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The derivative of the determinant along an invertible C¹ matrix curve: $(\det A)^{'} = \det A tr (A^{- 1} A^{'})$

Statement

Let $n \geq 1$ , let $I \subseteq R$ be a nonempty open interval, and let

$A : I ⟶ M_{n} (R), A (t) = (a_{i j} (t)),$

be a matrix-valued curve whose entries $a_{i j} : I \to R$ are differentiable at $t_{0} \in I$ (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral); write $A^{'} (t_{0}) : = (a_{i j}^{'} (t_{0}))$ (Finite rectangular matrices over a commutative ring, their entries, rows and columns). Suppose $A (t)$ is invertible for every $t \in I$ (A finite square real matrix is invertible if and only if its determinant is nonzero, If $\det (A)$ is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ ). Then $t \mapsto \det A (t)$ is differentiable at $t_{0}$ (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix) and

$(\det A)^{'} (t_{0}) = \det A (t_{0}) tr ⁣ (A (t_{0})^{- 1} A^{'} (t_{0})),$

with the trace of The trace $tr (A)$ as the sum of the diagonal entries. In particular, a $C^{1}$ matrix curve with invertible values has $(\det A)^{'} = \det A tr (A^{- 1} A^{'})$ on all of $I$ ( $C^{k}$ Euclidean maps and diffeomorphisms).

Facts & Assumptions

Given: Integers $n \geq 1$ ; an open interval $I$ ; a matrix curve $A$ with entries differentiable at $t_{0} \in I$ ; invertibility of $A (t)$ for all $t \in I$ .

[F1]

The Leibniz determinant $\det : M_{n} (R) \to R$ is column-multilinear, alternating and normalized, and $\det (X Y) = \det X \det Y$ for all $X, Y \in M_{n} (R)$ (The Leibniz determinant is column-multilinear, alternating and normalized over every commutative ring, For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ , For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[F2]

If $A$ is invertible, then $\det A \neq 0$ , $A^{- 1}$ exists, and $A^{- 1} = \det (A)^{- 1} adj (A)$ (A finite square real matrix is invertible if and only if its determinant is nonzero, If $\det (A)$ is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ ).

[F3]

For $0 \leq i < n$ , replacing the $i$ -th column of the identity by $w \in R^{n}$ produces a matrix whose determinant is the $i$ -th entry $w_{i}$ : the column expansion over the standard basis has one nonzero term, by alternation and normalization [F1]. Consequently, for invertible $B$ and $b \in R^{n}$ , the matrix $B^{(i)}$ obtained from $B$ by replacing its $i$ -th column by $b$ satisfies

$\det B^{(i)} = \det B (B^{- 1} b)_{i} .$

Indeed $B^{(i)} = B E$ where $E$ is the identity with $i$ -th column $B^{- 1} b$ , so [F1] gives $\det B^{(i)} = \det B \det E = \det B (B^{- 1} b)_{i}$ .

[F4]

The trace is linear, and $tr (X Y) = tr (Y X)$ for square matrices of matching size (Trace is a linear functional on $M_{n} (F)$ , For $A \in M_{m \times n} (F)$ and $B \in M_{n \times m} (F)$ , $tr (A B) = tr (B A)$ ).

[F5]

Differentiability of the entries means that for $h \to 0$ , $a_{i j} (t_{0} + h) = a_{i j} (t_{0}) + h a_{i j}^{'} (t_{0}) + o (h)$ ; equivalently $A (t_{0} + h) = B + h H + r (h)$ with $B = A (t_{0})$ , $H = A^{'} (t_{0})$ and $\max_{i, j} ∣ r_{i j} (h) ∣ / ∣ h ∣ \to 0$ (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

Proof

technique · direct; expand the determinant by column-multilinearity and keep the terms up to first order

1.1givenF2F5choose

Put $B : = A (t_{0})$ , $H : = A^{'} (t_{0})$ and, for small $h \neq 0$ with $t_{0} + h \in I$ , write $A (t_{0} + h) = B + h H + r (h)$ with the remainder $r (h)$ of [F5]. By [F2], $B$ is invertible. Since $r_{i j} (h) = o (h)$ entrywise, choose $0 < ε_{0} \leq 1$ such that $\max_{i, j} ∣ r_{i j} (h) ∣ \leq ∣ h ∣$ for $0 < ∣ h ∣ < ε_{0}$ , and fix such an $h$ .

