Invertible Linear Changes of Variables

Why Change Coordinates?

Some multivariable expressions are difficult to read in the coordinates in which they are first written. A numerator may contain a repeated linear expression, or a quadratic denominator may have cross terms. A linear substitution can replace those combinations with simpler coordinates.

The substitution is useful for limits only if it preserves the full approach to the point. An invertible linear map does: it changes distances by at most fixed factors, so approaching the original point is equivalent to approaching the transformed point. The new coordinates may make the calculation simpler, but they do not discard any approach direction.

Key idea: An invertible linear change of variables preserves the value and existence or nonexistence of a multivariable limit. Its usefulness comes from simplifying the expression, not from restricting how the point can be approached.

Linear Changes of Variables

Write a point in $\mathbb{R}^n$ as a column vector $\mathbf{x}$, and let $\mathbf{A}$ be an invertible $n\times n$ matrix. The substitution $\mathbf{x}=\mathbf{p}+\mathbf{A}\mathbf{u}$ sends the new coordinate $\mathbf{u}=\mathbf{0}$ to the point $\mathbf{p}$. Invertibility means that $\mathbf{A}^{-1}$ exists, so every displacement $\mathbf{x}-\mathbf{p}$ corresponds to exactly one $\mathbf{u}=\mathbf{A}^{-1}(\mathbf{x}-\mathbf{p})$.

Limits here use the Euclidean norm, denoted by $\|\cdot\|$. Suppose $f$ is defined on a set $D$, and $\mathbf{p}$ is an accumulation point of $D\setminus\{\mathbf{p}\}$. Define the transformed function $g(\mathbf{u})=f(\mathbf{p}+\mathbf{A}\mathbf{u})$ whenever $\mathbf{p}+\mathbf{A}\mathbf{u}$ belongs to $D$. The limit statement $\lim_{\mathbf{x}\to\mathbf{p}}f(\mathbf{x})=L$ means that, for every $\varepsilon>0$, there is a $\delta>0$ such that $\mathbf{x}\in D$ and $0<\|\mathbf{x}-\mathbf{p}\|<\delta$ imply $|f(\mathbf{x})-L|<\varepsilon$.

Theorem: If $\mathbf{A}$ is invertible, then $$ \lim_{\mathbf{x}\to\mathbf{p}} f(\mathbf{x})=L \quad\Longleftrightarrow\quad \lim_{\mathbf{u}\to\mathbf{0}} f(\mathbf{p}+\mathbf{A}\mathbf{u})=L. $$ The transformed function is considered on the corresponding transformed domain. Invertibility ensures that nonzero displacements from $\mathbf{p}$ correspond exactly to nonzero displacements from $\mathbf{0}$.

Why the Theorem Holds

A matrix and its inverse each have a finite bound on how much they can stretch a vector. For example, the Cauchy–Schwarz inequality applied to the rows of a matrix gives $\|\mathbf{A}\mathbf{v}\|\leq C\|\mathbf{v}\|$ for some finite constant $C>0$. Applying the same fact to $\mathbf{A}^{-1}$ gives another constant $K>0$. Thus, for every $\mathbf{u}$,

\|\mathbf{A}\mathbf{u}\|\leq C\|\mathbf{u}\|, \qquad \|\mathbf{u}\|=\|\mathbf{A}^{-1}\mathbf{A}\mathbf{u}\|\leq K\|\mathbf{A}\mathbf{u}\|.

These inequalities show that neither the map nor its inverse can turn arbitrarily small displacements into displacements bounded away from zero. Here is the limit argument in both directions.

From the original limit to the transformed limit.
Suppose $f(\mathbf{x})\to L$ as $\mathbf{x}\to\mathbf{p}$. Given $\varepsilon>0$, choose $\delta>0$ from the definition of that limit. If $0<\|\mathbf{u}\|<\delta/C$, then $\mathbf{u}\neq\mathbf{0}$ and $0<\|\mathbf{A}\mathbf{u}\|\leq C\|\mathbf{u}\|<\delta$. Therefore $|f(\mathbf{p}+\mathbf{A}\mathbf{u})-L|<\varepsilon$.

