Let \(T\) be a linear operator on \(F^n\). Define \[ D_T(\alpha_1, \ldots, \alpha_n) = \det(T\alpha_1, \ldots, T\alpha_n). \]
Show that \(D_T\) is an alternating \(n\)-linear function.
If \[ c = \det(T\epsilon_1, \ldots, T\epsilon_n) \] show that for any \(n\) vectors \(\alpha_1, \ldots, \alpha_n\) we have \[ \det(T\alpha_1, \ldots, T\alpha_n) = c\det(\alpha_1, \ldots, \alpha_n). \]
If \(\mathfrak{B}\) is any ordered basis for \(F^n\) and \(A\) is the matrix of \(T\) in the ordered basis \(\mathfrak{B}\), show that \(\det A = c\).
What do you think is a reasonable name for the scalar \(c\)?
Solution
Part (a).
1. \(D_T\) is \(n\)-linear.
\(\det\) is \(n\)-linear in its columns, and \(T\) is linear, so \(D_T(\alpha_1, \ldots, \alpha_n) = \det(T\alpha_1, \ldots, T\alpha_n)\) is linear in each \(\alpha_i\) separately.
2. \(D_T\) is alternating.
if \(\alpha_i = \alpha_j\) for \(i \neq j\), then \(T\alpha_i = T\alpha_j\), so \(\det(T\alpha_1, \ldots, T\alpha_n) = 0\) since \(\det\) is alternating.
3. Hence \(D_T\) is an alternating \(n\)-linear function.
1 and 2.
Part (b).
1. Every alternating \(n\)-linear function on \(F^n\) is a scalar multiple of \(\det\).
the space of alternating \(n\)-linear functions on \(F^n\) is \(1\)-dimensional, spanned by \(\det\).
2. Hence \(D_T = c \det\) for some scalar \(c\).
1.
3. \(c = D_T(\epsilon_1, \ldots, \epsilon_n) = \det(T\epsilon_1, \ldots, T\epsilon_n)\).
evaluate \(D_T = c\det\) at the standard basis \((\epsilon_1, \ldots, \epsilon_n)\), where \(\det(\epsilon_1, \ldots, \epsilon_n) = 1\).
4. Therefore \(\det(T\alpha_1, \ldots, T\alpha_n) = c\det(\alpha_1, \ldots, \alpha_n)\) for all \(\alpha_1, \ldots, \alpha_n\).
2 and 3.
Part (c).
1. \(\det A = c\).
1.1. The columns of \(A\) are \(T\beta_1, \ldots, T\beta_n\) expressed in the basis \(\mathfrak B = (\beta_1, \ldots, \beta_n)\).
definition of the matrix of \(T\) in the ordered basis \(\mathfrak B\).
1.2. \(\det A = \det(T\beta_1, \ldots, T\beta_n)\).
the determinant of a matrix is the determinant of its columns.
1.3. \(\det(T\beta_1, \ldots, T\beta_n) = c\det(\beta_1, \ldots, \beta_n) = c \cdot 1 = c\).
part (b) applied to the basis vectors, and \(\det(\beta_1, \ldots, \beta_n) = 1\) since \(\mathfrak B\) is a basis.
2. Hence \(\det A = c\).
1.
Part (d).
1. The scalar \(c\) is the determinant of \(T\).
by part (c), \(c = \det A\) for the matrix \(A\) of \(T\) in any basis, which is the definition of \(\det T\).
2. Q.E.D.
parts (a)–(d).