1.2F1givenalgebra

For fixed $h$ , expand $\det A (t_{0} + h)$ by column-multilinearity [F1]: the $i$ -th column is $b_{i} + h h_{i} + r_{i} (h)$ , so $\det A (t_{0} + h) = \sum_{S \subseteq {0, \dots, n - 1}} \det (c_{0}^{S}, \dots, c_{n - 1}^{S}), c_{i}^{S} = {\begin{cases} h h_{i} + r_{i} (h), & i \in S, \\ b_{i}, & i \notin S, \end{cases}$ the sum being over the $2^{n}$ subsets $S$ . The term $S = \emptyset$ is $\det B$ .

1.3F3F4algebra

For $S = {i}$ a single index, [F3] applied to $b : = h h_{i} + r_{i} (h)$ gives $\det B (B^{- 1} (h h_{i} + r_{i} (h)))_{i}$ . Summing over $0 \leq i < n$ and using linearity of $B^{- 1}$ and of the trace [F4], $\sum_{i} \det B (B^{- 1} (h h_{i} + r_{i} (h)))_{i} = \det B (h tr (B^{- 1} H) + tr (B^{- 1} r (h))) .$

2.1F1step 1.1algebra

For $∣ S ∣ \geq 2$ , the Leibniz formula [F1] expresses $\det (c^{S})$ as $n!$ signed products. Each product has $∣ S ∣$ entries from increment columns and $n - ∣ S ∣$ entries from $B$ . Put $M : = \max {1, \max_{i, j} ∣ H_{i j} ∣ + 1}$ and $P : = \max {1, \max_{i, j} ∣ B_{i j} ∣}$ . Step 1.1 bounds each increment entry by $∣ h ∣ M$ , so for $∣ h ∣ \leq 1$ each product has magnitude at most $∣ h ∣^{∣ S ∣} M^{∣ S ∣} P^{n - ∣ S ∣} \leq h^{2} (M P)^{n}$ . Summing over all $S$ and permutations gives $∣ \sum_{∣ S ∣ \geq 2} \det (c^{S}) ∣ \leq 2^{n} n! (M P)^{n} h^{2}$ . When $n = 1$ the sum is empty and equals zero.

3.1step 1.3step 2.1F4F5algebra∎

Combining the preceding steps, $\det A (t_{0} + h) - \det B - h \det B tr (B^{- 1} H) = \det B tr (B^{- 1} r (h)) + O (h^{2}) .$ The right-hand side is $o (h)$ : $∥ tr (B^{- 1} r (h)) ∥ \leq n \max_{i, j} ∣ (B^{- 1} r (h))_{i j} ∣ \leq C^{'} ∣ r (h) ∣_{\max} = o (∣ h ∣)$ by [F5], and $O (h^{2}) = o (h)$ . Hence $(\det A)^{'} (t_{0}) = \det B tr (B^{- 1} H)$ , which is the stated formula.

Remarks

The formula is exactly the Jacobi formula for the derivative of the determinant. Its use in this collection is for $n = 3$ and $A (t) = D_{X} χ_{t} (X)$ , where the chain rule identifies $A^{'} (t) = (D u (t, χ_{t} (X))) A (t)$ , so that $(\det A)^{'} = \det A div u$ ; that composition is carried out in the transport theorem, not here.
Only differentiability of the entries at the point $t_{0}$ is used; continuity of $t \mapsto tr (A (t)^{- 1} A^{'} (t))$ is not asserted by this lemma and is not needed for the pointwise derivative.

The derivative of the determinant along an invertible C¹ matrix curve: $(\det A)^{'} = \det A tr (A^{- 1} A^{'})$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

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Sources

The derivative of the determinant along an invertible C¹ matrix curve: (det⁡A)′=det⁡A tr⁡(A−1A′)(detA)′=detAtr(A−1A′)

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

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The derivative of the determinant along an invertible C¹ matrix curve: $(\det A)^{'} = \det A tr (A^{- 1} A^{'})$