From the transformed limit to the original limit.
Suppose $g(\mathbf{u})=f(\mathbf{p}+\mathbf{A}\mathbf{u})\to L$ as $\mathbf{u}\to\mathbf{0}$. Given $\varepsilon>0$, choose $\eta>0$ such that $0<\|\mathbf{u}\|<\eta$ implies $|g(\mathbf{u})-L|<\varepsilon$. Set $\delta=\eta/K$. If $0<\|\mathbf{x}-\mathbf{p}\|<\delta$, let $\mathbf{u}=\mathbf{A}^{-1}(\mathbf{x}-\mathbf{p})$. Then $0<\|\mathbf{u}\|\leq K\|\mathbf{x}-\mathbf{p}\|<\eta$, so $|f(\mathbf{x})-L|<\varepsilon$.

This proves the equivalence. The two norm bounds are the essential reason it works: they relate closeness in one coordinate system to closeness in the other.

Quadratic Expressions and Approach Directions

Linear substitutions are especially useful for quadratic expressions. If a quadratic form is written as $\mathbf{x}^{\top}\mathbf{B}\mathbf{x}$ and $\mathbf{x}=\mathbf{A}\mathbf{u}$, direct substitution gives

\mathbf{x}^{\top}\mathbf{B}\mathbf{x} =(\mathbf{A}\mathbf{u})^{\top}\mathbf{B}(\mathbf{A}\mathbf{u}) =\mathbf{u}^{\top}\mathbf{A}^{\top}\mathbf{B}\mathbf{A}\mathbf{u}.

Thus the expression may become simpler in the new coordinates. For example, combinations such as $x-y$ and $x+y$ can be treated as separate coordinates. Because the change is invertible, both new coordinates still range over all the values corresponding to nearby original points.

Directions transform just as systematically. A straight approach $\mathbf{x}=\mathbf{p}+t\mathbf{d}$, where $\mathbf{d}\neq\mathbf{0}$ and $t\to0$, becomes $\mathbf{u}=t\mathbf{A}^{-1}\mathbf{d}$. Conversely, a straight approach in $\mathbf{u}$-coordinates maps to a straight approach in $\mathbf{x}$-coordinates. A path that becomes a coordinate axis after substitution may therefore be easier to analyze. But checking one or several paths alone cannot prove that a limit exists; the full limit must account for every approach.

Worked Example: Simplifying a Limit with Sum-and-Difference Coordinates

Find the limit as $(x,y)\to(0,0)$ of $(x^2-y^2)/\sqrt{2(x^2+y^2)}$, with the function defined away from the origin. Introduce $u=x-y$ and $v=x+y$. These coordinates are invertible because $x=(u+v)/2$ and $y=(v-u)/2$.

The numerator factors as $x^2-y^2=(x-y)(x+y)=uv$. For the denominator, substitution gives $x^2+y^2=((u+v)^2+(v-u)^2)/4=(2u^2+2v^2)/4=(u^2+v^2)/2$. Hence $\sqrt{2(x^2+y^2)}=\sqrt{u^2+v^2}$. The transformed expression is therefore $uv/\sqrt{u^2+v^2}$.

Since $2|uv|\leq u^2+v^2$, for $(u,v)\neq(0,0)$ we have

\left|\frac{uv}{\sqrt{u^2+v^2}}\right| \leq \frac{\sqrt{u^2+v^2}}{2}.

As $(u,v)\to(0,0)$, the right-hand side tends to $0$, so the transformed expression tends to $0$. The substitution is invertible, so the original limit is also $0$. The bound verifies the result for every approach, not just a selected path.

