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Mathematics-Online course: Linear Algebra - Matrices - Determinants

Determinant Rules

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For $ n\times n$-matrices $ A$ and $ B$ the following holds true:

(i)
$ \operatorname{det} A = \operatorname{det} A^{\operatorname t}$,
(ii)
$ \operatorname{det}A = 0 \Leftrightarrow A$ is not invertible,
(iii)
$ \operatorname{det}(AB) = (\operatorname{det}A)(\operatorname{det}B)$.
(Authors: Burkhardt/Höllig/Hörner)
(i) Symmetry follows from the expansion of the determinant in a sum over permutations:

$\displaystyle \operatorname{det} A^{\operatorname t}= \sum_{i\in S_n} \sigma(i) a_{1,i_1}\cdots a_{n,i_n}\, . $

Let $ j$ be the inverse permutation to $ i$. Then, rearranging the factors according to $ j$ yields:

$\displaystyle a_{j_1,i_{j_1}} \cdots a_{j_n,i_{j_n}} = a_{j_1,1} \cdots a_{j_n,n}\, . $

Since $ \sigma(i)=\sigma(j)$ we obtain the same expansion as for $ \operatorname{det}A$.

For example,

$\displaystyle i = (2,3,1) \leftrightarrow j = (3,1,2) $

are mutually inverse permutations and we obtain

$\displaystyle a_{1,i_1}a_{2,i_2}a_{3,i_3} = a_{1,2}a_{2,3}a_{3,1} = a_{3,1}a_{1,2}a_{2,3} = a_{j_1,1}a_{j_2,2}a_{j_3,3}\, . $

The sign $ \sigma$ is positive since

$\displaystyle i = (1,3)\circ(1,2),\quad j = i^{-1} = (1,2)\circ(1,3)\,. $

and the sign of each transposition is $ -1$.

(ii) A square matrix $ A$ is singular (not invertible) if and only if the equation $ Ax=0$ has a non-trivial solution $ x$, that is, if there exists a non-trivial linear combination of the columns $ a_j$ of $ A$:

$\displaystyle x_1 a_1 + \cdots + x_n a_n = 0\, . $

Without loss of generality we can assume that $ x_1\ne 0$ (otherwise rearrange the columns appropriately), that is,

$\displaystyle a_1 = y_2 a_2 + \cdots + y_n a_n,\quad y_j = -x_j/x_1\, . $

By multilinearity it follows that

$\displaystyle \operatorname{det} A = \sum_{j=2}^n y_j \operatorname{det}(a_j,a_2,\ldots,a_n) = 0\, , $

since all determinants on the right hand side contain two equal columns, hence, they vanish because of antisymmetry.

It follows from the multiplicativity of the determinant, which will be proved below, that the determinant of an invertible matrix is unequal to zero.

(iii) If $ A$ is not invertible, then multiplicativity follows from (ii).

For $ \operatorname{det}A\ne 0$ we show that the mapping

$\displaystyle A \mapsto f(B) := \operatorname{det}(AB) / \operatorname{det}(A) $

has all defining properties of the determinant and, consequently, coincides with $ \operatorname{det}B$. The normalisation $ f(E_n)=1$ is obvious. In order to prove multilinearity and antisymmetry we observe that the mapping $ B\mapsto AB$ is linear with respect to the columns of $ B$, and that the interchange of two columns of $ B$ corresponds to the interchange of the respective columns of $ AB$.
(Authors: Burkhardt/Höllig/Hörner)
  automatically generated 4/21/2005