What is Determinants? — Mathematics

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Definition 2.16 University Mathematics — Year 2 · অধ্যায় 2 — Linear Algebra

The determinant of a family in a basis is $\det_{\mathcal{B}}(x_1, \dots, x_n)$ ; the determinant of a matrix $A$ is the determinant of its columns in the canonical basis — the permutation formula above; the determinant of an endomorphism $u$ is the scalar $\det u$ such that

\det{}_{\mathcal{B}}\bigl(u(x_1), \dots, u(x_n)\bigr) = \det u \cdot \det{}_{\mathcal{B}}(x_1, \dots, x_n) \quad \text{for all } x_i

(the left side is alternating $n$ -linear, hence a multiple of $\det_\mathcal{B}$ by Theorem 2.14; the factor does not depend on $\mathcal{B}$ ).

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Example 2.19 (A determinant by the rules)

Let $J \in \mathcal{M}_n(K)$ be the all-ones matrix and $a \in K$ ; we compute $\det(aI_n + J)$ with the tools just proved. Every column of $aI_n + J$ sums the same way: add all rows to the first (the determinant is unchanged — adding a multiple of one row to another adds a repeated-direction term, killed by alternation). The first row becomes $(a + n, a + n, \dots, a + n)$ ; factor out $a + n$ by linearity in that row, then subtract the first column from every other column: what remains is triangular with diagonal $(1, a, \dots, a)$ . Hence

\det(aI_n + J) = (a + n)\,a^{\,n-1}.

The closing insight: the roots $a = 0$ (multiplicity $n - 1$ ) and $a = -n$ say that $J$ has eigenvalue $0$ with multiplicity $n - 1$ and eigenvalue $n$ once — the spectrum of the rank-one matrix $J$ , one chapter early (Chapter 3 will make this systematic).

Example 2.20 (A determinant by the permutation formula)

For a matrix with many zeros the formula is practical by itself: in

A = \begin{pmatrix} 0 & a & 0 & 0\\ 0 & 0 & b & 0\\ 0 & 0 & 0 & c\\ d & 0 & 0 & 0 \end{pmatrix},

the only permutation picking nonzero entries is the $4$ -cycle $\sigma = (1\,2\,3\,4)$ mapping column $1 \to$ row $4$ , etc.; $\varepsilon(\sigma) = (-1)^3 = -1$ , so $\det A = -abcd$ . (Check via three column swaps to reach a diagonal matrix.)

Example 2.21 (A Vandermonde by the product formula)

For the nodes $0, 1, 2$ (used by quadrature rules like Exercise 2.4’s), the Vandermonde determinant of Exercise 2.11 evaluates in one glance:

\det\begin{pmatrix} 1 & 1 & 1\\ 0 & 1 & 2\\ 0 & 1 & 4 \end{pmatrix} = (1 - 0)(2 - 0)(2 - 1) = 2 ,

and by direct expansion along the first column: $1\cdot(4 - 2) = 2$ : agreement. Nonvanishing for distinct nodes is the whole theory of interpolation in one determinant: the evaluation forms $P \mapsto P(a_i)$ are a basis of the dual exactly when this determinant is nonzero, i.e. always for distinct $a_i$ — Example 2.2 quantified.

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