorname{tr}(adj(A)dA)\!

Here adj(A)\! refers to the adjugate of A\,. If A\, is invertible then,

d|A|\ = |A|\operatorname{tr}(A^{-1}dA)\!

Identities

Block matrices

Suppose, A, B, C, D\, are n\times n\,, n\times m\,, m\times n\,, m\times m matrices, respectively. Then

\det\begin{pmatrix}A& 0\\ C& D\end{pmatrix} = \det\begin{pmatrix}A& B\\ 0& D\end{pmatrix} = \det(A) \det(D)\,.

This can be seen by performing a Laplace expansion along, say, the first row. The first n\, sub-determinants are of the same form as the original matrix, their coefficients containing only elements of A\,. The process may be applied recursively until reduced to the calculation of the determinant of an (m+1)\times(m+1)\, matrix:

\det \begin{pmatrix}a& 0\\ c& D\end{pmatrix} = a \det(D)\,.

Furthermore, using[1]

\begin{pmatrix}A& B\\ C& D\end{pmatrix} = \begin{pmatrix}A& 0\\ C& 1\end{pmatrix} \begin{pmatrix}1& A^{-1} B\\ 0& D - C A^{-1} B\end{pmatrix}\,

leads to

\det\begin{pmatrix}A& B\\ C& D\end{pmatrix} = \det(A) \det(D - C A^{-1} B)\,.

References

This article incorporates material from Determinant on Wikipedia, which is licensed under the GFDL. [2]

  1. ↑ Mike Brookes (2005). Proofs Section 3: Matrix Properties. Matrix Reference Manual.
  2. ↑ Determinant, from Wikipedia, The free encyclopedia; Retrieved April 4, 2007.
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