# Spivak's Calculus on Manifolds problem 2-15(c)

2-15. Regard an $$n \times n$$ matrix as a point in the $$n$$-fold product $$\Bbb{R^n} \times \cdots\times \Bbb{R^n}$$ by considering each row as a member of $$\Bbb{R^n}$$.

(a) Prove that $$det: \Bbb{R^n} \times \cdots \times \Bbb{R^n} \to \Bbb{R}$$ is differentiable and $$D(\mathrm{det})(a_1,\ldots,a_n)(x_1,\ldots,x_n)=\sum_{n=i}^{n} \mathrm{det} \begin{bmatrix} a_1 \\ \vdots \\ x_i\\ \vdots\\ a_n \end{bmatrix}$$

(b) if $$a_{ij}: \Bbb{R} \to \Bbb{R}$$ are differentiable and $$f(t)=det(a_{ij}(t))$$, show that $$f'(t)= \sum_{j=1}^{n} det \begin{bmatrix} a_{11}(t),\ldots, a_{1n}(t)\\ \vdots\\ a_{j1}'(t),\ldots, a_{jn}'(t)\\ \vdots \\ a_{n1}(t),\ldots, a_{nn}(t) \end{bmatrix}$$

(c) if $$\mathrm{det} (a_{ij}(t)) \neq 0$$ for all $$t$$ and $$b_1,...,b_n: \Bbb{R} \to \Bbb{R}$$ are differentiable, let $$s_1,\ldots,s_n: \Bbb{R} \to \Bbb{R}$$ be the functions such that $$s_1(t),\ldots,s_n(t)$$ are the solutions of the equations $$\sum_{j=1}^{n} a_{ij}(t)s_j(t)=b_i(t)$$ $$i=1,\ldots,n$$.

show that $$s_i$$ is differentiable and find $$s_i'(t)$$.

Problem 2-40 asks to redo problem 2-15(c) using the implicit function theorem.

2-12 theorem (implicit function theorem). Suppose $$f: \Bbb{R^n} \times \Bbb{R^m} \to \Bbb{R^m}$$ is continuously differentiable in an open set containing $$(a,b)$$ and $$f(a,b) =0$$. Let be the $$m \times m$$ matrix. $$(D_{n+j}f^i(a,b))~, \quad 1 \leq i,~j \leq m~.$$ If $$\mathrm{det} M \neq 0$$, there is an open set $$A \subset R^n$$ containing $$a$$ and an open set $$B$$ subset $$R^m$$ containing $$b$$, with the following property: for each $$x \in A$$ there is a unique $$g(x)$$ in $$B$$ such that $$f(x,g(x)) =0$$. the function $$g$$ is differentiable.

Call $$i^{th}$$ row of $$(a_{ji}(t))$$ as $$R_i(t)$$. Let me define $$g(t)= \begin{bmatrix} s_1(t) \\ s_2 (t)\\ \vdots \\ s_n(t) \end{bmatrix}$$. thus $$R_i(t)g(t)=b_i(t)$$. If I define $$f:\Bbb{R^n} \times \Bbb{R^n} \to \Bbb{R^n}$$ such that $$f(R_i(t),g(t)) = R_i(t)g(t)=b_i(t)$$, the first difficulty I face is that the theorem will provide $$g(t)$$ is differentiable only if $$b_i(t)=0$$, which may not be true.

To overcome this difficulty, how should I define $$f$$ so that the theorem will provide me that $$g(t)$$ is differentiable?

• sure, will edit soon – Vinay Deshpande Apr 30 '19 at 11:04

We have the system of linear equations

$$\begin{pmatrix} a_{11}(t) & a_{12}(t) & \cdots & a_{1n}(t) \\ a_{21}(t) & a_{22}(t) & \cdots & a_{2n}(t) \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1}(t) & a_{n2}(t) & \cdots & a_{nn}(t) \\ \end{pmatrix} \begin{pmatrix} s_1(t) \\ s_2(t) \\ \vdots \\ s_n(t) \end{pmatrix}= \begin{pmatrix} b_1(t) \\ b_2(t) \\ \vdots \\ b_n(t) \end{pmatrix}$$

Which is convenient to write in matrix form

$$A(t)s(t) = b(t)$$.

The goal is to show that if the matrix $$A(t)$$ is invertible for all $$t$$, then solutions exist, are differentiable (and the book describes how one may go about computing the derivatives).

To set-up the application of the Implicit function theorem, we can define

$$f : \mathbb{R} \times \mathbb{R}^n \rightarrow \mathbb{R}^n$$

by

$$f(t,\, s) = A(t)s - b(t)$$.

This completes the set-up of the question, which is all you have asked, but I will continue for my own curiosity.

Now we should look at the matrix

$$(D_{1+j}f^i(t,\,s))_{1 \leq i,j \leq n}$$

We have

$$f^i(t,\,s) = \left(\sum_{k=1}^n a_{ik}(t)s_k\right) - b_i(t)$$

and

\begin{align*} D_{1+1}f^i(t,\,s) &= \frac{\partial f^i(t,\,s)}{\partial s_1} = a_{i1}(t) \\ D_{1+2}f^i(t,\,s) &= \frac{\partial f^i(t,\,s)}{\partial s_2} = a_{i2}(t) \\ \vdots \\ D_{1+n}f^i(t,\,s) &= \frac{\partial f^i(t,\,s)}{\partial s_n} = a_{in}(t) \end{align*}

So the matrix in the statement of the I.F.T. is just

$$(D_{1+j}f^i(t,\,s))_{1 \leq i,j \leq n} = A(t)$$.

By hypothesis on $$A(t)$$, this is invertible for all $$t$$. In particular, about any $$t_0$$, there is an open set $$I$$ containing $$t_0$$ and an open set $$B \subseteq \mathbb{R}^n$$ containing $$s_0$$, such that for each $$t \in I$$, there is a unique $$s(t) \in B$$ such that $$0 = f(t,\,s(t)) = A(t)s(t) - b(t)$$, i.e. $$s(t)$$ is a solution of the system of linear equations, and $$s(t)$$ is a differentiable function of $$t$$.

• it's only now that I notice that the inverse function theorem requires $f$ to be continuously differentiable while the $f$ you defined is not, since $A(t)$ and $b(t)$ are not given to be continuously differentiable – Vinay Deshpande May 8 '19 at 9:28
• You are right. But I would guess Spivak's intention is for us to suppose that in addition that $A(t)$ and $b(t)$ are continuously differentiable. Otherwise, I don't see how to make headway on the problem. :) – Jane Doé May 8 '19 at 17:56