Worked Example: A Quadratic Ratio with No Limit

Determine whether $(x-y)^2/(x^2+y^2)$ has a limit as $(x,y)\to(0,0)$, excluding the origin. Set $u=x-y$ and $v=x+y$, with inverse $x=(u+v)/2$, $y=(v-u)/2$. As above, $x^2+y^2=(u^2+v^2)/2$. The ratio becomes

\frac{(x-y)^2}{x^2+y^2} =\frac{2u^2}{u^2+v^2}.

Along $u=0$, $v\neq0$, the transformed ratio is $0$. In original coordinates this path is $x=y$, and direct substitution gives $0/(2x^2)=0$ for $x\neq0$. Along $v=0$, $u\neq0$, the transformed ratio is $2u^2/u^2=2$. This path is $x=-y$, and direct substitution gives $4x^2/(2x^2)=2$ for $x\neq0$.

Both paths approach the origin, but their values approach different constants, $0$ and $2$. Therefore the limit does not exist. The two directions are easy to see after the change of variables: they are the two coordinate axes.

Worked Example: Tracking an Oblique Approach

Evaluate the expression $ (x+y)^2/(x^2+y^2) $ along the path $y=2x$ as $(x,y)\to(0,0)$, with $x\neq0$. This calculation concerns only the stated path, not the full multivariable limit. Use $u=x+y$ and $v=x-y$, so the expression becomes $2u^2/(u^2+v^2)$.

On the path $y=2x$, $u=x+2x=3x$ and $v=x-2x=-x$. Substitution into the transformed expression gives

\frac{2u^2}{u^2+v^2} =\frac{2(3x)^2}{(3x)^2+(-x)^2} =\frac{18x^2}{9x^2+x^2} =\frac{18x^2}{10x^2} =\frac95.

Thus the value along this path is constantly $9/5$, so its path limit is $9/5$. Checking in the original coordinates gives $(x+2x)^2/(x^2+(2x)^2)=9x^2/(5x^2)=9/5$, confirming the calculation. The direction $(1,2)$ in the original coordinates has become $(u,v)=(3,-1)$, up to the scale factor $x$.

A Common Pitfall: Losing a Direction

A tempting but incorrect argument is to take the transformed ratio $2u^2/(u^2+v^2)$ from the previous example and say: “As $(u,v)\to(0,0)$, the numerator and denominator both tend to $0$, so the ratio tends to $0$.” This conclusion does not follow. A quotient of two quantities tending to zero need not tend to zero; their relative sizes matter. In fact, along $v=0$ the ratio is $2$, while along $u=0$ it is $0$.

A related error is to use a substitution that is not invertible and then assume it preserves the original limit. For example, setting $u=x-y$ but discarding $v=x+y$ does not uniquely determine $(x,y)$; many different points share the same $u$. An invertible substitution keeps all the information needed to correspond to every nearby point. Check that the inverse exists before applying the limit-preservation theorem.

When the Method Is Useful

Look for an invertible linear substitution when an expression repeatedly contains the same linear combinations of variables or when a quadratic expression becomes simpler in adapted coordinates. After making the substitution, verify the inverse and rewrite both the function and its domain. Then use the transformed variables to calculate a limit or to identify useful paths. For proving existence, a bound valid for all transformed points near the origin is still required; the change of variables does not replace that step.

Takeaway: An invertible linear map changes coordinates without changing which points approach the limit point. Its matrix and inverse provide the norm bounds that justify the change, while the new coordinates can make quadratic expressions and approach directions easier to understand.

Check Your Understanding

Try answering these questions, using equations where appropriate.

What property of a matrix is required for a linear change of variables to preserve all approach directions?
Why are bounds for both $\mathbf{A}$ and $\mathbf{A}^{-1}$ useful in proving limit preservation?
Under $u=x-y$, $v=x+y$, what are $x$ and $y$ in terms of $u$ and $v$?
Why do the paths $u=0$ and $v=0$ disprove the existence of the limit of $2u^2/(u^2+v^2)$ at the origin?
Does evaluating a function along one transformed straight path establish its full multivariable limit? Explain.