Inverse maps and implicit level sets
Identify when a nonlinear map has a local inverse and when an equation defines a smooth graph.
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The question for today
What you’ll understand
- Check the hypotheses and local conclusion of the inverse function theorem.
- Differentiate a level-set equation using a nonvanishing partial derivative.
- Separate local solvability from global injectivity and singular-point behavior.
Solving an equation explicitly is often unnecessary and sometimes impossible. Near a suitable point, a derivative can still certify that outputs determine inputs uniquely, or that one variable depends smoothly on the others. The inverse and implicit function theorems formalize these possibilities. Their power lies in converting a linear nonsingularity test into a nonlinear local conclusion, provided the derivative varies continuously on a neighborhood.
A nonsingular derivative yields a local inverse
Let F be continuously differentiable on an open subset of Euclidean n-space, and suppose its square derivative matrix at a is invertible. The inverse function theorem supplies neighborhoods of a and F(a) on which F is a one-to-one correspondence with a continuously differentiable inverse. The inverse derivative at F(a) is the inverse matrix of DF(a), as the chain rule suggests. The theorem supplies actual nearby neighborhoods, not just a formal rule for a hypothetical inverse.
The conclusion is local. A map can have an invertible derivative everywhere and still identify distant input points. For example F(x,y)=(eˣ cos y,eˣ sin y) has positive determinant e²ˣ everywhere but repeats when y increases by a full turn. Restricting to sufficiently small neighborhoods removes that ambiguity. Conversely, a zero derivative does not by itself rule out an inverse as a continuous map: x↦x³ is globally invertible, but its inverse fails to have a finite derivative at zero. Nonsingularity guarantees smooth inverse behavior, not every possible kind of inverse.
A level equation becomes a graph in a chosen direction
For a scalar C¹ function G(x,y), suppose G(a,b)=0 and the y partial derivative is nonzero there. Then near (a,b), the equation G(x,y)=0 defines a unique C¹ graph y=φ(x) through the point. Differentiating the identity G(x,φ(x))=0 gives φ′=−Gx/Gy, evaluated along the graph. The nonzero denominator is a geometric condition: the level set can be crossed by changing y, making y recoverable from x locally.
The same curve may require a different graph near another point. On a circle, the upper and lower semicircles are graphs over x away from their horizontal endpoints. Near a leftmost or rightmost point, solving instead for x as a function of y works. If the entire gradient vanishes, neither scalar criterion applies; the set might still be smooth, isolated, self-crossing, or more complicated. The theorem's failure to apply is a request for further analysis, not a verdict that the level set is singular.
Block derivatives reveal the multidimensional version
For several equations, divide variables into free coordinates x in n-space and dependent coordinates y in m-space. A C¹ map G into m-space defines a local graph y=φ(x) if the m-by-m block DyG is invertible at the base solution. Differentiation gives Dφ=−(DyG)⁻¹DxG. The choice of dependent variables matters: a different nonsingular block may describe the same solution set using another coordinate chart. This perspective explains how constraints can reduce dimension without requiring a global explicit formula.
In applications, an almost singular derivative deserves attention even when its determinant is nonzero. The inverse theorem still applies, but a large inverse norm means that small output errors may create large input changes. This is sensitivity rather than nonexistence. For numerical work, solving the corresponding linear system is usually preferable to explicitly forming an inverse matrix. Always separate the exact existence theorem, the size of the valid neighborhood, and the conditioning of the calculation used to locate a solution.
Work it through
Choose a branch before differentiating
On the level curve x²+xy+y²=3 near (1,1), find dy/dx and a first-order prediction for y when x increases to 1.02.
Set G=x²+xy+y²−3. At (1,1), G=0 and Gy=x+2y=3, which is nonzero. The implicit function theorem therefore provides a unique nearby graph through this particular point.
Gx=2x+y=3 at the same point, so the graph derivative is −3/3=−1. An x increase of 0.02 predicts a y change of −0.02 and hence y≈0.98.
The positive branch is y=(−x+√(12−3x²))/2. At x=1.02 it gives approximately 0.979866. This nearby value agrees with the linear prediction to first order; the other quadratic branch does not pass through (1,1).
The local slope is −1 and the estimate is 0.98. Selecting the base solution is essential because the same equation can have another y value at the same x.
The living notebook
Explore the model
Interactive lab · matrix
Invertibility versus sensitivity in a linear family
Reset, set a=1 and b=0, and inspect condition number 1. For a second run, Reset, set a=0.25 and b=2; predict continued invertibility because det A=0.25, together with much stronger distortion and a larger condition number. Playback interpolates the geometry; readouts describe the selected final matrix. This triangular family illustrates linear conditioning, not a nonlinear inverse neighborhood.
- Area scale · det A
- 1.4
- Singular values · σmax, σmin
- 1.597, 0.876
- Condition number · κ₂(A)
- 1.823
The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forNonsingularity guarantees a local linear inverse, while conditioning measures how sensitive that inverse is.
Pause & reason
Check your understanding
Your turn
Try it yourself
For the unit circle x²+y²=1 at (0,1), compute the local graph derivative y′(0). At (1,0), explain which variable should instead be dependent and compute the corresponding slope.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Check both partial derivatives of G=x²+y²−1 at each point before selecting a denominator.
Compare with the solution
At (0,1), Gy=2, so y can be written as φ(x) and y′=−x/y gives y′(0)=0. At (1,0), Gy=0 but Gx=2, so use x=ψ(y). Its derivative is x′=−y/x, which is zero at y=0. The circle is smooth at both points; the failure of one graph direction does not imply failure of every graph description.
Keep the idea
Take it with you
- A C¹ square map with nonsingular derivative has a C¹ inverse on suitable local neighborhoods.
- Implicit differentiation requires a nonvanishing selected derivative block and a specified base solution.
- Local invertibility, global uniqueness, and numerical conditioning are distinct properties.
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